- This Calendar refers to Sections in the 14th Edition of the textbook
## Textbook Information for the 2021 – 2022 Academic Year MATH 1350

**Title:***Calculus for Business, Economics, Life Sciences, and Social Sciences, Brief Version, 14*^{th}Edition**Authors:**Barnett, Ziegler, Byleen, Stocker**Publisher:**Pearson, 2019**ISBN 13:**978-0-13-486264-4**Remark:**The ISBN number listed above is for- the 14
^{th}Edition of the book - Brief Version
- Includes the access code for the
*MyLab Math*online homework system.

- the 14
- The
*Homework Assignments*mentioned in the Calendar were created in the*MyLab Math*system by Mark Barsamian at Ohio University. The entire list of all assignments is available on the Homework Web Page. The Homework Assignments are available to be copied by other instructors at other institutions. Contact Mark Barsamian for more information. - The Calendar contains links to
*Videos*that accompany the Homework Assignments. These are instructional videos created by Mark Barsamian, and are publicly accessible on*YouTube*. - For each video, there is an accompanying link called
*Video Notes*. This link will take you to a PDF file containing the notes produced in the Video. - The list of
*Meeting Topics*for each class meeting include*Presentations*that are assigned to*Students #1 – #20*. On the Blackboard page for each Section of MATH 1350, the students in that section are listed along with a*Student Number*from 1 to 20. (These numbers may be reassigned throughout the semester, as students add or drop the class. Students will be notified when their student numbers are changed.)**Students should keep an eye on the upcoming class presentations and be prepared to do the presentations that are assigned to them.**(The instructor will not be reminding students of upcoming class presentations.)

**Sections, Content, Homework, and Videos:**

- 2.1 Introduction to Limits (Link to Homework and Videos for Section 2.1)
- 2.2 Limits Involving Infinity (Link to Homework and Videos for Section 2.2)

- Name
- Where are you from?
- Tell the students your
*Office Location*,*Office Phone Number*, and*Office Hours*.

Point out that the Blackboard Site will only be used for a few things:

- On the Blackboard site, students can find a link to the
. (The link is prominent on the Blackboard site main page.)*MATH 1350 Web Page* - On the Blackboard site, students will get access to the
for the course. (See the*Digital Course Materials***[Digital Course Materials]**link in the menu along the left side of the Blackboard page.)- The
system, where they will work on Homework and take Quizzes and Exams.*MyLab Math* - The
for the course.*eText*

- The

Urge the students to go to be Blackboard site right away and follow the instructions for getting ** MyLab** configured, so that they can begin work on their homework.

Show them how they can always find their way to the MATH 1350 Web Page by going first to the Blackboard site and finding the link there.

Or they can bookmark the MATH 1350 Web Page in their web browser and navigate to it directly:
**Link to MATH 1350 Web Page**

Discuss the *Two Components of the Course Format*

**Course Component #1:***Instructional Videos***Course Component #2:***Monday, Wednesday, Friday Meetings*

Discuss the ** Grading**.

Discuss the ** Attendance Policy**.

Discuss the ** Calendar**.

- Point out that the Calendar lists the
for each week. These assignments are done in the MyLab system.*Homework Assignments* - Point out that the Calendar has a list of
for each meeting.*Meeting Topics*

Discuss the ** Required Computer Tools**.

Discuss the ** Study Routine for the Student**.

- Access MyLab Math and Start Doing Homework
- To see Instructions: Go to Blackboard → Link to Instructions for Setting Up and Accessing MyLab Math
- To Access MyLab: Go to Blackboard → Digital Course Materials → VitalSource Course Dashboard

**Instructor: **In Section 2.1 of the textbook, and in **Limits Videos A, B, C, D, E**, you will learn about *limits*. It is important to understand the distinction between the following two symbols:

- The symbol
$$f(a)=b$$
means that the graph of \(f(x)\) has a
*point*at \( (x,y)=(a,b) \). The symbol gives information about what is happening at*exactly*\(x=a\). - The symbol
$$\lim_{x \rightarrow a}f(x)=b$$
means that the graph of \(f(x)\)
*appears to be heading for the location*\( (x,y)=(a,b) \). The symbol gives information about a*trend*in the graph. The symbol*does not*give any information about what is happening in the graph at*exactly*\(x=a\).

This is made more precise in the official *Definition of Limit*:

**The Definition of Limit**

One kind of problem that shows up in the examples in that video and in your Homework H02 is the following type:

**Instructor Example #1: **(See the **[Example 1] in the Notes for Limits Video A**.) For the function shown in the graph shown below, find the following two things and explain how they are found.

**Instructor Example #2: **For the same function shown in the graph above, find the following two things and explain how they are found.

**Instructor: **Introduce the concept of ** limit from the left** and

**Instructor Example #3: **For the same function shown in the graph above, find the following two things and explain how they are found.

**Instructor: **Discuss
**Class Drill: Limits for a Function Given by a Graph**
.

**Instructor: **In Section 2.1 of the textbook, and in **Limits Videos A, B, C, D, E**, you learn about *limits*. In the previous class meeting, we discussed the distinction between the following two symbols:

- The symbol \(f(a)=b\) means that the graph of \(f(x)\) has a
*point*at \( (x,y)=(a,b) \). The symbol gives information about what is happening at*exactly*\(x=a\). - The symbol \( \lim_{x \rightarrow a}f(x)=b\) means that the graph of \(f(x)\)
*appears to be heading for the location*\( (x,y)=(a,b) \). The symbol gives information about a*trend*in the graph. The symbol*does not*give any information about what is happening in the graph at*exactly*\(x=a\).

This is made more precise in the official *Definition of Limit*:

**The Definition of Limit**

Two kind of problems that show up in the examples in the Videos and in your Homework for Section 2.1 are the following types:

From the Homework for Section 2.1, you know that, in general, the value of \( \lim_{x\rightarrow c}f(x) \) might not be the same as the value of \( f(c)\).

In
**Limits Video B [Example 1]**
, on page 4 of the notes, the discussion turns to problems of the following type:

The first example involves the function \(f(x)=-7x^2+13x-25\). First, the \(y\) value \(f(-2)\) is computed, and found to be

$$f(-2)=-79$$The second part of the example is about ** estimating** the value of the limit

To estimate the limit, tables of \(x,y\) values are made, with the \(x\) values getting *closer and closer* to \(-2\), but *not equal to * \(-2\). It looks as though the values of \(f(x)\) are getting closer and closer to \(-79\). Based on this observation, we *estimate* the limit:

Immediately following the example are two **Remarks**.

**Remark #1** is a comparison of the limit and the \(y\) value. An important point is made:

**Even though, in this first example of a function \(f(x)\) given by a formula, it happens that the value of the limit does match the \(y\) value, that does not always happen.**

**Remark #2** is about the *unsatisfying nature* of the method of *estimating* the value of a limit by making a table of \(x,y\) values. An important question is raised:

**Question: **Is there is a better way? That is, is there some way to analyze the formula for \(f(x)\) to determine the value of the limit *precisely*, without *estimating*?

**Answer: **There *are* analytical techniques, developed in higher-level math, that provide a way of analyzing the formulas for certain kinds of functions to determine their limits.

The analytical techniques, themselves, are beyond the level of an introductory Calculus course. But the general *results* of using the techniques can be presented as **Theorems** that can be used in our course. Three such **Theorems about Limits** are presented on **pages 10,11 of Limits Video B**. These Theorems are the *tools* used in an ** Analytical Approach to Finding Limits**.

**Instructor: **In the Homework for Section 2.1, you solve problems involving finding limits where the function \(f(x)\) is given by a *formula*:

we'll start with a couple examples of similar problems.

**Instructor do [Example 1] on Chalkboard ** (This example is similar to **Limits Video B [Example 2]**.) For the function

Find the following using the **Theorems about Limits** presented earlier.

- Find \(f(4)\)
- Find \(lim_{x\rightarrow 4} f(x)\)

**Instructor do [Example 2] on Chalkboard ** (This example is similar to **Limits Video B [Example 3]**.) For the function

Find the following using the **Theorems about Limits** presented earlier.

- Find \(f(2)\)
- Find \(\lim_{x\rightarrow 2} f(x)\)
- Find \(f(7)\)
- Find \(\lim_{x\rightarrow 7} f(x)\)

**Instructor: **In the Homework for Section 2.1, you begin studying *more advanced limits* where the function \(f(x)\) is given by a *formula* and where the value of \( \lim_{x\rightarrow c}f(x) \) is not always the same as the value of \( f(c)\). Some of the functions that exhibit this behavior are *rational functions*. That is, they are *ratios of polynomials*.

Before embarking on learning about the new, more advanced kind of limit, it is important to review three concepts involving *ratios*:.

- The
**first concept involving ratios**that we should review is the possible ratios of*zero*and*non-zero real numbers*, and their resulting values as determined by the rules of arithmetic. These are reviewed on**page 5 of Limits Video D**. - The
**second concept involving ratios**that we should review is the following question:

**When can one cancel terms in a ratio, and why?**

This question is discussed on**page 6 of Limits Video D**. Note that this question is the**most important concept of the first month of calculus**. - The
**third concept involving ratios**that we should review is actually a summary of what we know so far about computing the limits of quotient. That is, limits of the form $$\lim_{x \rightarrow c}\frac{f(x)}{g(x)}$$ A discussion of the two kinds of limits of quotients that we have considered so far is found on**pages 7 and 8 of Limits Video D**.**An example of the first kind of limit of a quotient**, the case where**the the limit of the denominator is not zero**, was actually discussed earlier in today's meeting: For the function $$f(x)=\frac{x^2-8x+12}{x^2-13x+42}=\frac{(x-2)(x-6)}{(x-7)(x-6)}$$ we found that \(lim_{x\rightarrow 2} f(x) = 0\). If we look at this result in light of**Theorem 2.7**, we would observe that- The limit of the numerator is \(\lim_{x\rightarrow 2} numerator = \lim_{x\rightarrow 2}(x-2)(x-6) = ((2)-2)((2)-6)=(0)(-4)=0\)
- The limit of the denominator is \(\lim_{x\rightarrow 2} numerator = \lim_{x\rightarrow 2}(x-7)(x-6) = ((2)-7)((2)-6)=(-5)(-4)=20\)

Since the limit of the denominator is*not zero*, Theorem 2.7 tells us that*the limit of the quotient is equal to the quotient of the limits*. That is, $$\lim_{x\rightarrow 2}f(x)=\lim_{x\rightarrow 2} \frac{numerator}{denominator} \underset{\text{ Thm }2.7}{=} \frac{\lim_{x\rightarrow 2} numerator}{\lim_{x\rightarrow 2} denominator}=\frac{0}{20}=0$$ Of course, in that example, presented earlier in this meeting, an even easier solution was used: Since the function \(f(x)\) is a*rational function*and the number \(x=2\) is in the*domain*of the rational function--that is, the number \(x=2\)*does not cause the denominator to be zero*--We can use**Theorem 3**which tells us that the value of the limits \(\lim_{x\rightarrow 2} f(x)\) will equal the \(y\) value \(f(2)\). There is often more than one correct way to compute a limit.**An example of the second kind of limit of a quotient**, the case where**the the limit of the numerator**, was also discussed earlier in today's meeting: For the function $$f(x)=\frac{x^2-8x+12}{x^2-13x+42}=\frac{(x-2)(x-6)}{(x-7)(x-6)}$$ When finding \(lim_{x\rightarrow } f(x) \), we observed that*is not*zero and the limit of the denominator*is*zero- The limit of the numerator is \(\lim_{x\rightarrow 7} numerator = \lim_{x\rightarrow 7}(x-2)(x-6) = ((7)-2)((7)-6)=(5)(1)=5\neq 0\)
- The limit of the denominator is \(\lim_{x\rightarrow 7} denominator = \lim_{x\rightarrow 7}(x-7)(x-6) = ((7)-7)((7)-6)=(0)(1)=0\)

*not zero*and the limit of the denominator*is zero*, Theorem 4 tells us that*the limit of the quotient does not exist*. That is, $$\lim_{x\rightarrow 2}f(x)=\lim_{x\rightarrow 2} \frac{x^2-8x+12}{x^2-13x+42} \ \ DNE$$

**Instructor: **A **third type of limit of a quotient**, one that we have not yet considered, is the case when the limit of the numerator and the limit of the denominator are *both zero*. That is,
$$\lim_{x\rightarrow c} \frac{numerator}{denominator}$$
where

- \(\lim_{x\rightarrow c} numerator = 0\)
- \(\lim_{x\rightarrow c} denominator =0\)

The next example involves a limit that is an *indeterminate form*. The example is based on problems that you work on in the Homework for Section 2.1, problems involving *rational functions*. A description of the problem type is

These H06 problems involve *the most important concept of the first month of Calculus*

This is central to the question

**Instructor: Do [Example 3] ** (This example is similar to **Limits Video D [Example 1]**.) For the *rational* function \(f(x)\) defined below

Show how to find the following:

- \(f(1)\)
- \(\lim_{x\rightarrow 1}f(x)\) (Observe that this limit is
*not*an*indeterminate form*.) - \(f(2)\)
- \(\lim_{x\rightarrow 2}f(x)\) (Observe that this limit
*is*an*indeterminate form*.)

**Remark: **As a general rule, if a limit \(\lim_{x\rightarrow c}f(x)\) is *not* an *indeterminate form*, then the value of the limit will be the same as the value of the \(y\) value, \(f(c)\). (That is, either both will have the **same numerical value**, or both will **not exist**.) On the other hand, if a limit \(\lim_{x\rightarrow c}f(x)\) *is* an *indeterminate form* then the limit may or may not exist. But the \(y\) value \(f(c)\) will definitely **not** exist. So limits of *indeterminate forms* are examples where the value of the *limit* of a function can differ from the \(y\) value of the function.

A shortcoming of our textbook and the MyLab system is that none of the exercises about limits ask you to both compute a \(y\) *value* and compute a *limit*, so you might not realize that those results can differ, and so you might not realize the significance of the limit. In the videos, discussion of limits of **indeterminate forms** is in **Limits Video D**. In that video, the examples *do* involve both computing \(y\) *values* and computing *limits*. Be sure to watch that video and study the accompanying notes, so that you understand the significance of the limits.

**Instructor: **On Wednesday, we discussed limits, and ended the meeting with a discussion of limits of particular kinds of **quotients**. These limits were **indeterminate forms**. That is, they were limits of quotients with the property that the limit of the numerator and the limit of the denominator are both zero. That is,
$$\lim_{x\rightarrow c} \frac{numerator}{denominator}$$
where

- \(\lim_{x\rightarrow c} numerator = 0\)
- \(\lim_{x\rightarrow c} denominator =0\)

Limits of *difference quotients* are difficult because of the number of steps involved.

- Given a formula for a function \(f(x)\), the student has to build an expression of the form \(f(c+h)\), where \(c\) is some given number.
- The student then has to use the expression \(f(c+h)\) to build the more complicated expression, called a
$$\frac{f(c+h)-f(c)}{h}$$*Difference Quotient* - Finally, the student must find the
$$\lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}$$*limit of the difference quotient*Correctly computing this limit will, at some point, involve knowing to cancel \(\frac{h}{h}\). This is another occurence of

*the most important concept of the first month of Calculus*When can one cancel \(\frac{a}{a}\), and why?

Students will encounter such limits in the homework for Section 2.1. They are discussed in **Textbook Section 2.1** and in **Limits Video E**. In our first ** Class Drill** of the semester, you will do an example of this type.

For the function $$f(x) = x^2+5x+7$$ the goal is to find the value of the difference quotient

$$\lim_{h\rightarrow 0}\frac{f(3+h)-f(3)}{h}$$Do this in two steps.

**Get the parts needed for the expression**- Find \(f(3)\)
- Find the
*empty version*of \(f(x)\). That is, find \(f( \ )\). - Find \(f(3+h)\).

**Use the parts to build the limit expression and then find the limit.**

**Instructor: ** Show the solution to the Class Drill that the class just worked on.
(For reference, see
**[Example 1] on page 6 of the Notes for Limits Video E**.)

**Instructor: **So far, when we have discussed limits, the symbol
$$\lim_{x \rightarrow a}f(x)=b$$
means that the graph of \(f(x)\) *appears to be heading for the location* \( (x,y)=(a,b) \). The symbol gives information about a *trend* in the graph.

This is made more precise in the official *Definition of Limit* in
**Limits Video A**, on page 4 of the notes.

Realize that in that definition of limit, the letters \(a\) and \(b\) represent *real numbers*. That definition of limit could actually be thought of as describing *two* trends:

**Trend in the \(x\) values:**the \(x\) values are getting closer and closer to the*real number*\(a\), but not equal to \(a\).**Trend in the \(y\) values:**the \(y\) values are getting closer and closer to the*real number*\(b\), and may even equal \(b\).

The definition of **limit** presented above is from **Textbook Section 2.1** (content discussed in **Limit Videos A,B,C,D,E**).

Starting in **Textbook Section 2.2** (content discussed in **Video H10**), the definition of limit is *changed*. The definition is *expanded*, so that the definition encompasses *more types of trends* in the behavior of the \(x\) and \(y\) values. The kinds of trends that will be included in the new, expanded definition of limit are

**Possible Trends in the \(x\) values:**- the \(x\) values are getting closer and closer to the
*real number*\(a\), but not equal to \(a\). - the \(x\) values are getting
*more and more*.**positive**, without bound - the \(x\) values are getting
*more and more*.**negative**, without bound

- the \(x\) values are getting closer and closer to the
**Possible Trends in the \(y\) values:**- the \(y\) values are getting closer and closer to the
*real number*\(b\), but not equal to \(b\). - the \(y\) values are getting
*more and more*.**positive**, without bound - the \(y\) values are getting
*more and more*.**negative**, without bound

- the \(y\) values are getting closer and closer to the

New terminology, and new symbols, will be used to abbreviate particular pairs of trends: a trend in the \(x\) values and the corresponding trend in the \(y\) values.

Here's the first new term that is part of the *expanded definition of limit*: **Infinite Limit**.

**Instructor: **Project **The Definition of Infinite Limits**, from Video10. Discuss the *graphical significance*, involving *vertical asymptotes*. And discuss the significance of the phrase *without bound*.

Notice that this definition describes the following pair of trends:

- the \(x\) values are getting closer and closer to the
*real number*\(c\), but not equal to \(c\). - the \(y\) values are getting
*more and more*.**positive**, without bound

There are many variations of the basic definition of **Infinite Limit**. Some variations are presented in the textbook, and in the videos and their notes. But not all possible variations are presented. That would be too tedious to read. The idea is that you need to be able to figure out how to read, write, speak, and illustrate the variations.

**Instructor: ** Do the following **[Example 1]** about the Basic Notation of Infinite Limits
(See the **Notes from the Video for H10**.)

- How is the symbol $$ \lim_{x \rightarrow -5} f(x) = -\infty $$ spoken?
- What does the symbol $$ \lim_{x \rightarrow -5} f(x) = -\infty $$ mean, in words?
- Present a picture of a graph that has the property that $$ \lim_{x \rightarrow -5} f(x) = -\infty $$ Explain the features of the picture that correspond to the components of the symbol.

**Instructor: ** Do the following **[Example 2]** about Infinite Limits
(See the **Notes from the Video for H10**).

Suppose that you are told that the graph of some function \(g(x)\) has the following behavior

- The graph of has a vertical asymptote with line equation \(x=7\)
- The graph goes
*up*along the*left*side of the asymptote. - The graph goes
*down*along the*right*side of the asymptote.

- Draw a graph that exhibits the behavior described above.
- Write a limit expression that conveys the fact that the graph goes
*up*along the*left*side of the asymptote. - Write another limit expression that conveys the fact that the graph goes
*down*along the*right*side of the asymptote. - For the function \(g(x)\) discussed above (and for which you drew a graph), consider the following limit: $$ \lim_{x \rightarrow 7} g(x) $$ Does the limit exist? Explain why or why not.

Here's another new term that is part of the *expanded definition of limit*: **Limit at Infinity**.

**Instructor: **Project **The Definition of Limits at Infinity**, from Video10. Discuss the *graphical significance*, involving *horizontal asymptotes*.

Notice that this definition describes the following pair of trends:

- the \(x\) values are getting
*more and more*.**positive**, without bound - the \(y\) values are getting closer and closer to the
*real number*\(b\).

As was the case with the definition of **Infinite Limit**, there are also variations of the basic definition of **Limit at Infinity**. You’ll explore these in a **Class Drill**.

- How is the symbol $$ \lim_{x \rightarrow -\infty} f(x) = 7 $$ spoken?
- What does the symbol $$ \lim_{x \rightarrow -\infty} f(x) = 7 $$ mean, in words?
- Draw a graph of a function \(f(x)\) that has the property that $$ \lim_{x \rightarrow -\infty} f(x) = 7 $$ Explain the features of the picture that correspond to the components of the symbol.

Suppose that you are told that the graph of some function \(g(x)\) has the following behavior

- The graph of \(g(x)\) has a horizontal asymptote on the
*left*with line equation \(y=5\). - The graph of \(g(x)\) has a horizontal asymptote on the
*right*with line equation \(y=-2\).

- Draw a graph of a function \(g(x)\) that exhibits the behavior described above.
- Write a limit expression that conveys the fact that the graph of \(g(x)\) has a horizontal asymptote on the
*left*with line equation \(y=5\) - Write another limit expression that conveys the fact that the graph of \(g(x)\) has a horizontal asymptote on the
*right*with line equation \(y=-2\).

**Instructor: ** Show the solution to the Class Drill that the class just worked on.
(For Reference, see the
**Notes from the Video for H10**.)

**Sections, Content, Homework, and Videos:**

- 2.2 Limits Involving Infinity (Link to Homework and Videos for Section 2.2)
- 2.3 Continuity (Link to Homework and Videos for Section 2.3)

**Instructor: **On Friday, we discussed an **Expanded Definition of Limit**, that involved the terminology and notation of *infinity*. The first kind of limit that we discussed as part of the expanded definition of limit was the **Infinite Limit**.

**Instructor: **Project **The Definition of Infinite Limits**.

Remember that this definition describes the following pair of trends:

- the \(x\) values are getting closer and closer to the
*real number*\(c\), but not equal to \(c\). - the \(y\) values are getting
*more and more*.**positive**, without bound

In Friday's meeting, we considered examples where function is given by a *graph*. That is, examples of the following form:

Finding an infinite limit is fairly easy when the function is given by a *graph*. But *computing* an infinite limit is difficult when the function is given by a *formula*. That is, problems of this type:

Some computations of this type are discussed in the Video for H11.

For the function

$$ f(x) = \frac{3}{x-2} $$We are interested now in finding the following limit: $$ \lim_{x \rightarrow 2^+} f(x) $$ But as discussed in the Video for H11, the computation of this limit is done in different ways, depending on whether one is using the word *limit* in the sense of *Section 2.1* (the *original* definition of limit, in which the concept of *infinity* **is not** used) or in the sense of *Section 2.2* (the *expanded* definition of limit, in which the concept of *infinity* **is** used)

Let \(f(x)\) be the function

$$ f(x) = \frac{-5}{x+3} $$Using the techniques of *Section 2.1* (the *original* definition of limit, in which the concept of *infinity* **is not** used, the value of the limit is

Explain how that result is obtained. (Show the calculation clearly, with explanations.)

**Instructor: ** Show the solution to the Class Drill that the class just worked on. (See the **Notes from the Video for H11**. In particular, study pages 1 - 4.)

**Instructor: ** Show the solution to the Class Drill that the class just worked on. (See the **Notes from the Video for H11**. In particular, study pages 1 - 8.)

**Instructor: **It is important to know how to use the correct solution methods for these limits involving infinity. It is common for students to have learned some kind of *shortcut* method for reaching an answer. Realize that these *shortcut* methods often give the wrong answer. For example, consider the *Invalid Solutions* presented in the **Notes from the Video for H11** on pages 11 & 12. Even if an invalid solution method happens to give the correct answer in some particular instance, the method is still invalid.

More importantly, a *shortcut* method for reaching an answer (whether that answer is right or wrong) does not convey any understanding of *why* the answer comes out the way it does. You won't really understand what you are doing, and you won't be able to *explain* what you did to anybody else.

It is also important to note that when computing limits of the sort computed in your **Class Drills**, it is important to know whether to use the *original* definition of limit (in which the concept of *infinity* **is not** used), or the *expanded* definition of limit (in which the concept of *infinity* **is** used). In the **Textbook** and **MyLab**, it can sometimes be ambiguous. In general, in the **Textbook** and **MyLab**,

- If one is solving a limit problem from Textbook Section 2.1 (MyLab Homeworks H02 - H08), one should use the
*original*definition of limit (in which the concept of*infinity***is not**used). - If one is solving a limit problem from Textbook Section 2.2 (MyLab Homeworks H09 - H15), one should use the
*expanded*definition of limit (in which the concept of*infinity***is**used).

**Questions for the instructor to pose to students. **

- What is meant by the
of a function?*end behavior* - Suppose that a function \(f(x) \) is known to have this limit behavior: $$ \lim_{x \rightarrow \infty} f(x) = -13$$ What does that tell us about the graph of the function?
- Suppose that the graph of a function \(f(x) \) has a
*horizontal asymptote*on the*left*with line equation \( y=5 \). Describe this behavior using*limit notation*. - Consider the
*polynomial*function $$ f(x) = -4x^6 + 2x^3 - 13x +5 $$- What should you examine in the formula for the function to determine the behavior of the
*right end*of the graph? (That is, to determine if the right end go up or down, without bound, or maybe levels off, or maybe does something else.) - Describe the
*right end behavior*in*words*. (Does the right end go up or down, without bound, or does it level off, or does it do something else?) - Describe the
*right end behavior*using*limit notation*. - Describe the
*left end behavior*in*words*. (Does the left end go up or down, without bound, or does it level off, or does it do something else?) - Describe the
*left end behavior*using*limit notation*.

- What should you examine in the formula for the function to determine the behavior of the
- Now consider the
*polynomial*function $$ g(x) = -4x^6 + 2x^7 - 13x +5 $$- Describe the
*right end behavior*in*words*. (Does the right end go up or down, without bound, or does it level off, or does it do something else?) - Describe the
*right end behavior*using*limit notation*. - Describe the
*left end behavior*in*words*. (Does the left end go up or down, without bound, or does it level off, or does it do something else?) - Describe the
*left end behavior*using*limit notation*.

- Describe the

**Instructor: ** Here are two important observations about the questions that you just answered.

**Observation 1: ** In formulating your answers about the limits of the *polynomial* functions you (hopefully) made up some new notation to abbreviate a new kind of limit involving infinity that we have not seen before. That is, the symbol
$$ \lim_{x \rightarrow \infty} f(x) = -\infty$$
describes the following pair of trends in the *right end* of the graph of \(f(x)\):

- the \(x\) values are getting
*more and more***positive**, without bound - the \(y\) values are getting
*more and more*.**negative**, without bound

(Of course, you also made up other new notation to describe variations on this idea.)

**Observation 2: **Realize that questions about *end behavior*, *horizontal asymptotes*, and *limits at infinity* are all related. They all involve
$$\lim_{x\rightarrow \infty}f(x) \ \ \text{or} \ \ \lim_{x\rightarrow -\infty}f(x)$$

**Instructor: **On Monday, we began discussing the idea of the *end behavior* of a function. The *right end behavior* is a description of whether the right end of the graph *goes up (or down) without bound*, or *levels off*, or *does something else*. If the right end of the graph *levels off*, approaching a height \(y=b\), we say that the graph has a *horizontal asymptote on the right, with line equation *\(y=b\).

On Monday, we discussed that questions about *end behavior*, *horizontal asymptotes*, and *limits at infinity* are all related. They all involve
$$\lim_{x\rightarrow \infty}f(x) \ \ \text{or} \ \ \lim_{x\rightarrow -\infty}f(x)$$

On Monday, we only considered the end behavior (and limits at infity) of *polynomial functions*. Today, we'll discuss the end behavior of *rational functions*. (Remember that these are functions that are *ratios of polynomials*.)

**Question for the instructor to pose to students:**

Consider *rational* function
$$ f(x) = \frac{2x^2-6x-8}{3x^2-3x-36} $$

- What should you examine in the formula for the function to determine the
*end behavior*? - Describe the
*end behavior*in*words*. (Describe the left end behavior and the right end behavior.) - Describe the
*end behavior*using*limit notation*. (You will need two limit expressions: one that describes the*left end behavior*and another that describes the*right end behavior*.)

**Instructor: **Discuss the following:

- Show how to find \( \lim_{x\rightarrow \infty}f(x) \) for the above function, using the technique described in the
**Video for H13**. - Explain what that tells us about the
*right end of the graph*. Give the*line equation*for the asymptote. - Show how to find \( \lim_{x\rightarrow -\infty}f(x) \) for the above function.
- Explain what that tells us about the
*left end of the graph*. Give the*line equation*for the asymptote.

**Note to Students: **In MATH 1350, on Quizzes and Exams, you will need to be *precise* in your descriptions of asymptotes. For instance, for the example just considered, it would not be sufficient to say that

Your description of asymptotic behavior needs to indicate whether the asymptote is *horizontal* or *vertical*. Furthermore,

- If the asyptote is
*horizontal*, then specify if the asymptote is on the*left*or*right*, and give the*line equation*for the asymptote in the form \(y=b\). - If the asyptote is
*vertical*, then give the*line equation*for the asymptote in the form \(x=c\) and indicate whether the graph goes*up*or*down*on the*left*side of the asymptote, and whether the graph goes*up*or*down*on the*right*side of the asymptote.

**Class Drill:** Finding limits at infinity.

**Instructor: ** Have students work on this **Class Drill**.

- For the
*rational*function $$ f(x) = \frac{13x^3-5x-4}{7x^5-4x-6} $$- Find \(lim_{x\rightarrow \infty}f(x)\)
- What does this result tell us about the behavior of the graph of \(f(x)\)?

- For the
*rational*function $$ f(x) = \frac{13x^3-5x-4}{7x^3-4x-6} $$- Find \(lim_{x\rightarrow \infty}f(x)\)
- What does this result tell us about the behavior of the graph of \(f(x)\)?

- For the
*rational*function $$ f(x) = \frac{13x^5-5x-4}{7x^3-4x-6} $$- Find \(lim_{x\rightarrow \infty}f(x)\)
- What does this result tell us about the behavior of the graph of \(f(x)\)?

**Instructor: **Show the solution to the Class Drill that the class just worked on. (See the **Notes from the Video for H13**.)

**Instructor: **We have seen that the limit behavior \(\lim_{x\rightarrow c}f(x) = \infty \) corresponds to the graph behavior that \(f(x)\) has a *vertical asymptote* with line equation \(x=c\), and that the graph goes *up* along both sides of that asymptote.

And we have seen that the limit behavior \(\lim_{x\rightarrow \infty}f(x) = b \) corresponds to the graph behavior that \(f(x)\) has a *horizontal asymptote on the right* with line equation \(y=b\)

The computation of the limit can be time consuming when you are first starting out. But after you do enough of those kinds of limits, you begin to see some patterns emerge. Once you begin to notice those patterns, you may be able to give the values of some limits, or find the line equations for some asymptotes, by simply *scrutinizing the formula* for a function, rather than by going through all of the steps to find a limit.

That is the idea behind questions that ask you to ** "find all the horizontal & vertical asymptotes"** for a rational function.

This concept is discussed indirectly in various places in the textbook Section 2.2. In the videos, the Video for H15 is specifically about the concept. Most importantly, *key results* are presented on
**pages 5 and 7 of the Notes from the Video for H13**.
(Instructor, project and discuss those pages.)

**Question for the instructor to pose to students:**

Find all horizontal and vertical asymptotes for the *rational* function

(Be sure to give *line equations* for the asymptotes. That is, give equations of the form \(x=a\) or \(y=b\). And be sure to say whether each asymptote is *horizontal* or *vertical*.)

**Quiz Q1** during the last 20 minutes of the Wed May 22 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The quiz material is taken from the material covered in Section 2.1 and Section 2.2. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

**A Suggestion for Studying: **When you work on your homework in MyLab, write down your complete solutions to each problem on paper before you type the answer into MyLab. Focus on the clarity and correctness of your written solution. Keep your written work organized in a notebook. Compare your written work to the written examples in the Videos. If possible, find another student, or a tutor, or your instructor, to look over your written work with you. Even though MyLab does not require that you write stuff down, you will learn a lot by focusing on your written work.

**Fri May 24 is the last day to drop without a "W" and the last day to drop without being charged for access to the MyLab Digital Course Materials.**

**Instructor: **Given a *graph* for a function \( f(x) \), it is easy to spot funky features like *holes*, *jumps*, and *points in the wrong place* on the graph. If a function \( f(x) \) is given by a *formula*, rather than a *graph*, it would be nice to have some way of *analyzing the formula for* \( f(x)\) to determine where its graph might have those funky features. The *Definition of Continuity* presents a test that can be performed on the formula for a function \( f(x)\).

**Definition of Continuity at a particular \(x\) value**

**Words:**Function \(f\) is continuous at \(x=c\)**Meaning:**Function \(f\) and \(x=c\) pass the following**three-part test**- \(f(c)\) exists.
- \(\lim_{x\rightarrow c}f(x)\) exists.
- The
**limit from the left**, \(\lim_{x\rightarrow c^-}f(x)\), exists. - The
**limit from the right**, \(\lim_{x\rightarrow c^+}f(x)\), exists. - The values from (a) and (b) match. That is, $$\lim_{x\rightarrow c^-}f(x)=\lim_{x\rightarrow c^+}f(x)$$ Their common value is the value of the limit \(\lim_{x\rightarrow c}f(x)\).

- The
- The values from (i) and (ii) match. That is, $$f(c)=\lim_{x\rightarrow c}f(x)$$

**Definition of Continuity on an interval**

**Words:**Function \(f\) is continuous on the interval \((a,b)\).**Meaning:**The function \(f\) is continuous at all \(x=c\), where \(a \lt c \lt b\)

**Instructor: **Have students work on the following **Class Drill:**

- Draw an example of a graph that
*fails Continuity Test (i)*at \( x=4 \) and*fails Continuity Test (ii)*at \( x=4 \). - Draw an example of a graph that
*fails Continuity Test (i)*at \( x=4 \) but*passes Continuity Test (ii)*at \( x=4 \). - Draw an example of a graph that
*passes Continuity Test (i)*at \( x=4 \) but*fails Continuity Test (ii)*at \( x=4 \). - Draw an example of a graph that
*passes Continuity Test (i)*at \( x=4 \) and*passes Continuity Test (ii)*at \( x=4 \) but*fails Continuity Test (iii)*at \( x=4 \).

**Instructor: **Present a solution to the **Class Drill** that the class just worked on.(See the **Notes from the Video for H16**.)

**Instructor: **Given a graph for a function \( f(x) \), it is easy to spot portions of the graph that are above or below the \(x\) axis.
(Show Page 2 of the **Notes from the Video for H20**.)

If a function \( f(x) \) is given by a *formula*, rather than a *graph*, then it would be nice to have some way of *analyzing the formula for* \( f(x)\) to determine where its graph is above or below the \(x\) axis. In other words, where the \( y \) values are *positive, negative, or zero*.

Two important observations are the following:

- A function that is
*continuous*can only change sign by touching the \(x\) axis and crossing it. At those \(x\) values, there is an \(x\) intercept. - But a function that is
*not always continuous*can also change sign by by jumping across the \(x\) axis at an \(x\) value where the function is discontinuous. At these \(x\) values, there is*not*an \(x\) intercept. (Show**page 4 of the Notes from the Video for H21**.)

Based on those two observations, we realize that to determine the sign behavior of a function given by a *formula*, it is helpful to first determine the \(x\) values where the function has an \(x\) intercept or is discontinuous. We give these \(x\) values a name:

**Definition: **A ** partition number** for a function \(f\) is an \(x\) value where \(f\) is discontinuous or \(f(x)=0\).

On the intervals between the *partition numbers* for \(f\), the sign of the function \(f\) does not change.

This allows us to articulate a *procedure* for determining the sign behavior of a function using what we call a *sign chart*.

**PROCEDURE: Constructing Sign Charts**

Given a function \(f\)

**Step 1: ** Find all partition numbers of \(f\):

- Find all numbers \(x\) such that \(f\) is
*discontinuous*at \(x\). (Rational functions ar discontinuous at values of \(x\) that make the denominator \(0\).) - Find all numbers \(x\) such that \(f(x)=0\). (For a rational function, this occurs where the numerator is \(0\) and the denominator is not \(0\).)

**Step 2: **Plot the numbers found in Step 1 on a real number line and indicate the sign behavior of \(f\) at each partition number. (Is \(f(x)=0\) or does \(f(x)\) *not exist*) Observe that this will divide the number line into intervals.

**Step 3: **Select a test number in each open interval determined in step 2 and evaluate \(f(x)\) at each test number to determine whether \(f(x)\) is positive (+) or negative (-) in each interval. Show each sign calculation clearly, and put the result (+) or (-) above the number line in the corresponding interval.

**Step 4: **Title your diagram: "Sign Chart for \(f(x)= formula\)."

**Instructor: **Have students work on the following **Class Drill:**

- Make a sign chart for the function \(f(x)=7x^2-56x+105=7(x-3)(x-5)\). Show details clearly. In particular
- Show how the signs are computed. (Use the
**factored**form of \(f(x)\)!!) - On your sign chart, be sure to indicate the sign behavior at each partition number.
- Label your sign chart.

- Show how the signs are computed. (Use the
- Make a sign chart for the function \(f(x)=7x^3-77x^2+273x-315=7(x-3)^2(x-5).\). Show details clearly. In particular
- Show how the signs are computed. (Use the
**factored**form of \(f(x)\)!!) - On your sign chart, be sure to indicate the sign behavior at each partition number.
- Label your sign chart.

- Show how the signs are computed. (Use the

**Instructor: ** Show the solution to the **Class Drill** that the class just worked on. (See the **Notes from the Video for H21**. In particular study pages 1 - 7.)

Observe the difference between the sign chart made in **Question 1** and the sign chart made in **Question 2**:

- In the sign chart made in
**Question 1**, the signs go (from left to right) +,0,-,0,+. Notice that the signs**alternate**between positive and negative. - In the sign chart made in
**Question 2**, the signs go (from left to right) -,0,-,0,+. Notice that the signs**do not alternate**between positive and negative.

Furthermore, notice that if the sign calculations are done clearly, it is possible to understand *why* the signs are changing or not changing.

Indeed, in the example that I just did,

- observe that the factor \((x-5)\) changes from
*negative*to*zero*to*positive*at \(x=5\). This causes the sign of \(f(x)\) to change at \(x=5\). - observe that the factor \((x-3)^2\) changes from
*postive*to*zero*and then back to*positive*at \(x=3\). This is the reason that the sign of \(f(x)\)*does not change*at \(x=3\). Rather, the function changes from*negative*to*zero*and then back to*negative*.

**Sections, Content, Homework, and Videos:**

- 2.3 Continuity (Link to Homework and Videos for Section 2.3)
- 2.4 The Derivative (Link to Homework and Videos for Section 2.4)

**Instructor: **Recall that on Friday, we discussed a procedure for determining the *sign behavior* of a function given by a *formula*. That is, a procedure for determining where the function is *positive*, *negative*, or *zero*. And recall that a key part of the procedure was the concept of ** partition numbers**.

**Definition: **A ** partition number** for a function \(f\) is an \(x\) value where \(f\) is discontinuous or \(f(x)=0\).

Remember that on the intervals between the *partition numbers* for \(f\), the sign of the function \(f\) does not change.

This allows us to articulate a *procedure* for determining the sign behavior of a function using what we call a *sign chart*.

**PROCEDURE: Constructing Sign Charts**

Given a function \(f\)

**Step 1: ** Find all partition numbers of \(f\):

- Find all numbers \(x\) such that \(f\) is
*discontinuous*at \(x\). (Rational functions ar discontinuous at values of \(x\) that make the denominator \(0\).) - Find all numbers \(x\) such that \(f(x)=0\). (For a rational function, this occurs where the numerator is \(0\) and the denominator is not \(0\).)

**Step 2: **Plot the numbers found in Step 1 on a real number line and indicate the sign behavior of \(f\) at each partition number. (Is \(f(x)=0\) or does \(f(x)\) *not exist*?) Observe that this will divide the number line into intervals.

**Step 3: **Select a test number in each open interval determined in step 2 and evaluate \(f(x)\) at each test number to determine whether \(f(x)\) is positive (+) or negative (-) in each interval. Show each sign calculation clearly, and put the result (+) or (-) above the number line in the corresponding interval.

**Step 4: **Title your diagram: "Sign Chart for \(f(x)= formula\)."

On Friday, we only used sign charts to determine the sign behavior of functions. Today we will again discuss sign charts, but as part of a larger problem: *solving inequalities*. Here is a **Class Drill** involving making a *sign chart* for a function and then using the sign chart to *solve an inequality*

**Instructor: **Have students work on the following **Class Drill**.

- Make a sign chart for the function \(f(x)=3x^2-21x+30=3(x-2)(x-5)\)
- Use your sign chart to solve the inequality
$$3x^2-21x+30 \lt 0$$
Present the solution set 3 ways:
- a picture using a number line
- Interval Notation
- Set Notation

- Use your sign chart to solve the inequality $$3x^2-21x+30 \geq 0$$ Again, present the solution set 3 ways.

**Instructor: **Present a solution to the **Class Drill** that the class just worked on. (See **[Example 1]** on page 7 of the **Notes from the Video for H21**.)

**Instructor: **Sometimes, inequalities can be more difficult to solve simply because the *partition numbers* are harder to determine. Finding *partition numbers* involves *solving equations*. There are skills involved, and also some common mistakes to be avoided. We'll discuss one.

**Instructor: ** Pose this question to the students.

Frick and Frack are trying to solve the equation $$ x^3 = 25x $$ Frick does the following: He divides by \( x \) to get the new equation $$ x^2 = 25 $$ Then he says that the solution is $$ x=5 $$ Frack does the following: He divides by \( x \) to get the new equation $$ x^2 = 25 $$ Then he says that the solution is $$ x=5, x=-5 $$ Is Frick right? Is Frack right? Explain.

**Instructor: **Inequalities are generally tricky things to solve. Sometimes, they can be made easier to solve by finding an equivalent inequality that has the number \(0\) on one side. Then the inequality can be solved by finding when the expression on the other side is *greater than* \( 0 \) or *less than* \(0\). The *sign chart* is a useful way to structure the solution of the problem. In this **Class Drill**, you will find the solution to an inequality by first finding an equivalent inequality that has the number \(0\) on one side and then solving that inequality.

**Instructor: **Have students work on the following **Class Drill**.

Solve the inequality $$ x^3 \lt 25x $$ Present the solution set 3 ways:

- a picture using a number line
- Interval Notation
- Set Notation

**Instructor: **Present a solution to the **Class Drill** that the class just worked on. (See **[Example 3] on page 10 of the Notes from the Video for H21**.)

**Instructor: **Section 2.4 is basically about the calculations of the slopes and equations of **secant lines** and **tangent lines**. The computations can be messy, and in some cases difficult, so that students might not really understand what those computations *mean*. Although textbook examples often include illustrations of the calculations, the homework problems in the **MyLab** system rarely ask the student to visualize or illustrate the meaning of the computations. Making illustrations of a computation is usually not nearly as hard as the computation, and the making of the illustration can improve students' understanding of the computation, so it is worthwhile to take some time to make a drawing. In a Class Drill, you will work on making a drawing to illustrate a calculation that you did in **Homework H25**.

**Instructor: **Have students work on the following **Class Drill**.

In Homework H25 problem [1](A), you were asked to compute a quantity of the following form: $$ \frac{f(b)-f(a)}{(b)-(a)} $$ This Class Drill is based on that problem.

Let \( f(x) = 7-x^2\).

- Show the calculation of $$ \frac{f(-1)-f(2)}{(-1)-(2)} $$
- The number that is the result of the calculation that you did in part (a) can be interpreted as the
*slope*of a*secant line*drawn on the graph of \( f(x)\). Sketch the graph of \(f(x)\). Make your graph large and neat. You can use as your guide the graph shown in the problem statement. On your graph, draw the*secant line*that corresponds to the slope calculation that you did in part (a). Label the line with its slope by writing $$ m = number $$ where \( number \) is the result of your calculation from (a).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on. (See **[Example 1](C)** on **page 4 of the Notes from the Video for H25**.)

**Instructor: **Most students are comfortable with *equations* like
$$y=7x^2-13x+27$$
but less comfortable with *function notation* like
$$f(x)=7x^2-13x+27$$
Calculus involves a lot of working with *functions* and *function notation*, and understanding the meaning of expressions that have something instead of just \(x\) inside the \(f(x)\). Throughout today's meeting, we'll encounter questions about building and simplifying expressions that involve something instead of just \(x\) inside the \(f(x)\).

**Instructor: ** Pose this question to the students.

For the function $$f(x)=x^2+3$$ Frick and Frack are trying to find $$f(x+2)$$ Frick does the following: \begin{eqnarray} f(x+2) &=& x^2+3+2 \\ &=& x^2 + 5 \end{eqnarray} Frack does the following: \begin{eqnarray} f(x+2) &=& (x+2)^2+3 \\ &=& x^2+2^2+3 \\ &=& x^2+4+3 \\ &=& x^2+7 \end{eqnarray}

**Question for the Class:** Is Frick right? Is Frack right? Explain.

**Answer:** Frick and Frack are both wrong!! Here is the correct calculation:

**Instructor: **An expression of the form
$$ \frac{f(2+h)-f(2)}{h} $$
is called a *difference quotient*. We will discuss *why* that name is used in a few minutes. But first, you will do a **Class Drill** involving *building* and *simplifying* a *difference quotient*

**Instructor: **Have students work on the following **Class Drill**.

For the function $$f(x)=3x^2-5x+1$$
**Instructor: **Find the following

- The
*empty version*of \( f \). That is, \( f( \ \ ) \). - \( f(2) \)
- \( f(2+h) \)
- Build the expression \( \frac{f(2+h)-f(2)}{h} \), but don't simplify it. This expression is a
*difference quotient*. - Simplify your expression \( \frac{f(2+h)-f(2)}{h} \) assuming that \( h\neq 0\).
- How is the fact that \( h\neq 0\) used in part (f)? Explain.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on. (See the **Notes from the Video for H24**.)

**Instructor: **Now we'll discuss why expressions like the expression
$$ \frac{f(2+h)-f(2)}{h} $$
that I just built and simplified are called *difference quotients*.

Let's back up and consider how we compute the slope of a line. Suppose that \((x_1,y_1)\) and \((x_2,y_2)\) are *distinct* (that is, *not the same*) points on a non-vertical line. The slope of the line is found by computing

Notice that this expression is a *quotient*. Also notice that the expression in the numerator is a *difference* of \(y\) values. And notice that the expression in the denominator is a *difference* of \(x\) values. So the expression is a *quotient of differences*. We could also say that the expression is a *difference quotient*.

Now consider the graph of a function \(f(x)\). The ordered pair \((2,f(2))\) represents a point on the graph of \(f(x)\), and the ordered pair \((2+h,f(2+h))\) represents a point on the graph of \(f(x)\). If \(h \neq 0\), then the expressions represent *distinct* (that is, *not the same*) points. In this case, there is exactly one line that passes through the two points. The slope of this line would be computed by the following computation

As discussed above, we could call this expression a *difference quotient*.

**Instructor: **Most students are comfortable with the concept of quantifying the *steepness* of a *line* by finding the *slope* of the line. A major concept in Calculus is learning how to quantify the *steepness* of a *curvy graph* at a given point on the graph. This is done by associating a *line* to the graph at the given point, and finding the slope of that *line*. The line is called the *tangent line*. You have learned about *tangent lines* in your reading and in your homework videos, and we will be discussing them in today's meeting. But before learning about *tangent lines*, it is important to review the terminology and computations involved when working with *slopes* and *equations* of *lines*. In this **Class Drill**, you will find the *point-slope* form and *slope-intercept form* of the equation of a line.

**Class Drill: ** (Material from Homework H23 involving Prerequisite Skills, so there is no video) For the line that passes through the point \( (3,7) \) and that has slope \( m=1 \), do the following.

- Write the equation of the line in
*point-slope*form. - Write the equation of the line in
*slope-intercept*form.

**Instructor: ** Show the solution to the **Class Drill** that students just worked on.

**Instructor: ** In the **Video for H25**, you were introduced to the ** Instantantaneuous Rate of Change**, the

**Definition of Instantaneous Rate of Change**

**words:**the instantaneous rate of change of \(f\) at \(a\)**alternate words:**the derivative of of \(f\) at \(a\)**symbol:**\(f'(a)\)**words:**the number $$m=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}$$**Remark:**The instantaneous rate of change \(f'(a)\)is a number.

**words:**the line tangent to the graph of \(f(x)\) at \(x=a\)**meaning:**the line that has the following two properties- The line contains the point \((x,y) = (a,f(a))\), which is called the
.*point of tangency* - The line has slope \(m=f'(a)\), called the
.*tangent line slope*

- The line contains the point \((x,y) = (a,f(a))\), which is called the

An important concept for the next month of the course is the relationship between the *derivative*, the *tangent line*, and *rate of change*:

- The value of \( f'(a) \)
- The slope \( m \) of the line
*tangent*to the graph of \( f(x) \) at \(x=a\) - The
*instantaneous rate of change*of \( f(x) \) at \(x=a\)

In a Homework problem, you are asked to compute the *slope* of the graph of a given function of the form \( f(x)=ax^2 \) at a particular given point. This Class Drill is based on that problem.

- For the function \(f(x)=3x^2\), find the slope of the graph at \( (1,f(1)) \).
- The number that is the result of the calculation that you did in part (a) can be interpreted as the
*slope*of a*tangent line*drawn on the graph of \(f(x)\). Sketch the graph of \(f(x)\). Make your graph large and neat. You can use as your guide the graph shown in the problem statement. On your graph, draw the*tangent line*that corresponds to the slope calculation that you did in part (a). Label the line with its slope by writing $$ m = number $$ where \( number \) is the result of your calculation from (a).

**Instructor: ** Show the solution to the **Class Drill** that the class just worked on. (See **[Example 1](E),(F)** on **pages 9 - 12 of the Notes from the Video for H25**.)

**Quiz Q2** during the last 20 minutes of the Fri May 31 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The quiz material is taken from the material covered in portions of Section 2.2 and Section 2.3. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

**Sections, Content, Homework, and Videos:**

- 2.4 The Derivative (Link to Homework and Videos for Section 2.4)
- 2.5 Basic Differentiation Properties (Link to Homework and Videos for Section 2.5)

There are a lot of difficult computational skills to learn in Section 2.4. But there is also a lot of new *terminology* and *notation*. Sometimes it is helpful to just focus on some of the *terminology* and *notation*, and think about what it means *without* having to do a difficult calculation. Here is a **Class Drill** that is just about recognizing what certain symbols and terms *mean*.

**[Drill for the Class] **(See the **Notes from the Video for H25**.) Shown below is a list of seven phrases and mathematical expressions that each represent a *number*. Each of these phrases and mathematical expressions is discussed in the Video for H25.

- the number \( m = \frac{f(b)-f(a)}{b-a} \)
- the slope \(m\) of the line is
*tangent*to the graph of \(f(x)\) at \(x=a\) - the number \( m = lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h} \)
- the
*average rate of change of*\(f\)*from*\(x=a\)*to*\(x=b\) - \(f'(a)\)
- the
*instantaneous rate of change of*\(f\)*at*\(x=a\) - the slope \(m\) of the
*secant line*that touches the graph of \(f(x)\) at the points \((a,f(a))\) and \((b,f(b))\)

**Question for the Class: **What are the members of the two groups?

One Group:

- the number \( m = \frac{f(b)-f(a)}{b-a} \)
- the
*average rate of change of*\(f\)*from*\(x=a\)*to*\(x=b\) - the slope \(m\) of the
*secant line*that touches the graph of \(f(x)\) at the points \((a,f(a))\) and \((b,f(b))\)

The Other Group:

- the number \( m = lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h} \)
- the
*instantaneous rate of change of*\(f\)*at*\(x=a\) - the slope \(m\) of the line is
*tangent*to the graph of \(f(x)\) at \(x=a\) - \(f'(a)\)

**Instructor: **In the **Video for H25**, you learned that the expression
$$m=\lim_{h\rightarrow0}\frac{f(a+h)-f(a)}{h}$$
computes a *number* \(m\) that is the ** slope of the line tangent to the graph of \(f(x)\) at \(x=a\)**. In this expression, \(a\) is a

In the **Video for H26**, you learned about replacing the *number* \(a\) with a *variable* \(x\). The resulting expression
$$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$$
would not represent a *number*, but rather would represent a *function* of \(x\). That function is called ** the derivative of \(f\)**. Here is the full definition:

**symbol:**\(f'(x)\)**spoken:**\(f\)*prime of*\(x\)**also spoken:***the derivative of*\(f\)*of*\(x\)**meaning:**the function $$m=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$$**Used For:**When an actual number \(x=a\) is substituted into the derivative function \(f'(x)\), the resulting number \(f'(a)\) can be interpreted as- the slope of the line tangent to the graph of \(f(x)\) at \(x=a\)
- the slope of the the graph of \(f(x)\) at \(x=a\)
- the instantaneous rate of change of \(f(x)\) at \(x=a\)
- if \(f(x)\) is a position function for a moving object, the number \(f'(a)\) is the velocity of the object at time \(x=a\).

Computing \(f'(x)\) using the *Definition of the Derivative* is difficult and messy.

Some of you may have studied calculus before and know about shortcuts to finding derivatives. The shortcuts are not nearly so difficult and not nearly so messy.

We learn the shortcuts in this course, too. They are called *Derivative Rules*. But they come later, starting in Section 2.5 of the book, with homework coverage starting in **Homework H29**. We'll begin discussing those easier methods next week. In this meeting, we will discuss two examples involving finding derivatives using the *Definition of the Derivative*. That is, the harder, messier method. Problems of this sort are among the most difficult problems that you will do in MATH 1350. The Examples will be done as **Class Drills**.

**Instructor: **Have students work on these two **Class Drills**.

Let \(f(x)=5x^2-7x+3\)

- Find \(f'(x)\) using the
*Definition of the Derivative* - Find \(f(2)\) and \(f'(2)\).
- What do the results of question
**b**tell you about the graph of \(f(x)\)?

**Instructor: **Show the solution to the problem that the class just worked on in the **Class Drill**. (See **[Example 1]**on page 9 of the **Notes from the Video for H26**.)

Let \(f(x)=5+\frac{7}{x}\)

- Find \(f'(x)\) using the
*Definition of the Derivative* - Find \(f(2)\) and \(f'(2)\).
- What do the results of question
**b**tell you about the graph of \(f(x)\)?

**Instructor: **Show the solution to the problem that the class just worked on in the **Class Drill**. (See **[Example 1]** on pages 4 - 10 of the **Notes from the Video for H27**.)

**Instructor: **Notice that the computation done in the second Class Drill involved a lot of steps, a lot more than the computation done in the first Class Drill. But notice that although the computation for the second Class Drill involved a lot of steps, there were no *tricks* in the computation. It was necessary to find a common denominator for two fractions, and that led to a mess of work. But finding a common denominator is not mysterious. You learned the basic idea of it in grade school.

**Instructor: **If there is enough time remaining in the meeting, present this **Example**.

In the two **Class Drills** that you just worked on, you used the *Definition of the Derivative* to find the derivative of a *polynomial* function and a \(1/x\) *type* function. A type of function that is simpler than either of these types is the *linear function*. That is, a function of the form
$$ y=mx+b$$
Because *linear functions* are relatively simple, one would expect that computing their *derivatives*, using the *Definition of the Derivative*, would also be relatively simple.

Well, it is true that the computation *is* relatively simple. But students are often confused by that very simplicity. Terms that students are used to seeing in the computation of the derivative of most functions are just not present in the computation of the derivative of a linear function. You have a problem involving finding the derivative of a *linear function* in your Homework for Section 2.4. You may find it confusing. For that reason, it is worthwhile for the instructor to do an example. (If there is not time in the meeting for the instructor to do this example, students can see the very similar **[Example 2]** on pages 14 - 17 of the **Notes from the Video for H26**.)

Let \(f(x)=5+\frac{7}{x}\)

- Find \(f'(x)\) using the
*Definition of the Derivative* - Use \(s'(x)\) to find \(s'(1), s'(2), s'(3)\).
*Illustrate*your results of \(s'(1), s'(2), s'(3)\) using a graph of \(s(x)\).

**Instructor: **Recall that on Monday, we discussed the **Definition of the Derivative**.

**symbol:**\(f'(x)\)**spoken:**\(f\)*prime of*\(x\)**also spoken:***the derivative of*\(f\)*of*\(x\)**meaning:**the function $$m=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$$**Used For:**When an actual number \(x=a\) is substituted into the derivative function \(f'(x)\), the resulting number \(f'(a)\) can be interpreted as- the slope of the line tangent to the graph of \(f(x)\) at \(x=a\)
- the slope of the the graph of \(f(x)\) at \(x=a\)
- the instantaneous rate of change of \(f(x)\) at \(x=a\)
- if \(f(x)\) is a position function for a moving object, the number \(f'(a)\) is the velocity of the object at time \(x=a\).

Also on Monday, you did two **Class Drills** in which *used* the **Definition of the Derivative** to compute derivatives

- Use the
**Definition of the Derivative**to compute the derivative of \(f(x)=5x^2-7x+3\). - Use the
**Definition of the Derivative**to compute the derivative of \(f(x)=5+\frac{7}{x}\)

We'll begin today's meeting with computation of a derivative using the **Definition of the Derivative**. You'll see that today's computation will be as messy as the computation of the derivative of \(f(x)=5+\frac{7}{x}\) on Monday. But not only is today's computation *messy*, it also involves a *trick*. You learned about the trick in your reading and in the Video for H27, and you use the trick in your solution to at least one of the homework problems from Section 2.4. Here is a review of the Trick.

Consider the limit
$$\lim_{h\rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$$
If we *try* substituting \(h=0\) into the expression, we obtain
$$\frac{\sqrt{x+(0)}-\sqrt{x}}{(0)}=\frac{\sqrt{x}-\sqrt{x}}{0}=\frac{0}{0}$$
This result **DOES NOT** tell us that the limit is undefined. Rather, it only tells us that the limit is an **indeterminate form**, and that we are *not allowed* to simply substitute \(h=0\). Instead, we must first do some work to eliminate the indeterminate form.

The work that we do is the following **trick**: We multiply the numerator and denominator of the fraction by the same particular expression, to obtain a new fraction. This doesn't change the meaning of the limit, because we are just multipling the fraction by \(1\). But if we simplify the new fraction, we will end up with a form that will allow us to cancel something, and thereby eliminate the indeterminacy.

Here are the details of the calculation: $$\begin{eqnarray} \lim_{h\rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} &\underset{\text{ Trick }}{=} & \lim_{h\rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\left(\sqrt{x+h}+\sqrt{x}\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)} \text{ (still indeterminate)}\\ &=& \lim_{h\rightarrow 0}\frac{\sqrt{x+h}\sqrt{x+h}-\sqrt{x}\sqrt{x+h}+\sqrt{x}\sqrt{x+h}-\sqrt{x}\sqrt{x}}{h\left(\sqrt{x+h}+\sqrt{x}\right)} \text{ (still indeterminate)}\\ &=&\lim_{h\rightarrow 0}\frac{(x+h)-(x)}{h\left(\sqrt{x+h}+\sqrt{x}\right)} \text{ (still indeterminate)}\\ &=& \lim_{h\rightarrow 0}\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)} \text{ (still indeterminate)} \\ & \ &\text{Because }h\rightarrow 0\text{, we know that }h \neq 0\text{, so we can cancel }\frac{h}{h} \\ &=& \lim_{h\rightarrow 0}\frac{1}{\sqrt{x+h}+\sqrt{x}} \text{ (no longer indeterminate, so we are allowed to just substitute in }h=0\text{)} \\ &=& \frac{1}{\sqrt{x+(0)}+\sqrt{x}} \\ &=& \frac{1}{\sqrt{x}+\sqrt{x}} \\ &=& \frac{1}{2\sqrt{x}} \end{eqnarray} $$

You'll do a similar computation in the following **Class Drill**.

**Instructor: **Have students work on the following **Class Drill**:

Let \(f(x)=5+7\sqrt{x}\)

- Find \(f'(x)\) using the
*Definition of the Derivative* - Find \(f(4)\) and \(f'(4)\).
- What do the results of question
**b**tell you about the graph of \(f(x)\)?

**Instructor: **Show the solution to the **Class Drill** that the class just worked on. (See **[Example 2]** on pages 11 - 15 of the **Notes from the Video for H27**.)

**Instructor: **The **Tangent Line** is introduced on **page 11 of the Notes from the Video for H25**. We discussed the Tangent Line during this past Friday's class meeting. Recall the definition:

**words:**the line tangent to the graph of \(f(x)\) at \(x=a\)**meaning:**the line that has the following two properties- The line contains the point \((x,y) = (a,f(a))\), which is called the
.*point of tangency* - The line has slope \(m=f'(a)\), called the
.*tangent line slope*

- The line contains the point \((x,y) = (a,f(a))\), which is called the

The example that the Instructor did on Friday was about finding the *slope* of the tangent line. This involved computing the number \(m=f'(1)\).

It is often useful to know not only the *slope* of the tangent line, but also the *equation* of the tangent line. (Since the *tangent line* is a *line*, its *equation* will be a *line equation*.) Problems involving the *equation of a tangent line* first appear in **Section 2.4** of the book, in your Homework for Section 2.4, and then they appear throughout the rest of Chapter 2 and in Chapter 3. Let's review the *form* of the *equation of a tangent line*.

Knowing the *two properties* of the tangent line introduced above, we are able to write down a *general form* for the ** equation for the tangent line**. It is discussed on

**words:**The equation for the line tangent to the graph of \(f(x)\) at \(x=a\)**meaning:**The equation $$(y-f(a))=f'(a)(x-a)$$ Observe that this equation is in. In the equation,*point slope form*- The symbol \(a\) represents the
*number*that is the \(x\) coordinate of the*point of tangency*. - The symbol \(f(a)\) represents the
*number*that is the \(y\) coordinate of the*point of tangency*. - In other words, the
*point of tangency*has coordinates \((a,f(a))\). - The symbol \(f'(a)\) represents the
*number*that is the*slope of the tangent line*.

- The symbol \(a\) represents the

- Write down the form \((y-f(a))=f'(a)(x-a)\) as a reminder of the equation that you need to build.
- Get the parts that you will need to populate the equation.
- Identify the value of \(a\). (the \(x\) coordinate of the
*point of tangency*) - Compute the value of \(f(a)\). (the \(y\) coordinate of the
*point of tangency*) - Compute the value of \(f'(a)\). (the slope \(m\) of the
*tangent line*)

- Identify the value of \(a\). (the \(x\) coordinate of the
- Substitute the parts \(a,f(a),f'(a)\) into the equation.
- Convert the equation to
*slope intercept form*.

In this **Class Drill**, you'll find the equation for a tangent line.

**Instructor: **Have students work in groups of on the following **Class Drill**:

The goal is to find the equation of the line tangent to the graph of \(f(x)=x^2-4x+8\) at \(x=3\).

- Identify the number \(a\). (It is the \(x\) coordinate of the
*point of tangency*. - Compute the number \(f(a)\). (It is the \(y\) coordinate of the
*point of tangency*.) - Compute the derivative \(f'(x)\) using the Definition of the Derivative. That is, compute this limit: $$f’(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ Show all steps clearly and use correct notation.
- Using the \(f'(x)\) that you found in (c), compute \(f'(a)\). (It is the slope \(m\) of the
*tangent line*.) - Using your \(a\),\(f(a)\),\(f'(a)\) from (a),(b),(d), build the point slope form of the equation of the tangent line. That is, build the equation $$(y-f(a))=f'(a)(x-a)$$
- Convert your tangent line equation to
*slope intercept form*.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on. (See **[Example 1]** on pages 11 - 15 of the **Notes from the Video for H25**.)

**Instructor: **If there is time remaining, disucuss the following Topic:

Some simple problems give you information about the values of \(a,f(a),f'(a)\) and ask you to give the equation of the tangent line. These problems are simpler in the sense that you don't have to do the computation of the parts \(f(a)\) and \(f'(a)\), because those quantities are given to you. But you still have to know the *general form* for the *equation for the tangent line*, and you have to understand how to substitute the given information about \(a,f(a),f'(a)\) into the general formula. Sometimes, this can actually be tricky. You will work on two such problems in this **Class Drill**.

- (Basic problem) Suppose that \(f(3)=5\) and \(f'(3)=7\).

What is the equation for the line tangent to the graph of \(f(x)\) at \(x=3\)?

Present your tangent line equation in slope intercept form. - (Harder problem) In Homework H25 problem [3], you were given information about a cyclist's progress in a bicycle race, and then asked some questions. Here is a similar problem:

Two hours after the start of a 200-kilometer bicycle race, a cyclist passes the 70 kilometer mark while riding at a velocity of 40 kilometers per hour.- Find the cyclist's average velocity during the first two hours of the race.
- Let \(x\) be the time (in hours) since the start of the race and let \(f(x)\) represent the distance traveled (in kilometers) at time \(x\). Find the slope of the
*secant line*through the points \((0,f(0))\) and \((2,f(2))\) on the graph of \(y=f(x)\). - Find the equation of the line tangent to the graph of \(y=f(x)\) at the point \((2,f(2))\).

**Instructor: **Show the solution to the **Class Drill** that the students just worked on. (See **[Example 2]** on
**pages 16 - 19 of the Notes from the Video for H25**
.)

**Instructor: **Now we come to discussing shortcuts to finding derivatives. The shortcuts are called *Derivative Rules*. They are presented in Section 2.5 and Chapter 3 of the book, with homework coverage beginning in the Homework for Section 2.5.

**Instructor: **(on the board) Introduce the new *notation* for derivatives. (Notation that is presented on
**page 8 of the Notes from the Video for H29**)

Our first derivative Rule is the ** Constant Function Rule**.

**The Constant Function Rule** This rule is used for finding the derivative of a

**Two equation form:**If \(f(x)=c\) then \(f'(x)=0\).**Single equation form:**\(\frac{d}{dx}c=0\)

**Instructor: **(on the board) Discuss why this rule makes sense, using the graph of a constant function \(f(x)=c\) and its derivative \(f'(x)=0\) to explain. (similar to discussion on
**pages 10,11 of the Notes from the Video for H29**)

Our next derivative Rule is the ** Power Rule**.

** Power Rule** This rule is used for finding the derivative of a

**Two equation form:**If \(f(x)=x^n\) then \(f'(x)=nx^{n-1}\).**Single equation form:**\(\frac{d}{dx}x^n=nx^{n-1}\)

Using the *Power Rule* can be complicated by the fact that functions are often not given in *power function form*. Using the *Power Rule* can be also complicated by calculations involving *fractional exponents*.

Furthermore, even though it is often necessary to rewrite functions in *power function form* in order to take a derivative, it will often happen that *after* finding a derivative, you will need to to rewrite the result in *positive exponent form*

**Instructor: Present this first basic Example of Using the Power Rule with Integer Exponents**

(See **[Examples 2,3]** on
**pages 13 - 15 of the Notes from the Video for H29**
.)

- Let \(f(x)=x^5\).
- Find \(f'(x)\) using the
*Power Rule*. - Find \(f(2)\) and \(f'(2)\) and simplify your result.

- Find \(f'(x)\) using the
- Let \(g(x)=\frac{1}{x^5}\).
- Rewrite \(g(x)\) in
*power function form*. - Find \(g'(x)\) using the
*Power Rule*. - Find \(g(2)\) and \(g'(2)\) and simplify your result.

- Rewrite \(g(x)\) in

**Instructor: **(on the board) Point out that the ** Power Rule** can only be used on functions that are written in

**Instructor: **(on the board) We will now move on to more difficult examples showing the use of the ** Power Rule**. In preparation for that, it will be useful to review a kind of calculation involving

**Instructor: **If there is time, have students work on the following **Class Drill**:

- Let \(f(x)=x^{1/3}\).
- Find \(f'(x)\) using the
*Power Rule*. Be sure to rewrite the result in*positive exponent form* - Find \(f(27)\) and \(f'(27)\) and simplify your result.

- Find \(f'(x)\) using the
- Let \(g(x)=\frac{1}{x^{1/3}}\).
- Rewrite \(g(x)\) in
*power function form*. - Find \(g'(x)\) using the
*Power Rule*. Be sure to rewrite the result in*positive exponent form* - Find \(g(27)\) and \(g'(27)\) and simplify your result.

- Rewrite \(g(x)\) in

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(See **[Examples 4,6]** on
**pages 15 - 17 of the Notes from the Video for H29**
.)

**Instructor: **If there is time, have students work on the following **Class Drill**:

- Let \(f(x)=\sqrt{x}\).
- Rewrite \(f(x)\) in
*power function form*. - Find \(f'(x)\) Rewrite the result using the
*square root symbol*. - Find \(f(9)\) and \(f'(9)\) and simplify your result.

- Rewrite \(f(x)\) in
- Let \(g(x)=\frac{1}{\sqrt{x}}\).
- Rewrite \(g(x)\) in
*power function form*. - Find \(g'(x)\) Rewrite the result in
*positive exponent form*. - Find \(g(9)\) and \(g'(9)\) and simplify your result.

- Rewrite \(g(x)\) in

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(See **[Examples 4,5,6]** on
**pages 15 - 17 of the Notes from the Video for H29**
.)

**Quiz Q3** during the last part of the Fri Jun 7 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The quiz material is taken from the material covered in Section 2.4. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

**Sections, Content, Homework, and Videos:**

- 2.5 Basic Differentiation Properties (Link to Homework and Videos for Section 2.5)
- 2.7 Marginal Analysis in Business and Economics (Link to Homework and Videos for Section 2.7)

**Instructor: **We will learn only one more derivative Rule in Chapter 2.

**The Sum and Constant Multiple Rule Rule **If \(f(x)\) and \(g(x)\) are

**Instructor [Example]** Use the basic derivative rules to find the following derivative.

Observe that this same derivative was the subject of a **Class Drill** during last Monday's meeting (Jun 3). On that day, the derivative was found used the ** Definition of the Derivative**. That is, by evaluating the following limit:
$$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$$
That method was much harder, but it gave the same result as using the

Using the *Sum and Constant Multiple Rule* can be complicated by the fact that functions are often not given in *power function form*. In two **Class Drills** you students will do examples of computing derivatives that require that the function be rewritten first, before taking the derivative.

**Instructor: **Have students work on the following **Class Drill**:

For the function $$f(x)=5+\frac{7}{x}$$ find \(f'(x)\) using the Derivative Rules.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(See **[Example 1]** on
**page 3 of the Notes from the Video for H31**
.)

**Instructor: **Observe that this same derivative was the subject of a **Class Drill** during last Wednesday's meeting (Jun 5). On that day, the derivative was found used the ** Definition of the Derivative**. That method was much harder, but it gave the same result as using the

**Instructor: **Have students work on the following **Class Drill**:

For the function $$f(x)=5+7\sqrt{x}$$ find \(f'(x)\) using the Derivative Rules.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(See **[Example 2]** on
**page 5 of the Notes from the Video for H31**
.)

**Instructor: **Observe that this same derivative was the subject of a **Class Drill** during last Wednesday's meeting (Jun 5). On that day, the derivative was found used the ** Definition of the Derivative**. That method was much harder, but it gave the same result as using the

**Instructor: **The skill of rewriting the function be *first*, *before* taking the derivative, will be one of the most important skills for the rest of the course. Please work on a **Class Drill** that will show you the general approach.

**Instructor: **Recall that we discussed the **Tangent Line** in a previous meeting. The *Tangent Line* is introduced on
**page 11 of the Notes from the Video for H25**. The key thing is to remember from the definition is the ** two properties of the tangent line**.

The line tangent to the graph of \(f(x)\) at \(x=a\) has the following two properties

- The line contains the point \((x,y) = (a,f(a))\), which is called the
.*point of tangency* - The line has slope \(m=f'(a)\), called the
.*tangent line slope*

Knowing those *two properties* of the tangent line, we are able to write down a *general form* for the ** equation for the tangent line**. It is discussed on

The general approach to finding the equation of the tangent line is

- Write down the form \((y-f(a))=f'(a)(x-a)\) as a reminder of the equation that you need to build.
- Get the parts that you will need to populate the equation.
- Identify the value of \(a\). (the \(x\) coordinate of the
*point of tangency*) - Compute the value of \(f(a)\). (the \(y\) coordinate of the
*point of tangency*) - Compute the value of \(f'(a)\). (the slope \(m\) of the
*tangent line*)

- Identify the value of \(a\). (the \(x\) coordinate of the
- Substitute the parts \(a,f(a),f'(a)\) into the equation.
- Convert the equation to
*slope intercept form*.

Some simple problems actually give you information about \(a,f(a),f'(a)\) and ask you to give the equation of the tangent line. These problems are simpler, because you don't have to do the computation of the parts \(f(a),f'(a)\). They are given to you. But you still have to know the *general form* for the *equation for the tangent line*, and you have to understand how to substitute the given information about \(a,f(a),f'(a)\) into the general formula.

**Instructor: **Have students work on the following **Class Drill**:

Suppose that a function \(f(x)\) has \(f(3)=5\) and \(f'(3)=-7\).

- What is the
*slope*of the line that is*tangent*to the graph of \(f(x)\) at \(x=3\)? - What is the
*equation*of the line that is*tangent*to the graph of \(f(x)\) at \(x=3\)?

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(To prepare for this presentation, study
**pages 11 - 14 of the Notes from the Video for H25**
.)

**Instructor: **Here is a much longer example, involving a variety of types of questions that can be asked about *tangent lines*.

(Instructor present this example. For a similar example, see
**[Example 2] on pages 7 - 11 of the notes for Video 32**.)

**Instructor [Example #1]: ** Let \(f(x)=x^{3}+3x^2-9x+5=(x+5)(x-1)^2\)

- Find \(f'(x)\).
- Find the slope of the line that is tangent to the graph of \(f(x)\) at \(x=-2\)
- Find the slope of the line that is tangent to the graph of \(f(x)\) at \(x=0\)
- Find the \(x\) coordinates of all points on the graph of \(f(x)\) that have horizontal tangent lines.
- Find the equation of the line that is tangent to the graph of \(f(x)\) at \(x=2\). Show all details clearly and present your equation in slope intercept form.
- Illustrate your results on a graph of \(f(x)\).

**Instructor Ask Question #1 for the Class:** Frick and Frack have been asked the following:

They are arguing about the result.

- Frick says that the slope is \( 3x^2 \) because the
*derivative*is the*tangent line*. - Frack that the the slope is $$ m=\frac{f(6)-f(5)}{6-5}=\frac{216-125}{1}=91$$

Frick and Frack are ** both** wrong!

Frick says that the derivative is the slope of the tangent line. But this is not correct. The *slope of the tangent line* is a *number*. The *derivative* is a *function*, not a *number*. (The *derivative* is a *function* that can be used to *find* the *number* that is the *slope of the tangent line*.)

Frack is also wrong. Frack computed the *slope of a secant line*.

The correct procedure to find the *slope* of the line *tangent* to the graph of \(f(x)=x^3\) at \(x=5\) is as follows.

**Step 1: **Find \(f'(x)\). The result is

**Step 2: **Substitute \(x=5\) into \(f'(x)\) to get \(m=f'(5)\). The result is

**Instructor Ask Question #2 for the Class:** Wacky Jack has been asked the following:

Their answer was $$y=2x^3-5x^2+4x-11$$ Is Wacky Jack's answer correct?

- Wacky Jack's answer is correct.
- Wacky Jack's answer is incorrect.
- Not enough information. One needs to know the function \(g(x)\) before being able to say whether Wacky Jack's answer is right or wrong.

At first, you might think that of course one would need more information before being able to say whether Wacky Jack's answer is right or wrong. But in fact, it is easy to see immediately that **Wacky Jack's answer is incorrect**.

The key is to remember that Wacky Jack was asked to find *the equation of a line*. That means that his result must be in the form
$$y=mx+b$$
where \(m\) and \(b\) are *numbers*. Since Wacky Jack's answer is not in that form, his answer is incorrect.

This example illustrates one kind of *quick check* on problems involving finding the equation of a tangent line. You will encounter problems of that sort where the calculations get quite messy. But the end result should always be an equation of the form \(y=mx+b\).

Instructor: If there is time, discuss another topic:

**Instructor ask Question #3 for the Class:** Suppose that an object is moving along a straight track with *position* function \(f(x)\), where \(x\) is the time in seconds and \(f(x)\) is the position in meters at time \(x\).

- What is the meaning of this quantity? $$ \frac{f(17)-f(13)}{17-13}$$
- What is the meaning of the quantity \(f'(8)\)?

Recall that the quantity
$$ \frac{f(17)-f(13)}{17-13}$$
is called the *average rate of change of \(f\) from \(x=13\) to \(x=17\)*.

And the quantity $$ f'(8)$$
is called the *instantaneous rate of change of \(f\) at \(x=8\)*.

In the special case that the variable \(x\) represents *time* and the function \(f(x)\) represents the *position, at time \(x\)*, of an object moving in one dimension, the quantity
$$ \frac{f(17)-f(13)}{17-13}$$
is called *the average velocity of the object from time \(x=13\) to time \(x=17\)*.

And the quantity $$ f'(8)$$
is called the *instantaneous velocity of the object at time \(x=8\)*.

The units of both of these velocities will be the ratio
$$\frac{units \ of \ position}{units \ of \ time}$$
So in our particular example, the units will be *meters per second*.

To understand why these are the units, remember that the quantities are *slopes* of lines on a graph of \(f(x)\) -vs- \(x\). The slope of a line is computed by the ratio
$$m=\frac{\Delta y}{\Delta x}$$
If the \(x\) and \(y\) quantities have units attached, then the units of the slope will work out to be
$$m=\frac{units \ of \ y}{units \ of \ x}$$
In the special case of *position* and *velocity*, the units will be
$$m=\frac{units \ of \ f(x)}{units \ of \ x}$$
That is
$$m=\frac{units \ of \ position}{units \ of \ time}$$

Instructor: If there is time, discuss another topic:

**Instructor: **Remember that in Mathematics, an *Application Problem* involves using math to solve some *real world problem*. An important part of an *application problem* is *interpreting the result*. What this means is the following: In an *application problem*, one uses *functions* and *equations* to model an actual real-world situation. One then does *abstract mathematical calculations* to reach some *abstract mathematical result*. It is important to write a clear conclusion that explains what the abstract mathematical result tells us about the real-world situation. This is what is meant by *interpreting the result*.

In Calculus, an *Application Problem* often uses the important fact that the following quantities are *equal*.

- the value of \(f'(a)\)
- the number \(m\) that is the slope of the line tangent to the graph of \(f(x)\) at the point \((a,f(a))\)
- the
*instantaneous rate of change of \(f(x)\) at \(x=a\)* - (In the special case that \(f(x)\) is a
*position function*for an object moving in one dimension, another way of saying the previous item is, the*velocity of the object at time \(x=a\)*.)

*Interpreting the result* will often involve explaining something about a *rate of change*.

Here is an example that asks the student to *interpret* some results.

**Instructor [Example #2]** (See
**pages 5-8 of the Notes from the Video for H33** for a similar example.)

A company introduced a new **Gas Powered Back Scratcher**. Total sales are described by the function \(S(t)\), where \(t\) is the time in months since the new Gas Powered Back Scratcher was introduced and \(S(t)\) is the total sales (in millions of dollars) at time \(t\).

- Suppose that \(S(6) = 3\).
*Interpret*that information. That is, explain what it tells us about sales of the Gas Powered Back Scratcher. - Suppose that \(S'(6) = 2\).
*Interpret*that information. That is, explain what it tells us about sales of the Gas Powered Back Scratcher. - What does the information in parts (a) and (b) tell us about the
*graph*of \(f(t)\)?

**Instructor: **So far, when we have discussed *applications* of Calculus, they have been applications of a sort that one would see in any Calculus course. Namely, the applications have been about the important fact that the following quantities are *equal*.

- the value of \(f'(a)\)
- the number \(m\) that is the slope of the line tangent to the graph of \(f(x)\) at the point \((a,f(a))\)
- the
*instantaneous rate of change of \(f(x)\) at \(x=a\)* - (In the special case that \(f(x)\) is a
*position function*for an object moving in one dimension, another way of saying the previous item is, the*velocity of the object at time \(x=a\)*.)

But MATH 1350, in addition to being a *Survey of Calculus* course, also fills the role of being the *Business Calculus* course for Ohio University. In that role, the course will consider applications of Calculus that have to do with *Business*.

In the Video for H34, you were introduced to some *Business Terminology*, namely *Demand* and *Cost*.

Instructor Project
**page 2 of the Notes from the Video for H34**.

Homework H34 is a very short homework assignment about computing *Cost*. The homework set contains only one problem, because the *MyLab* system contains only one problem of the sort. But the Video for H34 presents a more thorough example about computing cost. You will work on a **Class Drill** involving a similar example, which we will call the **[Bicycle Example]**.

In the Video for H35, you were introduced to some more *Business Terminology*, namely *Revenue* and *Profit*, and the concept of *Marginal Quantities*.

Remember that a ** Marginal Quantity** is simply the

In the Video for H36, you learned about using *Marginal Quantities* to *Estimate Change in Quantities* . The general idea is summarized as follows:

**Using Marginal Quantities to Approximate Change in Quantities**

In the following **Class Drill**, you will do an example called the **[Bicycle Example]**. In the example, you will first compute an ** Exact Change** in a Quantity. Then you will compute a

**Instructor: **Have students work on the following **Class Drill**:

The total cost of producing \(x\) bicycles is

$$C(x)=10,000+150x-0.2x^2 \ \ dollars$$**(A) **What is the cost of producing a batch of \(140\) bicycles? (Show all details clearly and simplify your answer.)

**(B) **What is the cost of producing a batch of \(141\) bicycles? (Show all details clearly and simplify your answer.)

**(C) **If batch size changes from \(x=140\) bicycles to \(x=141\) bicycles, what will be the change in cost of producing a batch of bicycles? That is, if \(x=140\) and \(\Delta x = 1 \), what is \(\Delta C\)? (exact value) (Show all details clearly and simplify your answer.) This quantity is an ** Exact Change in Cost**. (The book calls this quantity

**(D) **Find the *Marginal Cost*. (Show all details clearly and simplify your answer.)

(Hint: Since you are not told what method to use to find the Marginal Cost, use the *easy method*, involving the *Derivative Rules*, rather than the *hard method*, involving the *Definition of the Derivative*.)

**(E) **If batch size changes from \(x=140\) bicycles to \(x=141\) bicycles, use the marginal cost function to find an approximate value for the change in cost of producing a batch of bicycles. That is, use the *Marginal Cost Function* to find an *approximation* for \(\Delta C\). (Show all details clearly and simplify your answer.) The book calls this quantity *the approximate cost of producing the 141 ^{st} bicycle*.)

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For Reference, see the
**[Example 1] on page 3 of the Notes from the Video for H34**
and the
**[Example 1] on page 10 of the Notes from the Video for H36**.)

**Instructor: **Notice that the *approximate* result found in **(E)** is very close to the *exact* result found in **(C)**.

**Instructor: **In the Video for H35, you were introduced to *Revenue*.

(Instructor Project
**page 2 of the Notes from the Video for H35**.)

Not much was said there about *Revenue* except that \(R(x)\) is the amount of money that comes in from the sale of a batch of \(x\) items.

In the Video for H36, the concept of *Revenue* is developed further. Using the simplifying assumption that all the items in a batch of items are sold for the *same selling price*, we can introduce a new business term: *price*.

(Instructor Project
**page 13 of the Notes from the Video for H36**. Discuss the terms *Price*, *Price Demand Equation*, and *Price Function*. Also discuss the relationship between *Price* and *Revenue* shown on that page.)

You will explore *Price* and *Revenue* in an extended example in a **Class Drill**.

**Instructor: **Have students work on the following **Class Drill**:

A company makes *hoverboards*, a very dangerous toy that kids would not ride if they had a lick of sense. The *price demand equation* is
$$x=6,000-30p$$
where \(p\) is the selling price of a hoverboard, in dollars, and \(x\) is the number of hoverboards that will sell at that price.

- Find the
*price function*. That is, express the*price*\(p\) as a*function*of the*demand*\(x\). Graph the*price function*and state its domain. Make your graph large and clear, and label important stuff. Use you graph to explain why the domain is what you say it is. - Find the
*Revenue function*, \(R(x)\), and state its domain. The formula that you get for the*Revenue Function*should simplify to the following form: $$R(x)=200x-\frac{x^2}{30}$$ where \(R(x)\) is the amount of*Revenue*obtained when \(x\) hoverboards are sold. - If you can, make a graph of the
*Revenue function*on the domain that you have determined for the function. Make your graph large and clear, and explain why it looks the way it does. (Hint: Consider the*standard form*of the function \(R(x)\) to determine the shape of the graph. Use the*factored form*of the function \(R(x)\) to determine the \(x\)*intercepts*.) - Find the
*Marginal Revenue at a production level of 2250 hoverboards*and. (Use correct units in your explanation.)*interpret the results* - Find the
*Marginal Revenue at a production level of 3750 hoverboards*and. (Use correct units in your explanation.)*interpret the results*

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 2] on page 14 of the Notes from the Video for H36**
.)

**Exam X1** lasts the full duration of the Fri Jun 14 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The exam material is taken from the material covered in Chapter 2. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

**Sections, Content, Homework, and Videos:**

- 3.1 The Constant \(e\) and Continously-Compounded Interest (Link to Homework and Videos for Section 3.1)
- 3.2 Derivatives of Exponential and Logarithmic Functions (Link to Homework and Videos for Section 3.2)

**Instructor: **Book Section 3.1 begins with an introduction to the number \( e \) and the function \( e^{(x)} \). That introduction involves a lot of detailed discussion of limits that is better left to the video. (See the
Video for H37 and its accompanying notes.) But it is useful to discuss the *graphs* of basic exponential functions.

**Instructor: **Draw graphs of \( y=2^{(x)} \) and \( y=e^{(x)} \) \( y=3^{(x)} \) together on one set of axes. Point out that the graph of \( y=e^{(x)} \) is always *between* the other two because \( 2 \lt e \lt 3 \).)

**Questions for Class: **What are some important properties of these graphs? (Instructor: see
**page 11 of the Notes from the Video for H37** for a discussion of this topic.)

- Domain and Range
- Coordinates of some distinctive points
- End behavior (expressed in
*words*and expressed using*limit*terminology and notation.)

**Instructor: **Book Section 3.1 also explores ways of computing interest on a bank account. Three kinds of bank account interest are discussed:

- Simple Interest
- Periodically-Compounded Interest
- Continuously-Compounded Interest

(Instructor: See
**page 2 of the Notes from the Video for H38**
for a discussion of this topic.)

**Instructor: **Introduce how an account with ** Simple Interest** works. Write the formula for the balance of an account with

- \(y\) intercept
- slope

**Instructor do this [Example]: ** Suppose that $1000 is deposited into a bank account with \(5\%\) ** simple interest**. What will be the balance after 10 years?

(Instructor: See
**page 5 of the Notes from the Video for H38** for a discussion of this topic.)

**Instructor: **Introduce how an account with ** Periodically-Compounded Interest** works. Write the formula for the balance of an account with

- \(y\) intercept
- straight segments where the account is earning simple interest
- the points where the slope abruptly changes due to the interest being
*compounded*

**Instructor do this [Example]: ** Suppose that $1000 is deposited into a bank account with \(5\%\) interest ** compounded monthly**. What will be the balance after 10 years?

(Instructor: See
**page 11 in the Notes from the Video for H38** for a discussion of this topic.)

**Instructor: **Introduce how an account with ** Continuously-Compounded Interest** works. Write the formula for the balance of an account with

**Instructor: **The equation \(A=Pe^{(rt)}\) involves the four letters \(A,P,r,t\) and is solved for \(A\). The equation can be solved for each of the other letters \(P,r,t\). This is useful in solving a variety of problems that you will encounter in your homework.

**Instructor: **Show how to solve that equation for the other variables.

Here are examples of four kinds of problems whose solutions involve solving the equation \(A=Pe^{(rt)}\) for one of the letters \(P,r,t\):

- For an account with an
*interest rate*of \( 5 \% \)*compounded continuously*, how much money should be deposited if the goal is to have a balance of \( $1000 \) after \(10\) years? - For an account with an
*initial deposit*of \( $100 \) and an*interest rate*of \( 5 \% \)*compounded continuously*, how long until the balance is \( $150 \)? - For an account with an
*interest rate*of \( 5 \% \)*compounded continuously*, how long until the balance*doubles*? - If you want an account with
*continuously compounded interest*to*double*in value in \(20\) years, what*interest rate*will you need?

You will have problems of these types to do in your Homework H38. Examples of three of the four types are presented on
**pages 12 - 14 of the Notes from the Video for H38**. Observe the style of the solutions. They always follow this format:

- Write down any known values for the four variables \(A,P,r,t\), and identify which variable is to be found.
- Solve the equation \(A=Pe^{(rt)}\) for the variable that is to be found.
- Only then,
*after*the equation has been solved, are the known values substituted in for the other variables. - Simplify the expression to get an
.*exact, simplified answer* - Only then,
*after*anis found, are numbers typed into a calculator (or computer) to get a decimal approximation.*exact, simplified answer*

we'll see two basic examples of these types of problems in **Class Drills**.

**Instructor: **Have students work in groups of 2 or 3 on the following **Class Drill**:

Suppose that \($2000\) is deposited into an account with \(5\%\) interest *compounded continuously*. How long after the initial deposit will the balance have grown to \($3000\)? (Give an *exact answer*, in *symbols*, then a *decimal approximation*. Show all details clearly.)

**Student #2 Presentation CP2: **Show the solution to the **Class Drill** that the class just worked on.
(To prepare for this presentation, study
**pages 12 - 14 of the Notes from the Video for H38**
.)

**Instructor: **Have students work on the following **Class Drill**:

If you want an account with *continuously compounded interest* to double in 15 years, what interest
rate will you need? (The book would ask *at what nominal rate compounded continuously must the money be invested?*)
(Give an *exact answer*, in *symbols*, then a *decimal approximation*. Show all details clearly.)

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**pages 12 - 14 of the Notes from the Video for H38**
.)

**Instructor: **In this section, we will learn how to find the derivatives of *exponential functions*. Before discussing that, however, it is worthwhile to note the distinctions between *exponential functions* and *power functions*. Both types of functions have a *base* and an *exponent*. But there is an important difference: which parts are *constant* and which parts are *variable*.

In a *power function*,
$$f(x)=x^p$$
the *base* is a *variable* and the *exponent* is a *number*.

In an *exponential function*,
$$f(x)=e^{(x)} \ \text{ or } \ b^{(x)} \ \text{ or } \ e^{(kx)} \ \text{ or } \ b^{(kx)}$$
the *base* is a *number* and the *exponent* is a *variable* (or an expression involving a *variable*).

Recall the *Power Rule* for derivatives of *power functions*:

In book Section 3.2 and in the MATH 1350 Videos, we see the introduction of *Exponential Function Rules #1, #2, #3*, for finding derivatives of *exponential functions*:

We now have *four rules* for finding derivatives of expressions that involve exponents. When taking such a derivative, it is important to use the correct rule!

In your Homework for Section 3.2, you are asked to find various derivatives involving exponential functions, power functions, and constant functions. To succeed in these problems, you have to figure out which rule to use. Barsamian does similar examples in the Video for H40. (Link to Notes for that Video)

**Instructor: ** Do this **[Example: Computing Derivatives of Expressions that have Exponents]**.

Find the following derivatives. Use clear notation and indicate which rule you used in each step.

$$(a) \ \frac{d}{dx} 7x = $$ $$(b) \ \frac{d}{dx} 7x^3 = $$ $$(c) \ \frac{d}{dx} 7e^x = $$ $$(d) \ \frac{d}{dx} 7e^{13} = $$ $$(e) \ \frac{d}{dx} 7\cdot 13^x = $$In your Homework for Section 3.2, you are asked to solve a tangent line problem for a function \(f(x)\) that has a term that is an exponential function. Barsamian does a similar example in the Video for H41. (Link to Notes for that Video)

**Instructor: ** Do this **[Example: Computing a Tangent Line Equation for an Exponential Function]**.

For the function $$f(x)=3e^{(x)}-5x$$

- Find the equation of the line tangent to the graph of \(f(x)\) at \(x=0\).
- Sketch the graph of \(f(x)\) and draw the tangent line. Label important stuff.

**Instructor: ** Recall something that was first discussed during the Friday May 31 meeting and has been discussed in meetings since, and that is discussed throughout the Videos:

**An important concept for the second month of MATH 1350 is the relationship between the derivative, the tangent line, and rate of change:**

- The value of \( f'(a) \)
- The slope \( m \) of the line
*tangent*to the graph of \( f(x) \) at \(x=a\) - The
*instantaneous rate of change*of \( f(x) \) at \(x=a\)

Sometimes you will get questions that require you to understand the *connection* between those three quantities.

In your Homework for Section 3.2, you are asked to answer two questions about the ** instantaneous rate of change** of the value of an investment that is earning

**Instructor: **Have students work on the following **Class Drill**:

An Investment of \($5000\) earns interest at an annual rate of \(4 \% \) compounded continously.

- Find the
**amount in the account**after \(7\) years. - Find the
**instantaneous rate of change of the amount in the account**after \(7\) years. - Find the
**instantaneous rate of change of the amount in the account**when the amount is equal to \($7000\).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see **[Examples 2 and 3] in the Notes from the Video for H41**
.)

**Sections, Content, Homework, and Videos:**

- 3.2 Derivatives of Exponential and Logarithmic Functions (Link to Homework and Videos for Section 3.2)
- 3.3 Derivatives of Products and Quotients (Link to Homework and Videos for Section 3.3)

**Instructor: **Today, we'll be discussing *logarithmic functions* and their *derivatives*. We start by reviewing the *graph* of the simplest *logarithmic function*, the function

See the **Notes from the Video for H43**, and discuss the following questions:

- How does one get the graph of \(y=\ln{(x)}\) from the graph of \(y=e^{(x)}\)?
- What is the value of \(\ln{(1)}\)? Use the graph to explain.
- What is the value of \(\ln{(e)}\)? Use the graph to explain.

We find the *derivative* of the *natural logarithm* function \(y = \ln{(x)} \) by using a rather surprising new derivative rule:

**Logarithmic Function Rule #1: **
$$\frac{d}{dx}\ln{(x)} = \frac{1}{x} $$

We can consider why **Logarithmic Function Rule #1** makes sense by projecting and discussing
**page 7 of the Notes from the Video for H43**.)

We find the *derivative* of the more general *base \(b\) logarithm* function \(y = \log_b{(x)} \) by using a slight variation of the above rule:

**Logarithmic Function Rule #2: **
$$ \frac{d}{dx}\log_b{(x)} = \frac{1}{x\cdot\ln{(b)}} $$

**Instructor: **It is important to realize that both of the new *Logarithmic Function Rules* only work when the thing inside the logarithm is just the *variable*. What happens when the thing inside the logarithm is an expression that is not simply the variable? You'll tackle this issue in a **Class Drill**.

Derivatives with just a *single item* inside the logarithm

- Let \(f(x)=3\ln{(x)}\). Find \(f'(x)\).
- Let \(f(x)=5\log_7{(x)}\). Find \(f'(x)\).
- Let \(f(x)=11\ln{(13)}\). Find \(f'(x)\).
- Let \(f(x)=17\log_{19}{(23)}\). Find \(f'(x)\).

Derivatives with an *expression* inside the logarithm

- Let \(f(x)=\ln{(5x)}\). Find \(f'(x)\).
**Hint: Rewrite the function***before*finding the derivative! - Let \(f(x)=\ln{(x^5)}\). Find \(f'(x)\).
**Hint: Rewrite the function***before*finding the derivative!

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see the
**[Examples] on page 9 of the Notes from the Video for H43**
.)

**Instructor: **In your homework, you are asked to find the equations of lines tangent to graphs of functions involving *exponential functions* (in H41) and *logarithmic functions* (H44). These might at first look like very hard problems.

But understand that one part of the difficulty involves using **rules of exponents and logarithms**, which are rules from

Another bit of difficulty in the tangent line problems is that you have to find ** derivatives**. But the only derivatives involved in Homeworks H41 and H44 are the very basic rules:

- Sum and Constant Multiple Rule $$\frac{d}{dx}\left(af(x)+bg(x)\right)=a\frac{d}{dx}f(x)+b\frac{d}{dx}g(x)$$ Using prime notation, we could write $$\left(af(x)+bg(x)\right)' = af'(x)+bg'(x)$$
- Constant Function Rule
**Two equation form:**If \(f(x)=c\) then \(f'(x)=0\).**Single equation form:**\(\frac{d}{dx}c=0\)

- Exponential Function Rule #1 $$ \frac{d}{dx} e^{(x)} = e^{(x)}$$
- Logarithmic Function Rule #1 $$\frac{d}{dx}\ln{(x)} = \frac{1}{x} $$

A third bit of difficulty in the tangent line problems is ** understanding what is being asked for and knowing how to structure your solution to the problem**. That is worth reviewing here.

Recall that the *line tangent to the graph of \(f(x)\) at \(x=a\)* has the following two properties:

- The line contains the point \((x,y) = (a,f(a))\), which is called the
.*point of tangency* - The line has slope \(m=f'(a)\), called the
.*tangent line slope*

Knowing those *two properties* of the tangent line, we are able to write down a *general point slope form* for the ** equation for the tangent line**.

The general approach to finding the equation of the tangent line is

- Write down the
*general form*\((y-f(a))=f'(a)(x-a)\) as a reminder of the equation that you need to build. - Get the parts that you will need to populate the equation.
- Identify the value of \(a\). (the \(x\) coordinate of the
*point of tangency*) - Compute the value of \(f(a)\). (the \(y\) coordinate of the
*point of tangency*) - Compute the value of \(f'(a)\). (the slope \(m\) of the
*tangent line*)

- Identify the value of \(a\). (the \(x\) coordinate of the
- Substitute the parts \(a,f(a),f'(a)\) into the equation.
- Convert the equation to
*slope intercept form*.

Although problems about finding tangent line equations can seem daunting, remember that the general *form* of the solution is always the same, and that often, the underlying details involve *old concepts from high school* or *basic derivatives from calculus*.

**Instructor: **For the rest of the meeting, you will work on two **Class Drills** involving tangent lines.

Find the *equation of the line tangent to the graph of \( f(x)=5e^{(x)}-7x\) at \(x=1\)*.

(**Hint: **Remember that you should start by building the *point slope form of the equation of the tangent line*.
$$\left(y-f(a)\right)=f'(a)\left(x-a\right)$$
After finding the parts \(a\),\(f(a)\),\(f'(a)\) and using them to build the formula, you should convert your formula to *slope intercept form*. Simplify your answer.)

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**MyLab Homework H41 problem #1** and
**[Example 1] on page 3 of the Notes from the Video for H41**
.)

**Instructor: **If there is time, have students do another **Class Drill**:

(This Class Drill is based on one of the problems from the Homework for Section 3.2.)

Let

$$f(x)=7+\ln{(x^5)}$$- Rewrite \(f(x)\) using a rule of logarithms so that it does not involve a nested function.
- Find \(f'(x)\).
- Find the
of the line tangent to the graph of \(f(x)\) at \(x=1\). Simplify your answer.*slope* - Find the
of the line tangent to the graph of \(f(x)\) at \(x=1\).*equation*

(**Hint:**Remember that you should start by building the*point slope form of the equation of the tangent line*. $$\left(y-f(a)\right)=f'(a)\left(x-a\right)$$ After finding the parts \(a\),\(f(a)\),\(f'(a)\) and using them to build the formula, you should convert your formula to*slope intercept form*. Simplify your answer.) - Find the
of the line tangent to the graph of \(f(x)\) at \(x=e\). Simplify your answer.*slope* - Find the
of the line tangent to the graph of \(f(x)\) at \(x=e\).*equation* - Find the
of the line tangent to the graph of \(f(x)\) at \(x=e^3\). Simplify your answer.*slope* - Find the
of the line tangent to the graph of \(f(x)\) at \(x=e^3\).*equation*

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(See the
**[Example] on page 3 of the Notes from the Video for H44**
.)

**Instructor: **We'll start today's meeting by discussing *Products of Functions*. That is, functions of the form
$$f(x)=g(x)\cdot h(x)$$
A natural question is:

There is an **obvious way** that might occur to you, but **the obvious way is WRONG!!**
$$ \frac{d}{dx}\left( g(x) \cdot h(x) \right) \neq \left( \frac{d}{dx}g(x) \right)\cdot \left( \frac{d}{dx} h(x) \right) $$
That is,

The correct way to find the derivative of a product is to use the **Product Rule**.

Using prime notation, we would write the **Product Rule** this way:

**Instructor : ** Do this Example] (Similar to MyLab H45 Problem #4)

For the function
$$f(x)=e^{(x)}\left(7x^3-5x^2+13\right)$$

- Show how to use the
*Product Rule*to find \(f'(x)\). Simplify your answer. - Find \(f'(0)\) and simplify.
- Find \(f'(1)\) and simplify.

The class will do another basic example involving the ** Product Rule** as a

For the function $$f(x)=3x^5\ln{(x)}$$

- Show how to use the
*Product Rule*to find \(f'(x)\). Show all steps clearly and simplify your answer. - Find \(f'(1)\). Simplify your answer.
- Find \(f'(e)\). Simplify your answer.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 4] on page 9 of the Notes from the Video for H45**
.)

**Instructor: **Next we'll discuss *Quotients of Functions*. That is, functions of the form
$$f(x)=\frac{top(x)}{bottom(x)}$$
A natural question is:

There is an **obvious way** that might occur to you, but **the obvious way is WRONG!!**
$$\frac{d}{dx}\left(\frac{top(x)}{bottom(x)}\right) \neq \frac{\frac{d}{dx}top(x)}{\frac{d}{dx}bottom(x)} $$
That is,

The correct way to find the derivative of a quotient is to use the **Quotient Rule**.

Using prime notation, we would write the **Quotient Rule** this way:

**Instructor : ** Do this **[Example]** (Similar to one of the exercises on the Homework for Section 3.3)

For the function
$$f(x)=\frac{e^{(x)}}{7x^5+3}$$

- Show how to use the
*Quotient*to find \(f'(x)\). Simplify your answer. - Find \(f'(0)\) and simplify.
- Find \(f'(1)\) and simplify.

**Instructor: **The *Quotient Rule* is a very messy derivative rule, and there are lots of opportunities to make mistakes when using it. Therefore, it is important to recognize situations when the quotient rule is *unnecessary*. Sometimes a function that is presented as a *quotient* can be *rewritten* in a way that it is *not* a quotient. Then, *simpler* derivative rules can be used to find the derivative.

**Instructor: ** Do this **[Example]** involving the ** Very Important Trick**.

For the function $$f(x)=\frac{3}{x^5}$$ find \(f'(x)\) using two different methods.

- Use the
*Quotient Rule*to find \(f'(x)\). Show all steps clearly. - Start over. First, simplify \(f(x)\) by converting from
*positive exponent form*to*power function form*. (This is the.) Then, use*Very Important Trick**simpler*derivative rules to find \(f'(x)\). Show all steps clearly.

Observe that in the **[Example]**, both methods give the same result, but using the *Quotient Rule* involves more work and presents more opportunities to make mistakes!

The important conclusion from the **[Example]** is

**When finding the derivative of a function, one should always check to see if it is possible convert the function to power function form before taking the derivative.**

The class will do another problem involving the ** Very Important Trick** in a

For the function $$ y=\frac{2x^5-4x^3+2x}{x^3} $$ find \(f'(x)\) using two different methods.

- Use the
*Quotient Rule*to find \(f'(x)\). Show all steps clearly. - Start over. First, simplify \(f(x)\) by converting from
*positive exponent form*to*power function form*. (This is the.) Then, use*Very Important Trick**simpler*derivative rules to find \(f'(x)\). Show all steps clearly.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see the
**Notes from the Video for H47**
.)

(Note: no books, no notes, no calculators, no phones)

**Quiz Q4** during the last part of the Jun 26 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The quiz material is taken from the material covered in book Sections 3.1 and 3.2. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

**Instructor: **Last Wednesday, we discussed the **Quotient Rule**.

Using prime notation, we would write the **Quotient Rule** this way:

We'll start today's meeting with a problem involving finding the **equation for the tangent line** for function that is a quotient. You'll solve the problem as a **Class Drill**.

Let \(f(x)\) be the function
$$f(x)=\frac{50x}{x+4}$$
Find the *equation of the line tangent to the graph of \(f(x)\) at \(x=6\)*. Convert your equation to *slope intercept form*.

**Student #9 Presentation CP2: **Show the solution to the **Class Drill** that the class just worked on.
(To prepare for this presentation, study
**MyLab Homework H48 problem #1**
and
**[Example 1] in the Notes from the Video for H48**
.)

**Instructor: **An important concept for Sections 2.4, 2.5, 3.2, 3.3, 3.4 is the relationship between the *derivative*, the *tangent line*, and *rate of change*. In the videos, you have seen discussed the idea that the following three quantities are all *equal*:

- The value of \( f'(a) \)
- The slope \( m \) of the line
*tangent*to the graph of \( f(x) \) at \(x=a\) - The
*instantaneous rate of change*of \( f(x) \) at \(x=a\)

In the following **[Example]**, an *Application* problem is posed. Questions (a), and (b) are about calculating certain *abstract mathematical* quantities. Question (c) asks the reader to *interpret* those abstract results.

**[Example] ** Sales of a game are described by the function
$$ S(t)=\frac{150t}{t+6} $$
where \(t\) is the time (in months) since the game was introduced

and \(S(t)\) is the total sales (in thousands of games) at time \(t\).

Observe that there are three Questions. Questions (a) and (b) involve calculations. In a HW, Quiz, or Exam question, you would have to do these calculations. But here in this example, I've provided answers to those parts. You just have to answer (c).

- Find \(S(4)\).
**You don't have to do this. I'll give you the answer: Using basic arithmetic, one can find that \(S(4)=60\)** - Find \(S'(4)\).
**You don't have to do this. I'll give you the answer: Using***calculus*(more specifically, the*quotient rule*), one can find that \(S'(4)=9\) **Question for the Class:***Interpret*the results of parts (a) and (b). That is,*explain what those abstract mathematical results of part (a) and (b) tell us about sales of the game*. Use correct units in your answer.- The answer to (a) tells us that at time \(t=4\) months after the game was introduced, a total of \(60,000\) games have been sold.
- The answer to (b) tells us that at time \(t=4\) months after the game was introduced, the games are selling at a rate of \(9,000\) games per month.

**Remark:**Notice that to answer Question (c), we had to make the connection that the following two quantities are*equal*:- The value of \( f'(a) \)
- The
*instantaneous rate of change*of \( f(x) \) at \(x=a\)

**Sections, Content, Homework, and Videos:**

- 3.4 The Chain Rule (Link to Homework and Videos for Section 3.4)
- 4.1 First Derivative and Graphs (Link to Homework and Videos for Section 4.1)

**Instructor: Compositions of Functions** are functions of the form
$$f(x)=outer\left(inner(x)\right)$$
Another name for this kind of function could be

One finds the derivative of a *nested function* by using the **Chain Rule**.

**The Chain Rule** (Used for finding the derivative of a *composition of functions*. That is, *nested functions*.)

**Two Equation Form:**

**Single Equation Form:**

Notice that the presentation of the *Chain Rule* in the MATH 1350 Videos uses the terminology of *inner function* and *outer function*. And more specifically, notice that the *inner function* always has a *variable* inside, while the *outer function* is often expressed as an *empty function*, with empty parentheses not holding any variable. In the course of using the *Chain Rule*, the *derivatives* of these inner and outer functions are used, and the derivatives have the same type of variable as the original. In other words, you will be working with the following four functions:

The use of *empty functions* for the outer function and its derivative is central to the use of the *Chain Rule* in the MATH 1350 Videos. It is worthwhile for you to learn to use that kind of notation.

In this second part of the meeting, we’ll three see examples involving using the *Chain Rule* when the *outer function *is a *power function*. The first two of these examples will be done by the instructor; the third will be done as **Class Drill**.

**Instructor: ** Do this **[Chain Rule Example #1]**:
Use the ** Chain Rule** to find the derivative of
$$f(x)=3\left(6x^2-4x+5\right)^4$$
Be sure to use the terminology and notation of the

**Instructor: ** Do this **[Chain Rule Example #2]**:
Use the ** Chain Rule** to find the derivative of
$$f(x)=\frac{3}{\left(6x^2-4x+5\right)^4}$$
Be sure to use the terminology and notation of the

The next *Chain Rule* example will be done as a **Class Drill**.

Use the ** Chain Rule** to find the derivative of
$$f(x)=4\sqrt{6x^2-4x+5}$$
Be sure to use the terminology and notation of the

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For Reference, see
**[Example 3] on page 9 of the Notes from the Video for H49**
.)

So far, you have seen three examples where the *Chain Rule* is used to find derivatives of functions where the *outer function* is a *power function*. While no Chain Rule problem is *easy*, it is important to understand that if one is careful to acknowledge the *form* of a Chain Rule problem, then at least all Chain Rule problems will begin to look *familiar*. Indeed, if we look at the three examples that start on
**page 7 in the Notes from the Video for H49**,
we see that they all have *identical form*! Only the details of the derivative of the outer function, and the details of the simplifying, differ.

In this third part of the meeting, we’ll see two examples involving using the *Chain Rule* when the *outer function *is an *exponential function*. The first of these examples will be done by the instructor; the second will be done as **Class Drill**.

We start with a chain rule problem that proves a result that you have seen before, as one of the ** derivative rules**.

**Instructor: **Use the **Chain Rule** to prove the following ** derivative rule**:
$$\frac{d}{dx}e^{\left(kx\right)}=ke^{\left(kx\right)}$$
Be sure to use the terminology and notation of the

In the following **Class Drill**, you students will do a longer example that starts by using the Chain Rule to find the derivative of a function where the *outer function* is an *exponental function*.

for the function $$f(x)=e^{\left(6x^2-4x+5\right)}$$

- Use the
*Chain Rule*to find \(f'(x)\). Be sure to use the terminology and notation of the*inner function*and*outer function*, where the*outer function*is an*empty function*. - Find the \(x\) coordinates of all points on the graph of \(f(x)\) that have
*horizontal tangent lines*.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 2] on page 6 of the Notes from the Video for H50**
.)

**Instructor: ** on Monday, we discussed the *Chain Rule*.

**The Chain Rule** (Used for finding the derivative of a *composition of functions*. That is, *nested functions*.)

**Two Equation Form:**

**Single Equation Form:**

In Monday's meeting, you saw three examples where the *chain rule* was used to find derivatives of functions where the *outer function* is a *power function*.

- The Instructor used the chain rule to show that the derivative of $$f(x)=3\left(6x^2-4x+5\right)^4$$ is $$f'(x)=48\left(6x^2-4x+5\right)^3\left(3x-1\right)$$
- Then the Instructor used the chain rule to show that the derivative of $$f(x)=\frac{3}{\left(6x^2-4x+5\right)^4}$$ is $$f'(x)=-\frac{48(3x-1)}{\left(6x^2-4x+5\right)^5}$$
- Then, in a
**Class Drill**the class used the chain rule to show that the derivative of $$f(x)=4\sqrt{5x^2-7x+11}$$ is $$f'(x)=\frac{20x-14}{\sqrt{5x^2-7x+11}}$$

And you saw two examples where the *chain rule* was used to find derivatives of functions where the *outer function* is an *exponential function*.

- The Instructor used the chain rule to prove the derivative rule $$\frac{d}{dx}e^{\left(kx\right)}=ke^{\left(kx\right)}$$
- Then, in a
**Class Drill**the class used the chain rule to show that the derivative of $$f(x)=e^{\left(6x^2-4x+5\right)}$$ is $$f(x)=e^{\left(6x^2-4x+5\right)}\cdot(12x-4)$$

In the first part of today's meeting, the class will work on a **Class Drill** involving using the Chain Rule to find the derivative of a *nested function* where the *outer function* is an *logarithmic function*

for the function
$$f(x)=\ln\left(2x^2-8x+10\right)$$
use the *Chain Rule* to find \(f'(x)\). Be sure to use the terminology and notation of the *inner function* and *outer function*, where the *outer function* is an *empty function*.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 3] on page 11 of the Notes from the Video for H50**
.)

**Instructor: ** We have seen a number of examples that included questions about *slopes* of tangent lines and about *horizontal* tangent lines. It is worthwhile to pause here and review the slopes of tangent lines.

**[Example] **Answer the following questions about the function \(f(x)=2x^3-3x^2-12x+5\).

- What is the
*slope of the line tangent to the graph of \(f(x)\) at \(x=3\)*? - What is the
*slope of the line tangent to the graph of \(f(x)\) at \(x=0\)*? - At what \(x\) values does the graph of \(f(x)\) have a
*horizontal tangent line*?

**Solution:** These questions are all about the important fact that the following three things are all equal:

- The value of \( f'(a) \)
- The slope \( m \) of the line
*tangent*to the graph of \( f(x) \) at \(x=a\) - The
*instantaneous rate of change*of \( f(x) \) at \(x=a\)

The strategy for solving problems (a),(b) is therefore

- Find \( f'(x) \)
- For (a): Substitute in \(x=3\) to get \(m=f'(3)\).
- For (b): Substitute in \(x=0\) to get \(m=f'(0)\).

Those calculations are straightforward.

$$\begin{eqnarray} f'(x)&=&\frac{d}{dx} \left(2x^3-3x^2-12x+5\right) = \cdots = 6x^2-6x-12 \\ (a) \ m &=& f'(3) = 6(3)^2-6(3)-12=54-27-12=15 \\ (b) \ m &=& f'(0) = 6(0)^2-6(0)-12=-12 \end{eqnarray} $$The solution of (c) is more subtle. The key is to remember the following important fact:

**Horizontal lines have slope \(m=0\)**

The strategy for solving (c) is therefore

- Find \( f'(x) \)
- Set \( f'(x)=0 \) and solve for \(x\).

We already found \(f'(x)=6x^2-6x-12\). Setting \( f'(x)=0 \) and solving for \(x\) is straightforward. It helps to turn the equation around

$$0=f'(x)=6x^2-6x-12=6\left(x^2-x-2\right)=6(x+1)(x-2)$$The solutions to this equation are \(x=-1\) and \(x=2\). So those are the \(x\) values where the graph of \(f(x)\) has a horizontal tangent line. This is believable, because the graph of \(f(x)\) will be a cubic shape, with two *turning points*. Now we see that the turning points will be at \(x=-1\) and \(x=2\).

**End of [Example]**

An important takeaway from the example above is the important distinction between two calculations involving \(f'(x)\) and the number \(0\).

- The calculation \(f'(0)\) finds
*the slope of the line tangent to the graph of \(f(x)\) at \(x=0\)*. - Setting \(f'(x)=0\) and solving for \(x\) finds the \(x\) coordinates of all points on the graph that have
*horizontal tangent lines*.

**Question: **When is a fraction equal to zero?

**Answer: **When the numerator equals zero and the denominator is NOT zero.

Therefore, solving equations of the form $$\frac{numerator(x)}{denominator(x)}=0$$ involves finding the \(x\) values that cause \(numerator(x)=0\), and then checking to be sure that for those \(x\) values, \(denominator(x) \neq 0\).

This idea came up in a few presentations and examples in this and the last couple of meetings.

Now, the class will work on a **Class Drill** involving *Tangent Lines* for the *logarithmic* function that was studied in the earlier Class Drill

Earlier in today’s meeting, you found that for the function $$f(x)=\ln\left(2x^2-8x+10\right)$$ the derivative is $$f’(x)=\frac{4x-8}{\left(2x^2-8x+10\right)}$$

- Find the
of the tangent line at \(x=0\).*slope* - Find the \(x\) coordinates of all points on the graph of \(f(x)\) that have
.*horizontal tangent lines* - Find the
of the line tangent to graph of \(f(x)\) at \(x=3\).*equation*

Remember that you should start by building the, $$(y-f(a))=f’(a)\cdot(x-1)$$ After finding the parts \(a\), \(f(a)\), \(f’(a)\) and using them to build the equation, you should convert your equation to*point slope form of the equation of the the tangent line*and simplify your answer.*slope intercept form*

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 3] on page 11 of the Notes from the Video for H50**
.)

**Quiz Q5** during the last part of the Wed Jul 3 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The quiz material is taken from the material covered in book Sections 3.3 and 3.4. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

**Instructor: **In Chapter 2, starting in Section 2.4 of the book, you learned about the *Derivative*. In Chapter 3, you learned how to use the *Derivative Rules* to find the derivative of a variety of kinds of functions. During the weeks of the course that you were learning about the derivative, an important concept was the fact that the following three quantitites are all equal:

- The value of \( f'(a) \)
- The slope \( m \) of the line
*tangent*to the graph of \( f(x) \) at \(x=a\) - The
*instantaneous rate of change*of \( f(x) \) at \(x=a\)

In Chapter 4, you will learn to use the derivative in more ways. In the first few sections, you will use the derivative in a more thorough analysis of the *formula* for \(f(x)\) to determine more about the behavior of the *graph* of \(f(x)\). Section 4.1 is about using the derivative, \(f'(x)\), to determine the *increasing & decreasing* and *max & min* behavior of the graph of \(f(x)\).

An important concept in Section 4.1 (and beyond) is the distinction between *positive & negative* and *increasing & decreasing*. These words are used in various ways in *common English*, but they have particular definitions in *Math*.

**words:**\(f(c)\)*is positive***meaning:**\(f(c) \gt 0\)**behavior of graph:**The point \((c,f(c))\) is*above*the \(x\) axis.

**words:**\(f(c)\)*is negative***meaning:**\(f(c) \lt 0\)**behavior of graph:**The point \((c,f(c))\) is*below*the \(x\) axis.

**words:**\(f(c)\)*is zero***meaning:**\(f(c) = 0\)**behavior of graph:**The point \((c,f(c))\) is*on*the \(x\) axis.

**words:**\(f(x)\)*is increasing on the interval*\((a,b)\)**meaning:**If \(a \lt x_1 \lt x_2 \lt b\) then \(f(x_1) \lt f(x_2)\)**behavior of graph:**As you move from left to right in the interval \((a,b)\), the graph of \(f(x)\) goes*up*.

**words:**\(f(x)\)*is cecreasing on the interval*\((a,b)\)**meaning:**If \(a \lt x_1 \lt x_2 \lt b\) then \(f(x_1) \gt f(x_2)\)**behavior of graph:**As you move from left to right in the interval \((a,b)\), the graph of \(f(x)\) goes*down*.

**Instructor: **In our first **Class Drill** for today, you will practice drawing graphs with instructions that use the terminology of *positive & negative* and *increasing & decreasing*.

Draw four examples (four separate graphs) of functions \(f(x)\) on the domain \(1 \lt x \lt 5\)

**Example 1:**Draw \(f(x)\) with the property that \(f(x)\) is*positive*and*increasing*in the interval \(1 \lt x \lt 5\).**Example 2:**Draw \(f(x)\) with the property that \(f(x)\) is*positive*and*decreasing*in the interval \(1 \lt x \lt 5\).**Example 3:**Draw \(f(x)\) with the property that \(f(x)\) is*negative*and*increasing*in the interval \(1 \lt x \lt 5\).**Example 4:**Draw \(f(x)\) with the property that \(f(x)\) is*negative*and*decreasing*in the interval \(1 \lt x \lt 5\).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see **Notes from the Video for H53**
.)

**Instructor: **In the Video for H53, Barsamian presents this important **correspondence**:

- If \(f'(x)\) is
*positive*on an interval \( (a,b) \) then \(f(x)\) is*increasing*on the interval \( (a,b) \). - If \(f'(x)\) is
*negative*on an interval \( (a,b) \) then \(f(x)\) is*decreasing*on the interval \( (a,b) \). - If \(f'(x)\) is
*zero*on a whole interval \( (a,b) \) then \(f(x)\) is*constant*on the interval \( (a,b) \).

Here is another short **Class Drill** that will give you practice using the **correspondence** just introduced.

Draw four examples (four separate graphs) of functions \(f(x)\) on the domain \(1 \lt x \lt 5\)

**Example 1:**Draw \(f(x)\) with the property that \(f(x)\) is positive and \(f'(x)\) is positive in the interval \(1 \lt x \lt 5\).**Example 2:**Draw \(f(x)\) with the property that \(f(x)\) is positive and \(f'(x)\) is negative in the interval \(1 \lt x \lt 5\).**Example 3:**Draw \(f(x)\) with the property that \(f(x)\) is negative and \(f'(x)\) is positive in the interval \(1 \lt x \lt 5\).**Example 4:**Draw \(f(x)\) with the property that \(f(x)\) is negative and \(f'(x)\) is negative in the interval \(1 \lt x \lt 5\).

**Instructor: **In the Homework for Section 4.1, you work with functions that are described by a *formula*, not by a *graph*. You analyze the formula for \(f(x)\) and you also analyze the formula for \(f'(x)\) in order to determine certain things about \(f(x)\). Here is the ** Strategy for analyzing \(f(x)\) and \(f'(x)\)**:

- Part 1: Analyze \(f(x)\) to find the intervals on which \(f(x)\) is
*positive*or*negative*:- Find the
*partition numbers*for \(f(x)\), - then make a
*sign chart*for \(f(x)\), - then use the
*sign chart*for \(f(x)\) to make conclusions about the*positive*and*negative*behavior of \( f(x) \).

- Find the
- Part 2: Analyze \(f'(x)\) to find the intervals on which \(f(x)\) is
*increasing*or*decreasing*:- Find \(f'(x)\),
- then find the
*partition numbers*for \(f'(x)\), - then make a
*sign chart*for \(f'(x)\). - then use the
*sign chart*for \(f'(x)\) to make conclusions about the*increasing*and*decreasing*behavior of \( f(x) \).

**Question for the Class: **What is a *partition number*? Why are they part of the strategies described above? That is, how are *partition numbers* used?

**Answer: **Recall from Section 2.3 (about *Continuity*) the definiton of *partition number*.

A ** partition number** for a function \(f\) is an \(x\) value where \(f\) is discontinuous or \(f(x)=0\).

Partion numbers for \(f\) are significant because on the intervals between the *partition numbers* for \(f\), the sign of the function \(f\) does not change.

There is a slight variation of the definition of *partition number* when we are talking about *partition numbers for *\(f'(x)\)

A ** partition number** for \(f'\) is an \(x\) value where \(f'\) is undefined or \(f'(x)=0\).

Partion numbers for \(f'\) are significant because on the intervals between the *partition numbers* for \(f'\), the *sign* of \(f'\) does not change. This in turn means that the *increasing/decreasing behavior* of \(f\) does not change.

**Instructor: ** in **Video for H21**, Barsamian first discussed the ** Procedure for Constructing Sign Charts for a function \(f(x)\)**. His procedure was basically the book's procedure, but with some modifications. Here is Barsamian's version of the procedure (with his modifications shown in

**Procedure: Constructing Sign Charts for a function \(f(x)\)**

Given a function \(f(x)\),

- Find all
*partition numbers of*\(f(x)\). - Find all numbers \(x\) such that \(f(x)\) is
*discontinuous*at \(x\). (Rational functions are discontinuous at all values of \(x\) that make the denominator equal to \(0\).) - Find all numbers \(x\) such that \(f(x)=0\). (For rational functions, this occurs where the numerator is \(0\) and the denominator is not \(0\).)
- Plot the numbers found in Step 1 on a real number line, dividing the number line into intervals.
**Indicate the behavior of \(f(x)\) at each partition number.** - Select a test number in each interval determined in Step 2 and evaluate \(f(x)\) at each test number to determine whether \(f(x)\) is positive \((+)\) or negative \((-)\) in each interval.
**Title the diagram: Sign Chart for \(f(x)\)**

The ** Procedure for Constructing Sign Charts for a derivative \(f'(x)\)** is basically the same, but there are some slight differences. It is simple enough to show that slightly dissimilar procedure:

**Procedure: Constructing Sign Charts for \(f'(x)\)**

Given a function \(f(x)\),

- Find all
*partition numbers of*\(f'(x)\). - Find all numbers \(x\) such that \(f'(x)\) is
*undefined*at \(x\). (Rational functions are undefined at all values of \(x\) that make the denominator equal to \(0\).) - Find all numbers \(x\) such that \(f'(x)=0\). (For rational functions, this occurs where the numerator is \(0\) and the denominator is not \(0\).)
- Plot the numbers found in Step 1 on a real number line, dividing the number line into intervals.
**Indicate the behavior of \(f'(x)\) at each partition number.** - Select a test number in each interval determined in Step 2 and evaluate \(f'(x)\) at each test number to determine whether \(f'(x)\) is positive \((+)\) or negative \((-)\) in each interval.
**Title the diagram: Sign Chart for \(f'(x)\)**

**Instructor: **In our final **Class Drill** for today, you will construct *two* sign charts: A sign chart for \(f(x)\) and another sign chart for \(f'(x)\). You will use these sign charts to determine behavior of the graph of \(f(x)\).

Let \(f(x) = x^2-8x+15 \).

- Follow Part 1 of the strategy above to find the intervals on which \(f(x)\) is
*positive*or*negative*. - Follow Part 2 of the strategy above to find the intervals on which \(f(x)\) is
*increasing*or*decreasing*.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 1] on page 5 of the Notes from the Video for H54**
.)

**Instructor: **Notice that in Barsamian's version of the ** Procedure for Constructing Sign Charts**, his

- Title the diagram:
**Sign Chart for \(f(x)\)**or**Sign Chart for \(f'(x)\)**.

Now that we have seen the Presentation of **Student #12**, we see why that label on the sign chart is so important. We saw that in the Presentation of **Student #12**, *two* sign charts were made. These two sign charts were used for different things:

- The
**Sign Chart for \(f(x)\)**was used to make conclusions about the.*sign behavior of*\(f(x)\) - The
**Sign Chart for \(f'(x)\)**was used to make conclusions about the.*increasing/decreasing behavior of*\(f(x)\)

That's why it is so important to *label* the sign charts. Depending on what a sign chart is the sign chart *for*, it will get used for different things.

**Instructor Present this much harder [Example]: **
(Similar to
**[Example 2] on page 10 of the Notes from the Video for H54**)
Let \(f(x) = x^4-4x^3 \).

- Follow Part 1 of the strategy above to find the intervals on which \(f(x)\) is
*positive*or*negative*. - Follow Part 2 of the strategy above to find the intervals on which \(f(x)\) is
*increasing*or*decreasing*. - Sketch a graph of \(f(x)\), labeling all important features.

**Sections, Content, Homework, and Videos:**

- 4.1 First Derivative and Graphs (Link to Homework and Videos for Section 4.1)
- 4.2 Second Derivative and Graphs (Link to Homework and Videos for Section 4.2)
- 4.5 Absolute Maxima and Minima (Link to Homework and Videos for Section 4.5)

**Instructor: **Remember that Section 4.1 is about using the derivative to determine the *increasing & decreasing* and *max & min* behavior of the graph of \(f(x)\).

So far, we have discussed techniques for determining the *increasing & decreasing* behavior of \(f(x)\).

On Friday, we discussed the important correspondence between *sign behavior of * \(f'(x)\) on an interval \( (a,b) \) and *behavior of the graph of* \( f(x) \) on the interval \( (a,b) \).

- If \(f'(x)\) is
*positive*on an interval \( (a,b) \) then \(f(x)\) is*increasing*on the interval \( (a,b) \). - If \(f'(x)\) is
*negative*on an interval \( (a,b) \) then \(f(x)\) is*decreasing*on the interval \( (a,b) \). - If \(f'(x)\) is
*zero*on a whole interval \( (a,b) \) then \(f(x)\) is*constant*on the interval \( (a,b) \).

Also on Friday, we talked about *partition numbers* for a function \(f(x)\) and for its derivative \(f'(x)\). These were discussed earlier in this meeting. You can scroll up and see the earlier discussion.

Today, we'll discuss techniques for determining the *max & min* behavior of \(f(x)\).

An important tool in determining the *max & min* behavior of \(f(x)\) is what are called *critical numbers*. You learned about critical numbers for a function \(f(x)\) on
**page 6 of the Video for Homework 55**.

- A
*critical number*for \(f(x)\) is a number \(x=c\) that satisfies these two requirements:- The number \(x=c\) is a
*partition number*for \(f'(x)\). - The number \(x=c\) is a in the
*domain*of \(f(x)\).

- \(f'(c)=0\) or \(f'(c)\)
*does not exist*. - \(f(c)\)
*exists*.

- The number \(x=c\) is a

For a *polynomial function* \(f(x)\), finding the partition numbers for \(f'(x)\) and the *critical numbers* for \(f(x)\) is pretty straightforward. But for a function \(f(x)\) that is *not* a polynomial, it can be tricky. In your first class drill, you will work on an example where it is a little tricky.

For the function
$$f(x)=x+\frac{16}{x}$$
Show how to find the partition numbers for \(f'(x)\) and the *critical numbers* for \(f(x)\).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 5] in the Video for Homework 56**
.)

**Instructor: **On **page 5 of the Notes for the Video for H56**, you saw the definitions of *Local Maximum* and *Local Minimum*. (Project the notes from the video and show them the definitions.)

Then, on
**page 7 of the same video**, you learned about the *First-Derivative Test for Local Extrema*. (Project the notes from the video and show them the test.)

In the next class drill, you will practice using the *First Derivative Test* in an example where some helpful information about \(f(x)\) and \(f'(x)\) is *given*. That information will allow you to immediately use the *First Derivative Test*. (In harder problems, you will have to *analyze* both \(f(x)\) and \(f'(x)\) to determine the information that will allow you to use the *First Derivative Test*.)

A function \(f(x)\) is continuous on the interval \( (-\infty,\infty) \). The sign chart for \(f'(x)\) is shown below.

Show how to use the *First-Derivative Test* to find the \( x \) coordinates of all *local extrema* of \(f(x)\).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 1] on page 9 of the Notes for the Video for Homework 56**
.)

**Instructor: **Today we'll be discussing concepts from book Section 4.2, which is about * 2nd Derivatives and Graphs*.

The section begins with a definition of * concavity *and

**Words:**\(f\)*is*\((a,b)\).**concave up**on the interval**Graphical Definition:**For every \(x=c\), with \(a \lt c \lt b\), the graph of \(f\) has a tangent line at \(x=c\) and the graph of \(f\) stays*above*that tangent line for \(x\) values in the interval \((a,b)\).**Abstract Definition:**\(f'(x)\) is*increasing*on the interval \((a,b)\).

**Words:**\(f\)*is*\((a,b)\).**concave down**on the interval**Graphical Definition:**For every \(x=c\), with \(a \lt c \lt b\), the graph of \(f\) has a tangent line at \(x=c\) and the graph of \(f\) stays*below*that tangent line for \(x\) values in the interval \((a,b)\).**Abstract Definition:**\(f'(x)\) is*decreasing*on the interval \((a,b)\).

**Related terminology:**Anis a point on the graph of a function where the function is continuous and the concavity changes (from up to down or from down to up.)*inflection point*

In our first **Class Drill** for today, you will practice drawing graphs with instructions that use the terminology of *positive & negative*, *increasing & decreasing*, and , *concave up & down*.

Draw four examples (four separate graphs) of functions \(f(x)\) on the domain \(1 \lt x \lt 5\)

**Example 1:**Draw \(f(x)\) with the property that \(f(x)\) is*positive*,*decreasing*, and*concave up*in the interval \(1 \lt x \lt 5\).**Example 2:**Draw \(f(x)\) with the property that \(f(x)\) is*positive*,*increasing*, and*concave down*in the interval \(1 \lt x \lt 5\).**Example 3:**Draw \(f(x)\) with the property that \(f(x)\) is*negative*,*decreasing*, and*concave down*in the interval \(1 \lt x \lt 5\).**Example 4:**Draw \(f(x)\) with the property that \(f(x)\) is*negative*,*increasing*, and*concave up*in the interval \(1 \lt x \lt 5\).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**Notes from the Video for H57**
.)

**Instructor: **The study of *concavity of * \( f(x) \) leads us to investigate the *derivative of the derivative of * \( f(x) \). This is what is called the *second derivative of * \( f(x) \), and denoted \(f''(x)\). In the videos for the course, this definition is found in the
**Video for H57**. Here is the definition:

**Words:***the second derivative of*\(f(x)\)**Symbols:**\(f''(x)\), or \(\frac{d^2}{dx^2}f(x)\)**Meaning:**\(f''(x) = \frac{d}{dx}f'(x) =\frac{d}{dx}\frac{d}{dx}f(x)\)

Immediately following the above **definition** Video for H57, Barsamian presents this important **correspondence**:

- If \(f''(x)\) is
*positive*on an interval \( (a,b) \) then \(f'(x)\) is*increasing*on the interval \( (a,b) \), which in turn means that \(f(x)\) is*concave up*on the interval \((a,b)\). - If \(f''(x)\) is
*negative*on an interval \( (a,b) \) then \(f'(x)\) is*decreasing*on the interval \( (a,b) \), which in turn means that \(f(x)\) is*concave cown*on the interval \((a,b)\).

Here is another short **Class Drill** that will give you practice using the **correspondence** just introduced.

Draw four examples (four separate graphs) of functions \(f(x)\) on the domain \(1 \lt x \lt 5\)

**Example 1:**Draw \(f(x)\) with the property that \(f(x)\) is*positive*, \(f'(x)\) is*negative*, and \(f''(x)\) is*positive*on the interval \( 1 \lt x \lt 5 \).**Example 2:**Draw \(f(x)\) with the property that \(f(x)\) is*positive*, \(f'(x)\) is*positive*, and \(f''(x)\) is*negative*on the interval \( 1 \lt x \lt 5 \).**Example 3:**Draw \(f(x)\) with the property that \(f(x)\) is*negative*, \(f'(x)\) is*negative*, and \(f''(x)\) is*negative*on the interval \( 1 \lt x \lt 5 \).**Example 4:**Draw \(f(x)\) with the property that \(f(x)\) is*negative*, \(f'(x)\) is*positive*, and \(f''(x)\) is*positive*on the interval \( 1 \lt x \lt 5 \).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**Notes from the Video for H57**
.)

**Instructor: **Finding \(f''(x)\) can be very easy, but sometimes it is tricky. Here is a **Class Drill** where you will be asked to find \(f''(x)\).

- Let \( f(x) = -x^6 + 30x^5 - 30x + 19 \). Find \(f'(x)\) and \(f''(x)\). Show steps clearly. Notice that in this case, finding \(f''(x)\) is
*easier*than finding \(f'(x)\) because of the lower degree of the polynomial. - Let \( f(x) = \ln{(x^2-2x+5)} \). Find \(f'(x)\) and \(f''(x)\). Show steps clearly and simplify your answer. Notice that, in this case, finding \(f''(x)\) is
*harder*than finding \(f'(x)\) because the*quotient rule*is in general more difficult than the*chain rule*.

**Instructor: **Show the solution to the Class Drill.

**Instructor: **In the Homework for Section 4.2, you do problems involving *graphing*. A *Graphing Strategy* was presented in the book and in the
Video for H60.
(See page 7 of the Notes for the Video for H60).
Here is that *Graphing Strategy*:

**Step 1: Analyze \(f(x)\).** Find the domain and the axis intercepts. The \(x\) intercepts are the solutions of the equation \(f(x)=0\), and the \(y\) intercept is the value of \(f(0)\).

**Step 2: Analyze \(f'(x)\).** Find the partition numbers for \(f'\) and the critical numbers of \(f\). Construct a sign chart for \(f'\), determine the intervals on which \(f\) is increasing and decreasing, and find the local maxima and minima of \(f\).

**Step 3: Analyze \(f''(x)\).** Find the partition numbers for \(f''\). Construct a sign chart for \(f''\), determine the intervals on which \(f\) is concave up and concave down, and find the inflection points of \(f\).

**Step 4: Sketch the graph of \(f(x)\).** Locate axis intercepts, local maxima and minima, and inflection points. Sketch in what you know from steps 1-3. Plot additional points as needed to complete the sketch.

Problems involving using the *Graphing Strategy* can be quite long, because each of the four steps in the strategy is long. But there are some problems where you are *given* some information about \(f(x)\) and \(f'(x)\) and \(f''(x)\), and then asked to sketch a graph of \(f(x)\). Realize that such problems are still about the *Graphing Strategy*; they have simply been shortened by giving you a bunch of the information that you would have had to determine for yourself in *Step 1*, *Step 2*, and *Step 3*. Here is an example of that type.

**Instructor do this [Example]: **
Sketch the graph of a function \( f(x) \) that has the properties listed below. Make your graph large and neat, and label all important points (*axis intercepts, maxs and mins, inflection points*) with their \( (x,y) \) coordinates. Draw *horizontal tangent lines* where needed, and label them with \( m=0 \).

- \( f(-5)=0 \)
- \( f(0)=-7 \)
- \( f(5)=0 \)
- \( f(10)=7 \)
- \( f(15)=0 \)
- \( f'(0)=0 \)
- \( f'(10)=0 \)
- \( f'(x) \lt 0 \) on the intervals \( (-\infty,0) \) and \( (10,\infty) \)
- \( f'(x) \gt 0 \) on the interval \( (0,10) \)
- \( f''(5)=0 \)
- \( f''(x) \gt 0 \) on the intervals \( (-\infty,5) \)
- \( f''(x) \lt 0 \) on the interval \( (5,\infty) \)

**Instructor: **Today we'll be discussing concepts from book Section 4.5, which is about * Absolute Extrema*.

It is important to understand the distinction between * Absolute Extrema* (from Section 4.5 of the book) and

**Instructor: ** Write the definitions for * local max* and

Realize that the locations of and *absolute extrema* often don't match, and in many situations one or the other type of extremum may not even exist at all. (Show
**[Example 1] on page 5 of the Notes from the Video for H61**
involving a function \(f(x)\) given by a *graph*.)

As seen in the example that was just projected, a particular function \(f(x)\) may not have an *absolute min* or *absolute max* on a particular domain. But there is a situation in which some *absolute extrema* are *guaranteed* to occur, and to occur only at *certain possible locations*.

**Instructor: ** Write out * Theorem 1: Extreme Value Theorem* and

The two theorems just presented are the basis for what is called the .

**Instructor: **In the case of a function \(f(x)\) that is *continuous* on a *closed interval* \([a,b]\), the *absolute extrema* of \(f(x)\) on the interval \([a,b]\) are found using a procedure called the * Closed Interval Method*.

**Instructor: ** Write the steps for the * Procedure for Finding Absolute Extrema on a Closed Interval* on the board. (or refer to

In our first **Class Drill** for today, you will do a basic example involving this new method.

Show how to use the *Closed Interval Method* to find the absolute extrema of \(f(x)=2x^3-3x^2-12x+7\) on the interval \( [-2,4]\).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**pages 7 - 8 of the Notes from the Video for H62**
.)

**For Class Discussion**

To find the absolute extrema of \(f(x)=x^3-3x+4\) on the interval \( [-2,3]\), one starts by finding \( f'(x)=3x^2-12x+9=3(x+1)(x-1) \), which has partition numbers \(x=-1\) and \(x=1\).

**Question For the Class: **Why isn't \(x=3\) also a partition number for \(f'(x)\)? It looks to be a very important number in the factorization of \(f'(x)\). And why isn't \(x=0\) a partition number? Explain why neither \(x=3\) nor \(x=0\) is a partition number.

**Answer: **Remember that a *partition number* for \(f'(x)\) is defined to be an \(x\) value such that \(f'(x)\) is *undefined* or \(f'(x)=0\). Since \(f'(x)\) is a *polynomial*, we know that there are *no* \(x\) values that will cause \(f'(x)\) to be *undefined*. Therefore, the only *partition numbers* for \(f'(x)\) will be the \(x\) values such that \(f'(x)=0\).

It is easy to confirm that \(x=-1\) and \(x=1\) are *partition number* for \(f'(x)\) by simply substituting \(x=-1\) and \(x=1\) into \(f'(x)\) and computing the result. The computation is easiest if we use the *factored* form, \( f'(x)=3(x+1)(x-1) \).

- \( f'(-1)=3((-1)+1)((-1)-1)=3(0)(-2)=0\)
- \( f'(1)=3((1)+1)((1)-1)=3(2)(0)=0\)

Now observe what happens if we try substituting \(x=0\) and \(x=3\) into \(f'(x)\) and computing the result. Again, the computation is easiest if we use the *factored* form of \(f'(x)\).

- \( f'(3)=3((3)+1)((3)-1)=3(4)(2)=24 \neq 0\)
- \( f'(0)=3((0)+1)((0)-1)=3(1)(-1)=-3 \neq 0\)

It is important to realize that absolute extrema do not always happen at \(x\) values that are integers. In our second **Class Drill** for today, you will again use the * Closed Interval Method.*, but this time, you'll see an absolute extremum occur at a

Show how to use the *Closed Interval Method* to find the absolute extrema of \(f(x)=x^4-4x^2 +5\) on the interval \( [-1,2]\).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**pages 12 - 13 of the Notes from the Video for H62**
.)

**Quiz Q6** during the last part of the Fri Jul 12 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The quiz material is taken from the material covered in book Sections 3.4, 4.1, and 4.2. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

**Sections, Content, Homework, and Videos:**

- 4.5 Absolute Maxima and Minima (Link to Homework and Videos for Section 4.5)
- 4.6 Optimization (Link to Homework and Videos for Section 4.6)

**Instructor: **On Friday, we discussed *Absolute Maxima* and *Absolute Minima*. We also discussed the fact that for a function \(f(x)\) that is *continuous* on a *closed interval* \([a,b]\), the *absolute extrema* of \(f(x)\) on the interval \([a,b]\), it is * guaranteed* that the function will have both an absolute max and an absolute min on that interval. On Friday, we discussed that in this case, the absolute extrema are found using a procedure called the

Today, we will discuss the case of a function \(f(x)\) that is *continuous* on a domain that is *not a closed interval*. In this case, the function is *not guaranteed* to have any absolute extrema at all. Often, it is helpful to determine whether there will be an *absolute max* or *absolute min* before finding them precisely. In the following short **Class Drill**, you'll have to think about how that might be done.

- Without doing any calculations, explain how you know whether or not \(f(x)=5x^4-3x^2+17\) will have an
*absolute max*and/or*absolute min*on the interval \( (-\infty,\infty) \). - Without doing any calculations, explain how you know whether or not \(f(x)=4x^5-3x^2+17\) will have an
*absolute max*and/or*absolute min*on the interval \( (-\infty,\infty) \).

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 1] on page 7 of the Notes from the Video for H63**
.)

**Instructor: **For some functions that are not a familiar form, on domains that are not closed intervals, it is not possible to figure out whether there will be any absolute extrema without doing any calculations.

For example, suppose we want to find out if the function
$$f(x)=4x+\frac{128}{x^2}$$
will have any extrema on the interval \( (0,\infty) \). Observe that \(f(x)\) is continuous on the interval \( (0,\infty) \), but the interval is *not a closed interval*. We have seen that in situations where the domain is *not a closed interval*, a function \(f(x)\) might not even have an *absolute min* or *absolute max*. We'll discuss this example now.

**Instructor Do this Example: ** Find the absolute max and absolute min, if they exist, for the function
$$f(x)=4x+\frac{128}{x^2}$$
on the interval \( (0,\infty) \).
As part of the solution, introduce the ** Second Derivative Test**.
(For reference, see

**Instructor: **For the rest of today's meeting, we'll start a discussion of *Optimization*, from Section 4.6 of the book. Although the topic sounds like something new, realize that *Optimization* problems are just *Absolute Max/Min* problems, but with some complications.

- They may be presented as
*word problems*, about applications to real world situations. - You may have to figure out a
*mathematical model*, that is,*equations*that describe the situation described in the word problem. - The initial mathematical model may involve
*more than one variable*. If it does, then you will have to figure out how to reduce it to a function of one variable. - You will have to figure out the
*domain*of that function, and the domain*might not be a closed interval*.

Today, we'll be discussing Single Variable Optimization Problems about Maximizing Revenue and Profit

In the Video for H64, Barsamian discusses

That means that of the possible complications listed above you will have to deal with #1,2,4, but not #3. Our discussion about Optimization today will be about problems of that sort.

**[Example: Hoverboards]** This is the continuation of an example that was started in the Jun 12 meeting (Day 13).

A company makes *hoverboards*, a very dangerous toy that kids would not ride if they had a lick of sense. The *price demand equation* is

where \(p\) is the selling price of a hoverboard, in dollars, and \(x\) is the number of hoverboards that will sell at that price.

In the Jun 12 meeting, you found in a **Class Drill** that the *price function* is
$$p(x)= \left(-\frac{1}{30}\right)x+200$$
with domain
$$ [0,6000] $$
You saw that the *graph* of the *price function* will be a line segment with endpoints at \((0,200)\) and \((6000,0)\).

You also found the *Revenue function* is
$$R(x) = \left(-\frac{1}{30}\right)x^2 + 200x $$
with domain
$$[0,6000]$$
You saw that the *graph* of the *Revenue function* will be a chunk of a *parabola*

- The parabola is facing down.
- The parabola has a \(y\)
*intercept*at \((0,0)\) - The parabola has \(x\)
*intercepts*at \((0,0)\) and \((6000,0)\) - The domain is the interval \([0,6000]\).

In our first **Class Drill** for today, you will revisit this example, this time with the goal of *maximizing Revenue*

Use the revenue function \(R(x) = \left(-\frac{1}{30}\right)x^2 + 200x\) with *domain* \([0,6000]\) that was discussed above.

**(a)** If the goal is to maximize *Revenue*, what *price* should the company charge for the hoverboards, and how many hoverboards should be produced? Illustrate your solution with a graph of the *Revenue* function.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 1] on page 8 of the Notes from the Video for H64**.
.)

In our second **Class Drill** for today, you will continue this example, this time with the goal of *maximizing Profit*

You will be continuing the example from the previous **Class Drill**, using the *Revenue function* \(R(x) = \left(-\frac{1}{30}\right)x^2 + 200x\) with *domain* \([0,6000]\) that was discussed above.

Suppose also that the *Cost function* is \(C(x) = 60,000 + 20x\)

**(b)** Find the *Profit function*, \(P(x)\).

**(c)** If the goal is to maximize *Profit*, what *price* should the company charge for the hoverboards, and how many hoverboards should be produced? Illustrate your solution with a graph of the *Profit* function.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 1] on page 8 of the Notes from the Video for H64**.
.)

**Instructor: **On Monday, we began a discussion of *Optimization Problems*. Remember that *Optimization* problems are just *Absolute Max/Min* problems, but with some complications.

- They may be presented as
*word problems*, about applications to real world situations. - You may have to figure out a
*mathematical model*, that is,*equations*that describe the situation described in the word problem. - The initial mathematical model may involve
*more than one variable*. If it does, then you will have to figure out how to reduce it to a function of one variable. - You will have to figure out the
*domain*of that function, and the domain*might not be a closed interval*.

On Monday, we discussed Single Variable Optimization Problems about Maximizing Revenue and Profit

We'll start today's meeting by discussing another kind of problem of that sort.

Often, in problems about ** Revenue** and

**Students: **Work on the following
**Class Drill: When the Variable is Not the Demand**

**Instructor: **Show the solution to the Class Drill that the students just worked on.
(For reference, see
**[Example 2] on page 9 of the Notes from the Video for H64**
.)

The Video for H65 (link to accompanying notes) discusses

In these problems, of the possible complications discussed earlier (complications that Optimization Problems may or may not have), you will have to deal with #3,4, but not #1,2. In this second part of our meeting for today, we'll see an example of that sort.

**Instructor do this [Example]: **(This example is similar to a problem in the Homework for Section 4.6)

Find positive numbers \(x,y\) such that

- The sum \(2x+y=1200\)
- The product \(P=xy\) is maximized

Observe that this problem does involve *two variables*, but it is not about some real world situation. Hence, we could call it a *two variable abstract optimization problem*.

(For reference, see
**[Example 1] on page 8 of the Notes from the Video for H65**
.)

The Video for H66 (link to accompanying notes) discusses

In these problems, of the four possible complications listed earlier in the meeting outline, you will have to deal with all four! In our **Class Drill**. You will solve a problem of that sort in a Class Drill.

A farmer needs to build a fence to make a rectangular corral next to an adjacent pasture. He only needs to fence three sides, because the fourth side has already been fenced. He has 1200 feet of fencing. Find the *dimensions* of the pasture that will enclose the *largest possible area*.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 1] on page 3 of the Notes from the Video for H66**
.)

**Instructor: **Notice that the *mathematical model* that you arrived at in your **Class Drill** exactly matches the *two variable abstract optimization problem* that I solved in the preceeding *Example*.

**Exam X2** lasts the full duration of the Fri Jul 19 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The exam material is taken from the material covered in Chapters 3 and 4. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

**Sections, Content, Homework, and Videos:**

- 5.1 Antiderivatives and Indefinite Integrals (Link to Homework and Videos for Section 5.1)
- 5.2 Integration by Substitution (Link to Homework and Videos for Section 5.2)

**Instructor: **Today we'll be discussing concepts from book Section 5.1, which is about * Antiderivatives and Indefinite Integrals*.

We'll start with the definition of *Antiderivative*

**Words:**\(F\)*is an antiderivative of*\(f\).**Meaning:**\(f\)*is the derivative of*\(F\). That is, \(f=F'\).

Observe that to answer the question

the strategy is to

**Instructor Example: **Show how to use that strategy to answer these questions:
$$\begin{eqnarray}
\text{(a) Is } \ F(x) &=& \frac{x^3}{3} \ \text{ an antiderivative of } \ f(x) = x^2 \ \text{?} \\
\text{(b) Is } \ F(x) &=& \frac{(5x+7)^3}{3} \ \text{ an antiderivative of } \ f(x) = (5x+7)^2 \ \text{?}
\end{eqnarray}
$$

**Instructor: **You’ll work on more of these kinds of questions in the following **Class Drill**.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 3] on page 4 of the Notes from the Video for H68**
.)

**Instructor: **Questions **(c)** and **(d)** that you just worked on illustrate that a given function \( f(x) \) will have a whole family of *antiderivatives* that differ from each other by an additive constant. This is articulated as a **Theorem about the Collection of Antiderivatives of a Function**. (Instructor present the theorem and illustrate it by drawing the family of antiderivatives of \( f(x) = x^2 \).) (See
**pages 4-7 of the Notes from the Video for H69**
for reference.)

**Instructor: **Discuss the distinction between *Particular Antiderivative* and *General Antiderivative*. (See
**pages 8-9 of the Notes from the Video for H69**
for reference.)

**Instructor: **Introduce ** the Indefinite Integral of \( f(x) \)** (See

**Instructor: **Present the ** Power Rule**, the

**Instructor: **Show the details of the calculation of the following *indefinite integral*
$$ \int 7\sqrt x dx$$

**Instructor: **You’ll use the ** Indefinite Integral Rules** that I just wrote down when you work in the following

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 1] on page 22 of the Notes from the Video for H70**
.)

**Instructor: **It is important to pay attention to when the ** Power Rule for Indefinite Integrals** can be used and when it cannot be used. For example, we already know that

Here are three common incorrect indefinite integrals of \(\frac{1}{x}\). Observe that each of them involves using the **Power Rule**, which is the ** wrong rule**!

**Instructor: **It is also important to remember that finding *Indefinite Integral* is *not* the same as finding a *Derivative*.

For example, we know that $$\frac{d}{dx}\ln(x)=\frac{1}{x}$$ while $$ \int \ln(x) dx = x\ln(x)-x+C. $$ And we know that $$\frac{d}{dx}1=0$$ while $$ \int 1 dx = x+C. $$

**Instructor: **The definition of ** Antiderivative** was first introduced in the Video for H68 and is reviewed at the start of the Video for H71
(Project

Recall that for a given function \(f(x)\), there will be a *collection* of antiderivatives. This collection is called the *General Antiderivative of *\( f(x) \) and is also called the *Indefinite Integral of *\( f(x) \). And remember the relationship between the indefinite integral and \( f(x) \)
(page ahead to pages 4 and 5 of the notes for the Video for H71.)

In your reading of the book or watching of the videos, you learn a collection of *Indefinite Integral Rules*.

The ** Power Rule for Indefinite Integrals**:

The **\( \frac{1}{x} \) Rule for Indefinite Integrals**:

The **\( e^{(x)} \) Rule for Indefinite Integrals**:

The **\( \ln{(x)} \) Rule for Indefinite Integrals**:

The ** Constant Multiple Rule for Indefinite Integrals**:

The ** Sum Rule for Indefinite Integrals**:

The ** Sum and Constant Multiple Rule for Indefinite Integrals**:

**Instructor: **Look at the list of ** Indefinite Integral Rules**. Notice that there is no

Integrals where the integrand is a more general quotient have to be treated carefully. For some integrals where the integrand is a general quotient, it is possible to first *rewrite* the integrand in *power function form*, and then find the integral using the *sum and constant multiple rule* and the *power rule*.

**Instructor do this Example:**
Consider the indefinite integral
$$ \int \frac{4}{\sqrt[3]{x}} + \frac{5}{x^4} dx $$

- Rewrite the integrand in
*power function form*. - Then compute the indefinite integral. (Write your final answer in
*positive exponent form*.)

**Instructor: Pose this Question to the Class **Three students were assigned the problem of finding an indefinite integral. Here are their solutions:
$$\begin{eqnarray}
\text{Student 1 wrote: } \ \int \frac{3}{5x^2} dx &=& \frac{3}{5\cdot \frac{x^3}{3}} + C = \frac{9}{5x^3} + C \\
\text{Student 2 wrote: } \ \int \frac{3}{5x^2} dx &=& \int 3\cdot 5x^{-2} dx = \frac{3 \cdot 5x^{-1}}{-1} + C = -\frac{3}{5x} + C \\
\text{Student 3 wrote: } \ \int \frac{3}{5x^2} dx &=& \int \frac{3}{5} \cdot x^{-2} dx = \frac{3}{5} \cdot \frac{x^{-1}}{-1} + C = -\frac{3}{5x} + C
\end{eqnarray}$$
Which student(s) wrote a correct solution to the problem? Explain.

It is important to realize that some fractions in integrands need to be left as fractions. For example, if the integrand contains a \( \frac{1}{x} \), then that term should be left alone, not converted into the power function form \( x^{-1} \). The reason is that the indefinite integral
$$ \int x^{-1} dx $$
*cannot* be found using the *Power Rule*, while the indefinite integral
$$ \int \frac{1}{x} dx $$
can be found using the \( \frac{1}{x} \) *Rule*.
$$\int \frac{1}{x} dx = \ln(|x|) + C$$

In the following **Class Drill**, you will need to rewrite the integrand involving a quotient so that part of it is converted to ** power function form** while part of the integrand is left in

Consider the indefinite integral $$ \int \frac{1+x^3}{x} dx $$

- Rewrite the integrand in a form more suited to computing the indefinite integral.
- Then compute the indefinite integral.

**Instructor: **Show the solution to the **Class Drill** that the class just worked on.
(For reference, see
**[Example 1](F) on page 12 of the Notes from the Video for H71**
.)

**Instructor: **A type of problem that you will encounter in your Homework for Section 5.1 involves finding a *particular antiderivative* satisfying an *extra condition*.

The strategy for finding a *particular antiderivative* satisfying an *extra condition* is to

- First, find the
*general antiderivative*(by*integrating*). - Then, find the
*particular antiderivative*by requiring that the*extra condition*be satisfied.

**Instructor do this Example:**
Find the particular antiderivative of the derivative
$$ \frac{df}{dx} = 3e^{(x)} - 4 $$
that satisfies \( f(0) = 8 \).
(See
**[Example 3] on page 13 of the Notes from the Video for H72**
for reference.)

**Instructor: **Note that \(x\) is not always the letter used as the variable. And \(f\) is not always the letter used as the name of the function. In particular, it is common to use the variable \(t\), indicating *time*. And in *Engineering* and *Physics*, it is common to have a function called \(x(t)\) that gives the *position*, at time \(t\), of an object moving in one dimension. Using this kind of notation, the example problem that I just solved would be stated as follows:

**Find the particular antiderivative of the derivative
$$ \frac{dx}{dt} = 3e^{(t)} - 4 $$
that satisfies \( x(0) = 8 \). Show all steps clearly.
**

This wording of the problem is the same as the wording used on a related problem in your **Mylab Homework for Section 5.1**.

Realize that the solution would be exactly the same as the solution presented in the previous example, but with the variable changed from \(x\) to \(t\) and the function name changed from \(f\) to \(x\)

**Instructor: **Today, we'll be discussing the method of *Integration by Substitution*. This method is used to find an integral in some situations where the integrand involves a *nested function*, also known as a *composition of functions*. Different books present this method in different ways. In the videos for our course, a 5-Step procedure is presented.

**Step 1: Identify the inner function and call it \(u\).** Write the equation \(inner(x)=u\) to
introduce the single letter \(u\) to represent the inner function. Circle the equation.

**Step 2: Build the equation \(dx=\frac{1}{u'}du\).** To do this, first find \(\frac{du}{dx}\), then use it to build equation
\(dx=\frac{1}{u'}du\). Circle the equation.

**Step 3: Substitute, Cancel, Simplify.** In steps (1) and (2) you have two circled equations.
Substitute these into the integrand of your indefinite integral. Cancel as much as possible and
simplify by using the *Constant Multiple Rule*. The result should be a new basic integral
involving just the variable \(u\).

**Step 4: Integrate.** Find the new indefinite integral by using the indefinite integral rules. The
result should be a function involving just the variable \(u\) (with constant of integration \(+C\) ).

**Step 5 Substitute Back.** Substitute \(u=inner(x)\) into your function from Step (4). The result
will be a new function of just the variable \(x\). (Be sure to include the constant of integration \(+C\) )
in your result.) This is the \(F(x)\) that we seek.

The first step of the *Substitution Method* presented in the Video is to identify the *inner function* and call it \(u\). Once \(u\) has been identified, everything else in the procedure unfolds without any choices to be made.

**Question #1 for the Class: **For the integral
$$ \int 2x\left(x^2+4\right)^7dx $$
What should you choose as the \(u\)?

**Instructor: **As mentioned in the Video for H74, identifying \(u\) can be straightforward when the *inner function* is in parentheses and nothing else in the integrand is in parentheses. In some integrals, however, the choice of \(u\) is not so obvious. Furthermore, in some of the problems in our book and in MyLab, the choice of \(u\) is made harder by the weird and misleading typesetting used by the publisher. The publisher sometimes puts parentheses where they are not needed and omits them in places where it would be helpful to have them.

**Question #2 for the Class: **For the integral
$$ \int e^{5x}(5)dx $$
What should you choose as the \(u\)?

**Question #3 for the Class: **For the integral
$$ \int \sqrt{1+4x^2}(8x)dx $$
What should you choose as the \(u\)?

**Question #4 for the Class: **For the integral
$$ \int \frac{1}{4+2x^4}\left(8x^3\right)dx $$
What should you choose as the \(u\)?

**Question #5 for the Class: **For the integral
$$ \int \frac{1}{9x+5}dx $$
What should you choose as the \(u\)?

**Instructor: **As mentioned earlier, if you follow the procedure for the *Substitution Method* that is presented above, once you identify \(u\), there are no choices to be made in the procedure. (And note that the identification of \(u\) is not really a choice. Either \(u\) works, or it doesn't.) But some substitution problems are easier than others because of the way that all of the constants cancel nicely in the substitution step.

For example, In the next two class drill, you will finish the problems started as **Questions #4 and #5 for the class**, above.

**Instructor: ** Divide the class into two clubs: the ** Jets** and the

**Instructor: **Show the solutions to the problems that the ** Jets** and the

**Instructor: **Notice that the integral that the ** Jets** had to find looked harder than the integral that the

In the
**Video for H74**
and the
**Video for H75**
there is extensive discussion of more types of integrals involving the *Substitution Method*.
You'll work on a couple more types in the following **Class Drill**.

**Instructor: ** Have students work in pairs for 20 minutes on
**Two Class Drills on Indefinite Integrals (Sections 5.1 and 5.2)**
After 20 minutes, discuss the solutions.

**Sections, Content, Homework, and Videos:**

- 5.4 The Definite Integral (Link to Homework and Videos for Section 5.4)
- 5.5 The Fundamental Theorem of Calculus (Link to Homework and Videos for Section 5.5)

**Instructor: **Today we'll be discussing the *the area between the graph of a function \( f(x) \) and the \(x\) axis from \(x=a\) to \(x=b\)*. A picture of this can be found on
**page 2 of the Notes from the Video for H76**.

As discussed in the video, there are two kinds of area that we will be studying

All regions have positive area*Unsigned Area:*Regions*Signed Area:**above*the \(x\) axis have*positive*area. Regions*below*the \(x\) axis have*negative*area.

When a region is made up of *basic geometric shapes*, we can compute both kinds of areas using formulas from *geometry*.

**Instructor: ** You will now work on a problem about computing both kinds of areas using formulas from *geometry* in a **Class Drill**.

**Instructor: ** Hand out the multi-part **Class Drills on Area and Definite Integrals**.

**Students: ** work on **Class Drill: Finding Unsigned and Signed Areas Using Geometry**. (Just the **first one** of the four class drills on the handout.)

**Instructor: **Show the solutions to the Class Drill that the class just worked on. (For reference, see
**[Example 1] on page 3 of the Notes from the Video for H76**. You might want to **print out** the drill ahead of Monday's meeting, so that you can practice showing how the solution goes.

**Instructor: (Project
page 5 in the Notes from the Video for H76**) But when the region between the graph of \(f(x)\) and the \(x\) axis from \(x=a\) to \(x=b\) is

*The Area Problem* (Two Questions about Area)

- How can we
*define*the*signed area*between the graph of a function \(f(x)\) and the \(x\) axis from \(x=a\) to \(x=a\) for a general function \(f(x)\) whose graph is not made up basic shapes? - How can we find the
*value*of that*signed area*?

The solution of the *Area Problem* is a long, detailed process that you should learn about by reading the book and/or watching the Video for H76. In our meeting, we will only discuss some aspects of what is presented in detail in the book and in the video.

The approach to solving the *Area Problem* for a *curvy* region is to first consider an *approximation* of the region using a simpler region made up of *basic geometric shapes*. One can find the area of the simpler region using geometry.

**Instructor: ** We will now work together on a problem about *Estimating the Area Under a Graph* in **Class Drill**.

**Instructor: ** Using the Document Camera, lead the students in the solution of
**Class Drill: Estimating the Area Under a Graph Using Riemann Sums ** (Just the **second** of the four class drills on the handout.).

(For reference, see the
**Video for H76** and read
**pages 7 - 22 in the Notes from the Video for H76**. The video would have benefited from a *simple, visual example* presented *before* the difficult analytical material. (Mark B says *Sorry*!!) The Class Drill is that kind of simple, visual example.)

**Instructor: **On
**page 7 of the Notes from the Video for H76**, a Quest begins:

**Quest: Find the value of the signed area between the graph of
$$f(x)=5+\frac{x^2}{10}$$
and the \(x\) axis from \(x=2\) to \(x=12\).**

On page 9 of the notes, you can see the region in question shaded in *Blue* and sandwiched in between a *Green Region* and a *Red Region*. The *Blue Region* is *curvy*, and we don't know how to compute its area. But the *Green* and *Red Regions* are made up of simple geometric shapes (rectangles) whose area we can compute exactly. We can use those *Green* and *Red* areas to get *lower and upper bounds* on the area of the *Blue* region. (Note that the computations use the *formula* for \(f(x)\) to get the \(y\) values on the graph. This is in contrast to what the students did in their last class drill, where they just got the \(y\) values from the *graph*.)

The area of the *Green Region* is called the ** Left Sum with 5 Subintervals**, denoted \(L_5\); the area of the

In
**[Example 4] on page 26 of the video**, better lower and upper bounds for the unknown *Blue* area are found by finding *Left sums* \(L_n\) and *Right Sums* \(R_n\) with *larger and larger values of* \(n\). Two trends are observed:

- When \(n\) gets larger and larger (that is, as \(n\rightarrow \infty \), the values of \(L_n\) and \(R_n\) get closer and closer to each other.
- When \(n\) gets larger and larger (that is, as \(n\rightarrow \infty \), the values of \(L_n\) and \(R_n\) get closer and closer to a number around \(107.33\).

It turns out that the two trends observed in **[Example 4]** of the video will always be observed. On
**page 31 of the Notes**, the following *Big Fact* is presented:

**The Big Fact:** It is a fact from more advanced math that for a function \(f(x)\) that is continuous on a closed interval \([a,b]\), the values of \(L_n\) and \(R_n\) always approach some common number as \(n\rightarrow\infty\). That is, \(\lim_{n\rightarrow\infty}L_n\) and \(\lim_{n\rightarrow\infty}R_n\) both exist and \(\lim_{n\rightarrow\infty}L_n = \lim_{n\rightarrow\infty}R_n\).

The **Big Fact** gives us a way to *define* the signed area of the region between the graph of a general curvy function \(f(x)\) and the \(x\) axis on a closed interval \([a,b]\), and also *compute its value*. We can just *define* the signed area to be the number that is the limit of the Riemann Sums. Notice that this answers the *Two Questions about Area* that make up *The Area Problem*.

Because the definition is so important, it gets a name (and a symbol): the ** Definite Integral**. (See

**Instructor: (only if there is enough time) **Have Students work on this **Class Drill**.

**Students: ** Work on
**Class Drill: Using Properties of the Definite Integral**. (Just the **third** of the four class drills on the handout.)

**Instructor: **Show the solutions to the Class Drill that the class just worked on.

**Instructor: (only if there is enough time) **In the Class Drill just done, the graph of a function \(f(x)\) was given, along with the values of the *unsigned areas* of certain regions between the graph and the \(x\) axis. Those given values of the *unsigned areas* were used to give values for certain *definite integrals*.

In the last Class Drill for today, the graph of a function \(f(x)\) is given, but there are no given unsigned areas. Instead, you will have to figure out the areas of the regions using *geometry*.

**Instructor: (if there is enough time) **Have Students work on this **Class Drill**.

**Students: ** Work on
**Class Drill: Definite Integrals for a Simple Graph **. (Just the **fourth** of the four class drills on the handout.)

**Instructor: **Show the solutions to the Class Drill that the class just worked on.

**The Area Problem**

**Instructor: **Recall that on Monday, we discussed the *Area Problem*, or **"Two Questions about Area"**:

- How can we
*define*the*signed area*between the graph of a function \(f(x)\) and the \(x\) axis from \(x\) = \(a\) to \(x\) = \(b\) for a general function \(f(x)\) whose graph is not made up basic shapes? - How can we find the
*value*of that*signed area*?

After much discussion (See the
**Notes from the Video for H76**), we arrived at a *definition* the *Signed Area* that involved *Riemann Sums*. The resulting definition fo the *Signed Area* was given a new name: ** The Definite Integral**.

**Is There An Easier Way?**

Video H76 ended with some discussion about the difficulty of the computations involved in finding the value of a *definite integral* using the definition of the integral as a limit of *Riemann Sums*. A question was posed:

The answer to that question is ** yes**; the

The ** Fundamental Theorem of Calculus** expresses the relationship between

**Instructor: **Introduce *change in \(F\) notation* and give an example of its use. (For reference, see
**page 17 in the Notes from the Video for H78**.)

(the relationship between ** definite integrals** and

If \(f(x)\) is continuous on the interval \([a,b]\), then

$$\int_a^bf(x)dx\underset{\text{FTC}}{=}\left. \left(\int f(x)dx\right)\right\vert_a^b$$This notation may help you better understand the statement of the theorem, and better remember how to use the theorem.

**Instructor Do this Example (Using the notation as described above and introduced in the
Video for H78).** Show why

This result is interesting because it is an *exact* result, not particularly hard to obtain, and it confirms the *estimate* that was obtained in Video H76. That is:

*Estimate*obtained using*Riemann Sums done by Computer*in Video H76: \(SA \approx 107.33\)*Exact Result*obtained using the*Fundamental Theorem of Calculus*in Video H78: \(SA = \frac{322}{3} = 107.\overline{333}\)

**Instructor: **In the *MyLab* system, problems about finding the value of a *definite integral* using the *Fundamental Theorem of Calculus* are not very well written. They simply ask you to type in the *value* of the definite integral. They don't ask you to show any of the steps that lead to that value. This style of problem is easy for the publisher to produce, but it does not test your *understanding* of the problem. And, if you *don't* understand the problem, those online questions that just ask you to enter the final answer will not help you *gain* an understanding.

Focusing on the steps that *lead to* the answer is the best way to improve your understanding of the *Fundamental Theorem*. For that reason, today's Class Drills and Presentations are all about showing the *steps* that lead to an answer. In fact, the answers will be *given* to you. Your job will be to show the *steps*.

**Instructor: **Have students work on the following **Drill problem #1: **A definite integral is shown below, along with the correct answer. Show how the answer is obtained. (Show the steps.)

**Instructor: **Show the solution of **Drill problem #1: ** (For reference, see
**[Example 4] on page 20 of the Notes from the Video for H78**.)

**Instructor: **Have students work on the following **Drill problem #2: **A definite integral is shown below, along with the correct answer. Show how the answer is obtained. (Show the steps.)

**Instructor: **Show the solution of **Drill problem #2: ** (For reference, see
**[Example 5] on page 22 of the Notes from the Video for H78**.)

**Instructor: **Have students work on the following **Drill problem #3: **A definite integral is shown below, along with the correct answer. Show how the answer is obtained. (Show the steps.)

**Instructor: **Show the solution of **Drill problem #3: ** (For reference, see
**[Example 7] on page 24 of the Notes from the Video for H78**.)

**Instructor: **Have students work on the following **Drill problem #4: **A definite integral is shown below, along with the correct answer. Show how the answer is obtained. (Show the steps.)

**Instructor: **Show the solution of **Drill problem #4: ** (For reference, see
**[Example 9] on page 26 of the Notes from the Video for H78**.)

**Instructor: **Have students work on the following **Drill problem #5: **A definite integral is shown below, along with the correct answer. Show how the answer is obtained. (Show the steps.)

**Instructor: **Show the solution of **Drill problem #5: ** (For reference, see
**[Example 6] on page 23 of the Notes from the Video for H78**.)

**Instructor:** For **Drill problem #5** just presented, *Wolfram Alpha* says that the answer is
$$ \int_{7}^{28}\frac{3}{x} \ dx = \log{(64)}$$
But if you type \( \log(64) \) into a *MyLab* answer box for this problem, it will be marked ** incorrect**!

**Questions for the Class:**

- Why is the
*Wolfram Alpha*answer different from the answer in the**presentation**? - Isn't
*Wolfram Alpha*always*right*? Why would*MyLab*say that the*Wolfram Alpha*answer is?*incorrect*

**Answer:** First, note that the correct answer can be presented in different forms:
$$ \int_{7}^{28}\frac{3}{x} \ dx = 3\ln{(4)} = \ln{(4^3)} = \ln{(64)}$$
Indeed, if you enter \(3\ln{(4)}\) or \(\ln{(4^3)}\) or \(\ln{(64)}\) into a *MyLab* answer box for this problem, it will be marked ** correct**!

Next note that

- In
*Wolfram Alpha*the symbol \( \log{(x)} \) denotes the*natural logarithm*. That is, the*base \(e\) logarithm*. - In our textbook and in
*MyLab*the symbol \( \log{(x)} \) denotes the*base \(10\) logarithm*.

Because of this, when *Wolfram Alpha* gives an answer \( \log(64) \), it means *the natural logarithm of \(64\)*.

But if you enter \( \log(64) \) into a *MyLab* answer box for this problem, *MyLab* will interpret your answer as *the base \(10\) logarithm of \(64\)*. That answer is ** incorrect**.

This illustrates that using handheld or online calculators makes some computations easier, but also requires that you be aware of complications that may arise.

**Instructor: **Have students work on the following **Drill problem #6: **A definite integral is shown below, along with the correct answer. Show how the answer is obtained. (Show the steps.)

**Instructor: **Show the solution of **Drill problem #6: ** (For reference, see
**[Example 8] on page 25 of the Notes from the Video for H78**.)

**Instructor:** Sometimes, computations of *Definite Integrals* will involve associated *Indefinite Integrals* that must be found using the *Method of Substitution*.

**Instructor: **Have students work on the following **Drill problem #7: **A definite integral is shown below, along with the correct answer. Show how the answer is obtained. (Show the steps.)

**Instructor: **Show the solution of **Drill problem #7: ** (For reference, see
**[Example 1] on page 4 of the Notes from the Video for H79**.)

**Instructor: **Our first topic for today is ** Total Change Problems**.

We begin by reviewing an ** old concept** and introducing a

**The Old Concept:** A very important concept for the

- the
*instantaneous rate of change of \(f(x)\) at \(x=a\).* - the
*slope of the line tangent to the graph of \(f(x)\) at \(x=a\).* - the
*derivative of \(f(x)\) at \(x=a\).*(abbreviated \(f'(a)\) )

**The New Concept:** An important concept in the

- the
*change in \(F(x)\) from \(x=a\) to \(x=b\).* - the
*signed area between the graph of \(F'(x)\) and the \(x\) axis from \(x=a\) to \(x=b\).* - the
*definite integral of \(F'(x)\) from \(x=a\) to \(x=b\).*(abbreviated \(\int_{a}^{b}F'(x)dx \) )

For now, we will discuss the equality of the *first* and *third* items on the list above. Later in this meeting, we will discuss the *second* item on the list.

In Section 5.5, on **Homework H80 Total Change Problems**, you are asked to find the total change in \(F(x)\) from \(x=a\) to \(x=b\) when the the derivative \(F'(x)\) is given. The point of those exercises is that you had to make the connection that the ** change in \(F(x)\)** (the

To understand why that is the approach, it is worth reviewing the different forms of the *Fundamental Theorem of Calculus* that Barsamian discusses on
**pages 3 - 5 of the Notes from the Video for H80**
. Barsamian concludes the discussion on page 5 by presenting a *green box* describing ** Total Change Problems**.

Our first **Class Drill** will be an example of a *Total Change Problem*.

A company manufactures electric scooters, a very dangerous gadget that people would not ride if they had a lick of sense. The Marginal Cost is
$$C'(x)=400-\frac{x}{3}\text{ for }0\leq x\leq 900$$
where the variable \(x\) represents the number of scooters made per month (the *quantity*) and \(C'(x)\) is in dollars.

Find the increase in cost going from a production level of 600 scooters per month to 630 scooters per month. Show all details of the calculation clearly.

**Instructor: **Show the solution of the **Class Drill**. (For reference, see
**[Example 1] on page 6 of the Notes from the Video for H80**
.)

**Instructor: **Next, we'll discuss the ** Average Value of a Function Over an Interval **.

In the *English language*, the word *average* has a variety of meanings. But in *Math & Science*, the word *average* gets assigned certain precise meanings in different situations. One use of the word *average* is in the phrase *Average Value of a Function Over an Interval*. On
**pages 4 - 5 of the Notes from the Video for H81**
, Barsamian poses a *geometric question*, and then answers the question. The answer to the question is presented in an official definition of the *Average Value of a Function Over an Interval*.

**Words:***The Average Value of \(f(x)\) over the Interval \([a,b]\)***Usage:**The function \(f(x)\) is continuous on the interval \([a,b]\)**Meaning:**The number \(h\) given by this formula: $$h=\frac{1}{b-a}\int_a^bf(x)dx$$**Graphical Interpretation:**The number \(h\) is the height of a rectangle sitting on the interval \([a,b]\) that would enclose a signed area that is equal to the signed area of the graph of \(f(x)\) on the same interval.

Our next **Class Drill** will be an example of computing the *The Average Value of a Function over an Interval*.

A drug is administered to a patient by a pill. The drug concentration in the bloodstream is described by the function $$C(t)=\frac{2t}{t^2+4}\text{ for }0\leq t \leq 12$$ where \(t\) is the time in hours after the pill is taken and \(C(t)\) is the drug concentration in the bloodstream (in micrograms/liter) at time \(t\).

Find the average drug concentration in the bloodstream over the first \(2\) hours. Give an exact answer and a decimal approximation. Show all details of the calculation clearly.

**Instructor: **Show the solution of the **Class Drill**. (For reference, see
**[Example 3](a) on pages 12-13 of the Notes from the Video for H81**.)

**Sections, Content, Homework, and Videos:**

- 6.1 Area Between Curves (Link to Homework and Videos for Section 6.1)
- 6.2 Applications in Business and Economics (Link to Homework and Videos for Section 6.2)

**Instructor: **Today, we'll begin discussing *Area Between Curves* (Concepts from Section 6.1).

In Chapter 5, we discussed the *signed area between the graph of a function \(f(x)\) and the \(x\) axis*, and we learned how to compute the *value* of the *signed area* using the *definite integral*.

It is important to note that the term *Area Between Curves*, as used in Chapter 6, refers to an *unsigned area*. We will eventually find the *value* of the *unsigned area* using a calculation involving *definite integrals*, but we must first be careful to set up the *correct* integral calculation. The process of setting up the correct integral calculation is discussed on
**pages 4 and 5 of the Notes from the Video for H82**
. As explained there, the key is to scrutinize the *graphs* of the curves, and determine the *simple regions*.

In today's first **Class Drill** you'll work on *setting up* a *definite integral* to compute the area between two curves.

**Instructor: **Show the solution of the **Class Drill** that the class just worked on.
(For reference, see Book Exercise 6.1#37, which is MyLab H83 problem [1], and
**[Example 1] on page 4 of the Notes from the Video for H83**
.)

**Instructor: **In the **Class Drill** just finished, the graphs of the functions \(f(x)\) and \(g(x)\) were given, so the setting up of the correct definite integral calculation amounted to simply identifying the *endpoints* of the *simple regions*, then identifying the *top* and *bottom* functions for each region, and finally, setting up the definite integrals.

In practice, however, one is only given *formulas* for the functions \(f(x)\) and \(g(x)\), *not* their *graphs*. The problem of finding the area between the curves is then made significantly harder because one must first *graph* the functions \(f(x)\) and \(g(x)\). Then one must use the *graphs* of the curves to determine the simple regions and set up the correct *definite integral calculation*. Then, one must find the *value* of the *definite integral calculation*. These are difficult problems, as hard as anything in MATH 1350.

You will work on problems of this type in two **Class Drills**.

- Set up a definite integral to find the area bounded by the graphs of the equations \(y=2x+4\) and \(y=x^2+1\) over the interval \(0 \leq x \leq 2\), and find the value of the integral. Give an exact, simplified answer, and a decimal approximation rounded to 3 decimal places
- Again, set up a definite integral to find the area bounded by the graphs of the equations \(y=2x+4\) and \(y=x^2+1\), but this time, the interval is not specified. (
**You have to figure out the interval.**). Then, find the value of the integral. Give an exact, simplified answer, and a decimal approximation rounded to 3 decimal places

**Instructor: **Show the solution of the **Class Drill** that the class just worked on.
(For reference, see
**[Example 2] on page 5 of the Notes from the Video for H83**
and
**[Example 4] on page 9 of the Notes from the Video for H83**
.)

Set up a definite integral to find the area bounded by the graphs of the equations \(y=e^x\) and \(y=-\frac{1}{x}\) over the interval \(1 \leq x \leq 3\). Then, find the value of the integral. Give an exact, simplified answer. Then, give a decimal approximation rounded to 3 decimal places

**Instructor: **Show the solution of the **Class Drill** that the class just worked on.
(For reference, see
**[Example 3] on page 7 of the Notes from the Video for H83**.
.)

**If there is time remaining: Instructor do this [Example]: **

Set up a definite integral to find the area bounded by the graphs of the equations \(y=27\sqrt{x}\) and \(y=x^2\). Then, find the value of the definite integral. Give an exact, simplified answer. Then give a decimal approximation rounded to 3 decimal places.

**Instructor: **An important concept in the final month of the course is the *equality* of three quantities related to *area*:

- the
*change in \(F(x)\) from \(x=a\) to \(x=b\).* - the
*signed area between the graph of \(F'(x)\) and the \(x\) axis from \(x=a\) to \(x=b\).* - the
*definite integral of \(F'(x)\) from \(x=a\) to \(x=b\).*(abbreviated \(\int_{a}^{b}F'(x)dx \) )

On **Homework H80 Total Change Problems**, you are asked to find the ** total change in \(F(x)\) from \(x=a\) to \(x=b\)** when the the derivative \(F'(x)\) is given. The point of those exercises is that you had to make the connection that the

In Section 6.1, on **Homework H84 Total Change Problems as Area Problems**, you are asked to compute ** areas between graphs of \(F'(t)\) and the \(t\) axis over a time interval \(t=a\) to \(t=b\)**, and then to

An online group has formed consisting of people who have been injured in electric scooter accidents. In its early months of existence, the group is growing at a rate $$N'(t) = 1000e^{0.5t} \ \ users \ \ per \ \ month$$

- Make a graph of the function $$N'(t) = 1000e^{0.5t}$$ on the interval \(4 \leq t \leq 8\)
- On your graph, shade the area between the graph of \(N'(t)\) and the \(t\) axis over the interval \(4 \leq t \leq 8\).
- Find the area between the graph of \(N'(t)\) and the \(t\) axis over the interval \(4 \leq t \leq 8\). (Give an exact answer and a decimal approximation.)
**Interpret**the results of part (c). That is, explain what the**numerical result from part (c)**tells us about the**online group**.- Make a graph of the function $$N(t) = 2000e^{0.5t}+C$$ on the interval \(4 \leq t \leq 8\)
- On the vertical axis of your graph, highlight the segment between \(N(4)\) and \(N(8)\). Explain what this segment length has to do with the
**interpretation**that you provided in part**(d)**.

**Student #15 Presentation CP3: **Show the solution of the **Class Drill** that the class just worked on. To prepare for this presentation, study

- Book Exercise 6.1#89 (which is MyLab Homework H84 problem [1]
**[Example 1] on page 7 of the Notes from the Video for H84**

**Instructor: **The degree of income inequality in the United States is a common subject in conversation and in the news. There is a quantitative measure of income inequality that is based on measuring the **area between curves**. The quantitative measure is called the ** Gini Index**.

The Gini Index is computed using a definite integral involving a function called a *Lorenz Curve*. In order to understand the Gini Index, we must first learn about that curve. The curve is introduced on
**pages 4 and 5 of the Video for H85**. In the video, three fictional countries are introduced, with details of their household incomes. The three countries have obviously different degrees of income inequality.

The *Lorenz Curves* for the three countries are shown on
**page 10 of the Video**, along with two observations:

- All three of the curves go through the points \( (0,0) \) and \( (1,1) \)
- When household income is very concentrated, very unequally distributed, the Lorenz curve is very bowed; When household income is less concentrated, more equally distributed, the Lorenz curve is less bowed and stays close to the line \(y=x\), which is shown as a dotted line.

One straightforward way to quantify the income concentration is to simply measure the area of the region between the *Lorenz Curve* \(f(x)\) and the *Line of Absolute Equality* \(y=x\). In the video, it is pointed out that this region is a *simple region*, whose top curve is the *Line of Absolute Equality* and whose bottom curve is the *Lorenz Curve*:
$$top(x)=x \\ bottom(x)=f(x)$$

That is the idea behind the *Gini Index*, defined as follows:

**Definition of the Gini Index of Income Concentration**

The ** Gini Index (GI)** for a country is defined to be twice the area of the region bounded by the

**Observations:**

- A
*Gini Index*\(GI=0\) would mean that the area of the region bounded by the*Lorenz Curve*for the country and the Line of*Absolute Equality*is*zero*. This would indicate that income is*equally distributed*among all households of the country. -
*Gini Index*\(GI=1\) would indicate that all of the household income in the country is earned by a single household.

In general,

You'll work in pairs on a **Class Drill** about the **Gini Index**.

A country called *Funland* has a *Lorenz Curve* \(f(x)=x^4\). A country called *Greatland* has a *Lorenz Curve* \(g(x)=x^5\).

- On a single, large pair of axes, draw both the graph of \(f(x)=x^4\) and \(g(x)=x^5\) on the interval \([0,1]\).
- Oh your graph, shade the region between the graph of \(f(x)=x^4\) and the \(x\) axis on the interval \([0,1]\).
- Compute the
*Gini Index*for each country. Show the details of the calculation clearly. - In which country is income more equally distributed? Explain, referring to your graph to illustrate.

**Student #16 Presentation CP3: **Show the solution of the**Class Drill** that the class just worked on. To prepare for this presentation, study

- Book problems 6.1#83 and #85, which are on MyLab Homework H85
**The second [Example 1], on page 13 of the Notes from the Video for H85**

**Instructor: **Return Exam X3 and discuss common mistakes.

**Instructor: **Last week, we discussed how to use a **definite integral** to compute the **total change in a quantity** in situations where the **derivative of the quantity** is known. These kinds of problems were called **Total Change Problems**.

Then, on Monday of this week, we discused how to use a **definite integral** to compute the **area between curves**.

And then, on Wednesday of this week, we put those two ideas together, using the **area between curves** to help *illustrate* **Total Change Problems**.

Remember that the topics for those three discussions were all about an important concept in the *final month* of the course: the *equality* of three quantities related to *area*:

- the
*change in \(F(x)\) from \(x=a\) to \(x=b\).* - the
*signed area between the graph of \(F'(x)\) and the \(x\) axis from \(x=a\) to \(x=b\).* - the
*definite integral of \(F'(x)\) from \(x=a\) to \(x=b\).*(abbreviated \(\int_{a}^{b}F'(x)dx \) )

Today, and for the rest of the semester, we'll be discussing topics from Section 6.2 of our textbook. That section is about *Applications of the Definite Integral in Business and Economics*. Today, we will discuss applications of the *Area Between Curves* to a particular problem in Business, computing what is called the ** Total Income for a Continuous Income Stream**. We will see that this kind of problem is actually just another example of a

The expression ** Continuous Income Stream** means that money is flowing into an account with

A simple question to ask about such a flow is the following:

The answer to the question is that the ** Total Income** is found by

Realize that because the value of the **total income (TI)** is obtained by **integrating the flow rate \(f(t)\)**, the quantity **TI** can be visualized as an **area between the graph of the flow rate \(f(t)\) and the \(t\) axis**. Such an illustration is shown on pages 10 and 16 of the same notes.

You'll work in pairs on a **Class Drill** about the **Total Income from a Continuous Income Stream**.

A *continuous income stream* has flow rate
$$f(t) = 2000e^{(0.02t)} \ \text{dollars per year}$$

- Find the total income produced by the income stream in the first \(10\) years (Give an
*exact answer*, in symbols, and a*decimal approximation*, rounded to the nearest dollar.) - Illustrate the total income by showing the corresponding
*area*between a graph of \(f(t)\) and the \(t\)*axis*.

**Student #17 Presentation CP3: **Show the solution of the **Class Drill** that the class just worked on. To prepare for this presentation, study

- Book problem 6.2#41, which is MyLab Homework H86 problem [2]
**[Example 2] in the Notes from the Video for H86**

**Instructor: **A more complicated situation arises when an income stream is flowing into an account where it will start earning *interest* upon arrival. This situation is discussed on
**pages 5 - 8 of the Notes from the Video for H87**. An illustration of money flowing into a bucket is again used there. A natural question is

*If a continuous income stream flows into an account that earns continuously compounded interest at a rate \(r\), what will be the balance of the account at time \(T\) years? (Note the capital \(T\).) This balance is called the Future Value at time \(T\) years. (abbreviated \(FV\))*

The answer to the question is that the *Future Value* is found by the following integral calculation.

The integral above is rather complicated. For that reason, Barsamian presents the result of the integral in two useful common cases, which are all that is needed for our course.

**Particular Result: Future Value of a Continuous Income Stream with Constant Flow** If a continuous income stream with flow rate \(f(t)=c\) dollars per year (*constant flow*) flows into an account that earns continuously compounded interest at rate \(r\), then the balance of the account at time \(T\) years (that is, Future Value (\(FV\)) at time \(T\) years) will be

**Particular Result: Future Value of a Continuous Income Stream with Exponential Flow** If a continuous income stream with flow rate \(f(t)=c \cdot e^{(kt)}\) dollars per year (*exponential flow*) flows into an account that earns continuously compounded interest at rate \(r\), then the balance of the account at time \(T\) years (that is, Future Value (\(FV\)) at time \(T\) years) will be

**Instructor: Present this [Example] Future Value of a Continuous Income Stream** (For reference, see

- How much money will be in the account \(50\) years later, when you retire at age \(75\)? (Give an
*exact answer*, in symbols, and a*decimal approximation*, rounded to the nearest dollar.) - How much of the final amount is interest? (Just give a decimal approximation, rounded to the nearest dollar.)

(Note: no books, no notes, no calculators, no phones)

**Sections, Content, Homework, and Videos:**

- 6.2 Applications in Business and Economics (Link to Homework and Videos for Section 6.2)

**Students: **

- Note that the last date to complete any
**MyLab**homework will be 11:59pm on Tue Apr 30. - Remember that all of the Athens Campus Sections of MATH 1350 will have a Combined Final Exam on
**Wednesday, May 1, 2024 from 4:40pm – 6:40pm**. - The Final Exam will be given in various rooms in
**Morton Hall**.**You will NOT be taking your Final Exam in the room where your regular class meetings are held.**Room assignments will be posted in the Calendar, in the entry for Wed May 1, in the next couple of days. - More information about the Final Exam will be posted in the Calendar, in the entry for Wed May 1, in the next couple of days.

**Instructor: **Review the **Grading** for the course on the Course Web Page.

**Instructor: **In this last week of the semester, we'll be continuing to discuss topics from Section 6.2 of our textbook. That section is about *Applications of the Definite Integral in Business and Economics*. The applications that we will be discussing this week are called *Consumers' Surplus, Producers' Surplus, and Equilibrium Price*.

Today, we will discuss **Consumers' Surplus**

Before discussing Consumers' Surplus, however, it is worthwhile to review the Business Terminology of *Demand* and *Price*.

**Instructor: **Review ** Demand** and

**Instructor: **Introduce the ** Consumers' Surplus (CS)** For reference

- Refer to
**pages 5 - 10 of the Notes from the Video for H88** - If you want to use it, there is a review of the
here:*Consumers' Surplus (CS)*

The **Consumers' Surplus**, denoted \(CS\), is a quantity that is the total amount that all consumers who are willing to buy the item at the price \(\overline{p}\) will feel like they saved if the selling price is \(\overline{p}\). The *Consumers' Surplus (CS)* can be *visualized* as the ** area between curves** on a graph of the

The *value* of the *Consumers' Surplus (CS)* is given by the integral calculation

In this integral,

- The function \(D(x)\) is the
*Demand Price*curve, which describes the relationship between*selling price \(p\)*per item and \(x\), which represents the*number of items that consumers are willing to buy*at that price. (This number of items, \(x\) is called the*Demand Quantity*.) - The ordered pair \( ( \overline{x},\overline{p}) \) is a point on the
*Demand Price*curve \(D(x)\).

Observe that this integral is of the standard form used for computing *area between curves*. That is,

**Instructor: **For the rest of the meeting, you will work on two **Class Drills** about Consumers' Surplus. Students #15 and #16 will present solutions to the drills after you work on them for awhile..

Suppose that

- The
*Demand Price Function*is \(p=D(x)=120-0.02x\). - The selling price has been established as \( \overline{p}=60 \).

The goal is to find the value of the *Consumers' Surplus (CS)* by two methods.

**Method 1 (Using Geometry) **Do the following:

- For the
*selling price*\( \overline{p}=60 \), find the corresponding*Demand*, \( \overline{x}\). - Graph the
*Demand Price Function*\(D(x)\). What is the domain of this function? Why? - Plot the point \(( \overline{x}, \overline{p})\) on the graph of the
*Demand Price Function*that you just made. - Shade the region corresponding to the
*Consumers' Surplus*. - Use a simple geometric formula for
*area*to find the area of the region that you shaded on your graph. This is the*Consumers' Surplus*, computed using*Geometry*.

**Method 2 (Using Calculus) **Do the following:

- Set up the
*definite integral*that calculates the*Consumers' Surplus*. That is, set up the integral $$CS=\int_{0}^{\overline{x}}[D(x)-\overline{p} \ ]dx$$ - Find the
*value*of the*Consumers' Surplus*by evaluating the definite integral that you set up in part (f). This is the*Consumers' Surplus*, computed using*Calculus*.

**Question for the Groups: **Do the results obtained by the two methods match?

**Student #18 Presentation CP3: **Show the solution of **Drill Problem #1** that the class just worked on.
To prepare for this presentation,

- Study book problem 6.2#69, which is MyLab Homework H88 problem [1].
- Study
**[Example 1] on page 11 of the Notes from the Video for H88**.

Suppose that

- The
*Demand Price Function*is \(p=D(x)=9-x^2\). - The selling price has been established as \( \overline{p}=5 \).

The goal is to *illustrate* the *Consumers' Surplus (CS)* using a *graph*, and then find its *value* using *calculus*.

**Part 1: Illustrate the Consumers' Surplus **Do the following:

- For the
*selling price*\( \overline{p}=5 \), find the corresponding*Demand*, \( \overline{x}\). - Graph the
*Demand Price Function*\(D(x)\). What is the domain of this function? Why? - Plot the point \(( \overline{x}, \overline{p})\) on the graph of the
*Demand Price Function*that you just made. - Shade the region corresponding to the
*Consumers' Surplus*. Notice that the region is not made up of simple geometric shapes, so it is not possible to find the area of the region using geometric formulas.

**Part 2: Find the Value of the Consumers' Surplus **Do the following:

- Set up the
*definite integral*that calculates the*Consumers' Surplus*. That is, set up the integral $$CS=\int_{0}^{\overline{x}}[D(x)-\overline{p} \ ]dx$$ - Find the
*value*of the*Consumers' Surplus*by evaluating the definite integral that you set up in part (e). This is the*Consumers' Surplus*, computed using*Calculus*.

**Student #19 Presentation CP3: **Show the solution of **Drill Problem #2** that the class just worked on.
To prepare for this presentation,

- Study book problem 6.2#69, which is MyLab Homework H88 problem [1].
- Study
**[Example 1] on page 11 of the Notes from the Video for H88**.

**Instructor: **In this last week of the semester, we are discussing topics from Section 6.2 of our textbook. That section is about *Applications of the Definite Integral in Business and Economics*. The applications that we will are discussing are called *Consumers' Surplus, Producers' Surplus, and Equilibrium Price*. On Monday, we discussed *Consumers' Surplus*

Today, we will discuss **Producers' Surplus**

We start by discussing the Business Terminology of *Supply* and *Price*.

**Instructor: **Discuss ** Supply** and

Now that we understand the terminology of *Supply* and *Price*, we are ready to discuss the *Producers' Surplus*.

**Instructor: **Introduce the ** Producers' Surplus (PS)** For reference

- Refer to
**pages 8-13 of the Notes from the Video for H89** - If you want to use it, there is a review of the
here:*Producers' Surplus (CS)*

The **Producers' Surplus**, denoted \(PS\), is a quantity that is the total amount that all producers who are willing to supply the item at the price \(\overline{p}\) will feel like they saved if the selling price is \(\overline{p}\). The *Producers' Surplus (PS)* can be visualized as the ** area between curves** on a graph of the

The *value* of the *Producers' Surplus (PS)* is given by the integral calculation

In this integral,

- The function \(S(x)\) is the
*Supply Price*curve, which describes the relationship between*selling price \(p\)*per item and \(x\), which represents the*number of items that producers are willing to supply*at that price. (This number of items, \(x\) is called the*Supply Quantity*.) - The ordered pair \( ( \overline{x},\overline{p}) \) is a point on the
*Supply Price*curve \(S(x)\).

Observe that this integral is of the standard form used for computing *area between curves*. That is,

**Instructor: **For the rest of the meeting, you will work on two **Class Drills** about Producers' Surplus.

Suppose that

- The
*Supply Price Function*is \(p=S(x)=0.01x\). - The selling price has been established as \( \overline{p}=30 \).

The goal is to find the value of the *Producers' Surplus (PS)* by two methods.

**Method 1 (Using Geometry) **Do the following:

- For the
*selling price*\( \overline{p}=30 \), find the corresponding*Supply*, \( \overline{x}\). - Graph the
*Supply Price Function*\(S(x)\) on the domain \( [0,\overline{x}]\) - Plot the point \(( \overline{x}, \overline{p})\) on the graph of the
*Supply Price Function*that you just made. - Shade the region corresponding to the
*Producers' Surplus*. - Use a simple geometric formula for
*area*to find the area of the region that you shaded on your graph. This is the*Producers' Surplus*, computed using*Geometry*.

**Method 2 (Using Calculus) **Do the following:

- Set up the
*definite integral*that calculates the*Producers' Surplus*. That is, set up the integral $$PS=\int_{0}^{\overline{x}}[ \ \overline{p}-S(x)]dx$$ - Find the
*value*of the*Producers' Surplus*by evaluating the definite integral that you set up in part (f). This is the*Producers' Surplus*, computed using*Calculus*.

**Question for the Groups: **Do the results obtained by the two methods match?

**Student #20 Presentation CP3: **Show the solution of **Class Drill** that the class just worked on.
To prepare for this presentation,

- Study book problem book problem 6.2#74, which is MyLab Homework H89 problem [1].
- Study
**[Example 1] on page 14 of the Notes from the Video for H89**.

Suppose that

- The
*Supply Price Function*is \(p=D(x)=3x^2\). - The selling price has been established as \( \overline{p}=75 \).

The goal is to *illustrate* the *Producers' Surplus (PS)* using a *graph*, and then find its *value* using *calculus*.

**Part 1: Illustrate the Producers' Surplus **Do the following:

- For the
*selling price*\( \overline{p}=75 \), find the corresponding*Supply*, \( \overline{x}\). - Graph the
*Supply Price Function*\(S(x)\) on the domain \( [0,\overline{x}]\) - Plot the point \(( \overline{x}, \overline{p})\) on the graph of the
*Supply Price Function*that you just made. - Shade the region corresponding to the
*Producers' Surplus*. Notice that the region is not made up of simple geometric shapes, so it is not possible to find the area of the region using geometric formulas.

**Part 2: Find the Value of the Producers' Surplus **Do the following:

- Set up the
*definite integral*that calculates the*Producers' Surplus*. That is, set up the integral $$PS=\int_{0}^{\overline{x}}[ \ \overline{p}-S(x)]dx$$ - Find the
*value*of the*Producers' Surplus*by evaluating the definite integral that you set up in part (e). This is the*Producers' Surplus*, computed using*Calculus*.

**Instructor: **Show the solution of the **Class Drill** that the class just worked on.
(For reference, see
**[Example 2] on page 17 of the Notes from the Video for H89** .)

**Instructor: **In this last week of the semester, we are discussing topics from Section 6.2 of our textbook. That section is about *Applications of the Definite Integral in Business and Economics*. The applications that we will are discussing are called *Consumers' Surplus, Producers' Surplus, and Equilibrium Price*. On Monday, we discussed *Consumers' Surplus*; on Wednesday, we discussed *Producers' Surplus*.

Today, we will discuss **Equilibrium Price and Quantity**

**Instructor: **Review the Business Terminology of *Price* and *Demand* and *Supply*.

- Review
and*Demand*, and discuss why the*Price*curve is a*Demand Price**decreasing*curve. (For reference, see**pages 2 - 4 of the Notes from the Video for H88**) - Review
and*Supply*, and discuss why the*Price*curve is an*Supply Price**increasing*curve. (For reference, see**pages 5 - 7 of the Notes from the Video for H89**)

**Instructor: **Introduce ** Equilibrium Price and Quantity** (For reference, see

**Instructor: **For the rest of the meeting, you will work on two **Class Drills** about Equilibrium Price and Quantity. Students #19 and #20 will present solutions to the drills after you work on them for awhile.

Suppose that

- The
*Demand Price Function*is \(p=D(x)=120-0.02x\). - The
*Supply Price Function*is \(p=S(x)=0.01x\).

The goal is to find the *Equilibrium Price and Quantity*, and then find the value of the *Consumers' Surplus (CS)* and the *Producers' Surplus (PS)*.

Do the following:

- Graph the functions \(D(x)\) and \(S(x)\) on the same set of axes.
- Find the
*Equilibrium Price Point*, \( ( \overline{x},\overline{p}) \). It is the point where the curves \(D(x)\) and \(S(x)\) cross. (Find it using algebra.) Label the*Equilibrium Price Point*with its \( ( \overline{x},\overline{p}) \) coordinates. - Add the horizontal line \(y=\overline{p}\) to the graph.
- Shade the region corresponding to the
*Consumers' Surplus*at the*equilibrium price point*. - In a different color, shade the region corresponding to the
*Producers' Surplus*at the*equilibrium price point*. - Use a simple geometric formula for
*area*to find the areas of the two regions that you shaded on your graph. The results will be is the*Consumers' Surplus*and the*Producers' Surplus*, computed using*Geometry*.

**Instructor: **Show the solution of **Class Drill** that the class just worked on.
(For Reference, see
**[Example 1] on page 10 of the Notes from the Video for H90**.)

Suppose that

- The
*Demand Price Function*is \(p=D(x)=-x^2-4x+260\). - The
*Supply Price Function*is \(p=S(x)=x^2+2x\).

The goal is to find the *Equilibrium Price and Quantity*, and then find the value of the *Consumers' Surplus (CS)* and the *Producers' Surplus (PS)*.

Do the following:

- Graph the functions \(D(x)\) and \(S(x)\) on the same set of axes.
- Find the
*Equilibrium Price Point*, \( ( \overline{x},\overline{p}) \). It is the point where the curves \(D(x)\) and \(S(x)\) cross. (Find it using algebra.) Label the*Equilibrium Price Point*with its \( ( \overline{x},\overline{p}) \) coordinates. - Add the horizontal line \(y=\overline{p}\) to the graph.
- Shade the region corresponding to the
*Consumers' Surplus*at the*equilibrium price point*. - In a different color, shade the region corresponding to the
*Producers' Surplus*at the*equilibrium price point*. - Find the
*Consumers' Surplus*and the*Producers' Surplus*. (You'll have to use*integrals*, because these are not simple geometric regions.)

**Instructor: **Show the solution of **Class Drill** that the class just worked on.
(For Reference, see
**[Example 2] on page 16 of the Notes from the Video for H90**.)

**To the Students: **We hope that you have found MATH 1350 to be not just *challenging* but also *rewarding*. We wish you success on the **Final Exam** next week, and in your future time at Ohio University and beyond!

From your Instructor and from Mark Barsamian, the MATH 1350 Coordinator.

**Exam X3** lasts the full duration of the Fri Aug 16 class meeting

**No Books, no Notes, no Calculators, no Phones, no Smart Watches**

The exam material is taken from the material covered in Chapters 5 and 6. See the *List of Homework Assignments* for links to the corresponding Instructional Videos.

page maintained by Mark Barsamian, last updated May 17, 2024