Sample Lecture Outline for 2019 2020 Spring …Sample Lecture Outline for 2019 – 2020 Spring Semester MATH 1350 Day 1 X Discuss Course Organization Show them the Blackboard Page

Sample Lecture Outline for 2019 – 2020 Spring Semester MATH 1350

Day 1

X Discuss Course Organization

Show them the Blackboard Page

Discuss the Course Organization

Show them the textbook

Today: Discuss Introduction to Functions from Section 2 of the book

Definition of function

• Symbol: 𝑓: 𝐴 → 𝐵

• Spoken: “𝑓 is a function from 𝐴 to 𝐵.”, or “𝑓 maps 𝐴 to 𝐵.”

• Usage: 𝐴 and 𝐵 are sets, called the domain and range. In this course, they will usually be sets of

real numbers.

• Meaning: 𝑓 could be thought of as a machine that takes as input any one element of set 𝐴, and

produces as output a single element of set 𝐵.

• Additional notation: if the element 𝑎 ∈ 𝐴 is used as input, then the symbol 𝑓(𝑎) denotes the

resulting output. Notice that 𝑓(𝑎) ∈ 𝐵. Also note that the name of the function is just 𝑓. You

don’t have to include the variable when giving the name of the function.

• Machine Diagram: It is often useful to draw a diagram

that conveys the idea of a function as a machine:

Difference between equations and functions

Examples:

𝑦 = 2𝑥 + 3 is an equation that can be thought of a function. Using function notation, 𝑓(𝑥) = 2𝑥 + 3.

2𝑥 − 𝑦 + 3 = 0 is also an equation that can be thought of as a function, the same as previous.

𝑥2 + 𝑦2 = 1 is an equation, but it cannot be thought of as a function. Why not?

The Natural Domain of functions

Consider 𝑓(𝑥) = 𝑥3 and 𝑔(𝑥) = √𝑥 and ℎ(𝑥) =1

𝑥−5 For which of x do these functions produce outputs?

Graph of a function

Use horizontal axis for the 𝑥 axis; vertical axis for the 𝑦 axis

input 𝑎 ∈ 𝐴 produces an ouput 𝑓(𝑎) ↔ there is dot on the graph at location (𝑥, 𝑦) = (𝑎, 𝑓(𝑎)).

Discuss implication: the vertical line test

• No vertical line can touch the graph more than once (because the would correspond to a single

input that has more than one output.

• Every vertical line of the form 𝑥 = 𝑎, where 𝑎 ∈ 𝐴, does touch the graph (exactly once), because

each input in the domain has to produce an output.

Consider graphs of 𝑓(𝑥) = 𝑥3 and 𝑔(𝑥) = √𝑥 and ℎ(𝑥) =1

𝑥−5 and 𝑥2 + 𝑦2 = 1, considering vertical

line test for each one.

Simplest kinds of Functions: Functions that come from line equations. Discuss kinds of line equations

and their graphs. Discuss that non-vertical line equations can be though of as functions.

Lecture Outline for Day 1 Continues on the Next Page ➔

f f(a) a

input is

an element

of set A

output is

an element

of set B

Day 1 Continued

X Analyzing functions. That is, making observations about them that will help us better understand them.

y intercept: Usually the easiest question to ask about a function is, what is 𝑓(0)? Easy to spot on graph.

Obvious correspondence : 𝑓(0) = 𝑏 ↔ graph has y-intercept at (𝑥, 𝑦) = (0, 𝑏).

x-intercept: Another easy behavior to spot on a graph is an x-intercept.

Obvious correspondence correspondence : 𝑓(𝑎) = 0 ↔ graph has x-intercept at (𝑥, 𝑦) = (𝑎, 0).

But finding x-intercept from formula for a function is a bit harder. The formula for a function is an

equation involving x and y, solved for y in terms of x. In order to find the x-intercepts, we must find the

values of x that cause 𝑓(𝑥) = 0. That means that we must set 𝑦 = 0 in the equation and solve for x.

Definition of root of a function

• Words: 𝑎 is a root of the function 𝑓.

• Meaning: 𝑎 is a real number such that 𝑓(𝑎) = 0

• Graphical significance: The graph of 𝑓 has an 𝑥 intercept at (𝑥, 𝑦) = (𝑎, 0).

• Observation: A function can have many roots, and a function can also have no roots.

• Additional terminology that we won’t use: Some people use the word zero instead of root. That

is, they say that “a is a zero of the function f. ” I find this leads to confusion. The number a causes

y to be zero. The number a will usually not have the value zero.

Examples:

Linear function 𝑦 = 5𝑥 + 3 has one root.

Polynomial function: 𝑦 = 𝑥2 − 4𝑥 + 3 has two roots.

Polynomial function: 𝑦 = 𝑥2 + 5 has no roots. Observe: graph has no x-intercepts!

Rational function: 𝑓(𝑥) =𝑥2−4𝑥+3

𝑥2−6𝑥+5=

(𝑥−1)(𝑥−3)

(𝑥−1)(𝑥−5) has only one root. Why? Recall that the only way for a

fraction 𝑦 =𝑎

𝑏 to be zero is for numerator 𝑎 = 0 AND denominator 𝑏 ≠ 0!

one-to-one functions: Another easy behavior to spot on a graph is when there is more than one point on

the graph with the same y-value. That is, an imaginary horizontal line touches the graph more than once.

For example, for the graph of 𝑓(𝑥) = 𝑥2, the x values 𝑥 = −2 and 𝑥 = 2 both produce an output 𝑦 = 4.

So (𝑥, 𝑦) = (−2,4) and (𝑥, 𝑦) = (−2,4) are two points on the graph with the same y-value. And the

imaginary horizontal line 𝑦 = 4 touches the graph more than once.

Definition of one-to-one function

• Words: function 𝑓 is one-to-one.

• Meaning in words: Different inputs always produce different outputs.

• Meaning in symbols: If 𝑥1 ≠ 𝑥2, then 𝑓(𝑥1) ≠ 𝑓(𝑥2).

• Graphical significance: No horizontal line touches the graph of 𝑓 more than once.

• Additional terminology: When no horizontal line touches the graph of 𝑓 more than once, we say

that the graph passes the horizontal line test. In this situation, function 𝑓 is one-to-one. On the

other hand, there is a horizontal line that does touch the graph of 𝑓 more than once, we say that

the graph fails the horizontal line test. In this situation, function 𝑓 is not one-to-one.


Day 1, continued

X Equality of Functions

When are two functions “equal”? For example, consider 𝑓(𝑥) = (𝑥 + 3)2 and 𝑔(𝑥) = 𝑥2 + 6𝑥 + 9. Are

they the same? They don’t look the same. But note that for any input, 𝑓, 𝑔 always produce same output.

Two functions are defined to be equal if they have the same domain and they produce the same output

for every input in that domain.

So the formulas 𝑓(𝑥) = (𝑥 + 3)2 and 𝑔(𝑥) = 𝑥2 + 6𝑥 + 9 represent the same function, even though the

formulas look different.

Now consider the functions 𝑓(𝑥) =𝑥2−2𝑥−3

𝑥−3 and 𝑔(𝑥) = 𝑥 + 1.

Are they the same?

They certainly don’t look the same.

But notice that we can factor 𝑓(𝑥) as 𝑓(𝑥) =(𝑥+1)(𝑥−3)

(𝑥−3).

Now consider again: are 𝑓 and 𝑔 the same?

Make table of output values for 𝑓 and 𝑔 for 𝑥 = 0,1,2,3,4

Observe that most of the time, a number that appears in both the numerator and denominator cancels, and

the two functions give the same output. That is, when the input is any 𝑥 ≠ 3, the resulting outputs are the

same 𝑓(𝑥) = 𝑔(𝑥).

But when 𝑥 = 3, we find 𝑔(3) = 4 while 𝑓(3) 𝐷𝑁𝐸. Cannot cancel 0

0

Observe that the domain of 𝑔 is all real numbers, while the domain of 𝑓 is all 𝑥 ≠ 3.

Since 𝑓 and 𝑔 do not have the same domains, they are not the same function.

Draw graphs of the two functions. On graph of 𝑓, put a hole at (𝑥, 𝑦) = (3,4).

Tell students to work on exercises in Blackboard

.

Day 2

X Continuing Chapter 2

Finish Leftover Topics from Day 1

First New Topic for Day 2: Composition of Functions (Book Section 2.3)

Definition of Composition of functions

• Symbol: 𝑔 ∘ 𝑓

• Spoken: “The composition of 𝑔 with 𝑓.”

• Usage: 𝑓 and 𝑔 are functions with 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐶. That is, the range of 𝑓 is the set 𝐵,

which is also the domain of 𝑔.

• Meaning: 𝑔 ∘ 𝑓 is the function 𝑔 ∘ 𝑓: 𝐴 → 𝐶 defined by (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥))

• Machine Diagram:

• Observation: Note the symbol 𝑔 ∘ 𝑓 has 𝑔 on the left, but the machine diagram below has 𝑔 on the

right. One way to make sense of this is to keep in mind that the symbol for the input to a function

gets put to the right of the symbol for the function. So the symbol (𝑔 ∘ 𝑓)(𝑥) indicates that 𝑥 is

going to be fed as input into the function 𝑔 ∘ 𝑓. In the symbol (𝑔 ∘ 𝑓)(𝑥), the 𝑓 is closest to the

input 𝑥. Notice that in the machine diagram, the function 𝑓 is what receives the input 𝑥.

Helpful concept: The empty version of a function.

function with variable 𝑥: 𝑓(𝑥) = 5𝑥2 − 3𝑥 + 7

same function with variable 𝑡: 𝑓(𝑡) = 5𝑡2 − 3𝑡 + 7

empty version: 𝑓( ) = 5( )2 − 3( ) + 7

The empty version is very useful when working with compositions.

Example 1: Let 𝑓(𝑥) = 4𝑥2 − 5𝑥 + 7 and 𝑔(𝑥) = 𝑥 + 3.

(a) Find (𝑔 ∘ 𝑓)(𝑥) (b) Find (𝑔 ∘ 𝑓)(5) (c) Find (𝑓 ∘ 𝑔)(𝑥) (d) Find (𝑓 ∘ 𝑔)(5)

Notice that the functions 𝑔 ∘ 𝑓 and 𝑓 ∘ 𝑔 are not the same function. Their formulas don’t look the same,

and, more importantly, we have an example where they both receive the same input 𝑥 = 2 but produce

different output.


𝑓 𝑓(𝑥) 𝑥

input is

an element

of set A

output of f is

an element of

set B, and it

gets fed into g.

𝑔 output is

an element

of set C

𝑔(𝑓(𝑥))

𝑔 ∘ 𝑓

Day 2 Continued

X Continuing Day 2

Another Example

Example 2: Let 𝑓(𝑥) =𝑥−3

𝑥−5 and 𝑔(𝑥) =

5𝑥−3

𝑥−1.

Same questions: (a) Find (𝑔 ∘ 𝑓)(5) (b) Find (𝑔 ∘ 𝑓)(𝑥) (c) Find (𝑓 ∘ 𝑔)(5) (d) Find (𝑓 ∘ 𝑔)(𝑥)

Notice that the functions 𝑔 ∘ 𝑓 and 𝑓 ∘ 𝑔 have formulas that do look the same. Does that mean the

functions 𝑔 ∘ 𝑓 and 𝑓 ∘ 𝑔 are the same? No! we have an example where they both receive the same input

𝑥 = 5 and produce different output. So the answer is no, functions 𝑔 ∘ 𝑓 and 𝑓 ∘ 𝑔 are not the same

function. The reason is that he functions have different domains. Consider the diagram for 𝑔 ∘ 𝑓. An

input 𝑥 = 𝑎 this function first gets fed into the function 𝑓. The input has to produce an output 𝑓(𝑎) that

in turns get fed into the function 𝑔. So the number 𝑥 = 𝑎 has to be a number that is in the domain of 𝑓.

Observe that (𝑔 ∘ 𝑓)(5) is undefined! The domain of 𝑓 is all 𝑥 ≠ 5. Therefore a more thorough

description of function 𝑔 ∘ 𝑓 is (𝑔 ∘ 𝑓)(𝑥) = 𝑥 for all 𝑥 ≠ 5

Put another way, the domain of the function 𝑔 ∘ 𝑓 is the set {𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑥 ≠ 5}.

That is, 𝑔 ∘ 𝑓: {𝑅𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑥 ≠ 5} → 𝑅𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 defined by the formula (𝑔 ∘ 𝑓)(𝑥) = 𝑥.

On the other hand, the domain of the function 𝑓 ∘ 𝑔 is the set {𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑥 ≠ 1}, because the input

has to be an element of the domain of 𝑔.

That is, 𝑓 ∘ 𝑔: {𝑅𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑥 ≠ 1} → 𝑅𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 defined by the formula (𝑓 ∘ 𝑔)(𝑥) = 𝑥.

So even though 𝑔 ∘ 𝑓 and 𝑓 ∘ 𝑔 have the same formula, they are not the same function because they have

different domains.

Inverse Functions (Book Section 2.3)

In this course, we think of functions as machines that receive an input and produce an output.

If the input is 𝑥, the resulting output is denoted by the symbol 𝑓(𝑥).

We could think of the output as 𝑦, so 𝑓 is a machine that receives as input some given 𝑥 and produces as

output the resulting 𝑦.

It is often desirable to want to run this machine backwards. That is, given some desired value of 𝑦, we

might want to know what input value of 𝑥 would be needed to produce that desired output 𝑦.

For some functions 𝑓, it is possible to run the function in reverse.

For example, for 𝑓(𝑥) = 𝑥3, an input of 𝑥 = 5 produces an output 𝑦 = 53 = 125.

If we desire to get an output of 𝑦 = 17, for example, we should use the input 𝑥 = √173

= 171/3.

Then we would get 𝑦 = 𝑓(171/3) = (171/3)3

= 17.

So for any desired output 𝑦, we can find a special input 𝑥 that will work.

We could think of the process 𝑓 running in reverse as an actual function. It takes as input the number 𝑦

and produces as output the number 𝑥.

Observe:

The function 𝑓 is the cubing function

The function 𝑓 running in reverse is the cube root function.

For some functions, though, if we run the function in reverse, the resulting process is not a function.

For 𝑔(𝑥) = 𝑥2, if we desire to get an output of 𝑦 = 25, there are two possible inputs: 𝑥 = 5 and 𝑥 = −5.

So the process 𝑔 running in reverse is not a function. If we feed it as input the number 𝑦, it can produce

as output more than one 𝑥. This is not allowed in a function.

The problem with function 𝑔 is that it is not one-to-one. So the process g running in reverse is not a

function.

For functions like 𝑓 that are one-to-one, the process f running in reverse will be a function, called the

inverse function.


Day 2 Continued

X Continuing Day 2

Definition of inverse function

• Symbol: 𝑓−1

• Spoken: “𝑓 inverse”

• Usage: 𝑓: 𝐴 → 𝐵 is a one-to-one function

• Informal Meaning: 𝑓−1 the process 𝑓 running in reverse.

• More precise meaning: 𝑓−1 is the function 𝑓−1: 𝐵 → 𝐴 defined as follows: 𝑓−1(𝑦) is the value of

𝑥 that causes 𝑓(𝑥) = 𝑦


• Observation: Notice that 𝑓−1(𝑓(𝑥)) = 𝑥 for all 𝑥 in set 𝐴. That is, (𝑓−1 ∘ 𝑓)(𝑥) = 𝑥

And notice that 𝑓(𝑓−1(𝑦)) = 𝑦 for all 𝑦 in set 𝐵. That is, (𝑓 ∘ 𝑓−1)(𝑦) = 𝑦

Sometimes, we can encounter an inverse function without it being labeled as 𝑓−1.

Here is a more general definition of inverse function

• Words: 𝑓 and 𝑔 are inverses of each other.

• Meaning: 𝑓: 𝐴 → 𝐵 and 𝑔: 𝐵 → 𝐴 are both one-to-one functions and they have these properties:

▪ 𝑔(𝑓(𝑥)) = 𝑥 for all 𝑥 in set 𝐴. That is, (𝑔 ∘ 𝑓)(𝑥) = 𝑥

▪ 𝑓(𝑔−1(𝑦)) = 𝑦 for all 𝑦 in set 𝐵. That is, (𝑓 ∘ 𝑔)(𝑦) = 𝑦


Example #2: Earlier we studied 𝑓(𝑥) =

𝑥−3

𝑥−5 and 𝑔(𝑥) =

5𝑥−3

𝑥−1 and found that

(𝑔 ∘ 𝑓)(𝑥) = 𝑥 for all 𝑥 ≠ 5 (𝑓 ∘ 𝑔)(𝑥) = 𝑥 for all 𝑥 ≠ 1

We see that 𝑓 and 𝑔 are inverses of each other. We could write: 𝑓−1(𝑥) =5𝑥−3

𝑥−1.

Given a formula for a one-to-one function 𝑓, how does one find the formula for the inverse function 𝑓−1?

Given formula for 𝑓, that is, 𝑓(𝑥) = 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑖𝑛𝑣𝑜𝑙𝑣𝑖𝑛𝑔 𝑥

think of it as an equation involving x and y that is solved for y in terms of x.

𝑦 = 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑖𝑛𝑣𝑜𝑙𝑣𝑖𝑛𝑔 𝑥.

Solve this equation for x in terms of y. The result will be 𝑥 = 𝑛𝑒𝑤 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑖𝑛𝑣𝑜𝑙𝑣𝑖𝑛𝑔 𝑦.

Think of this new formula a the inverse function: 𝑓−1(𝑦) = 𝑛𝑒𝑤 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑖𝑛𝑣𝑜𝑙𝑣𝑖𝑛𝑔 𝑦.

Apply this to the earlier example 𝑓(𝑥) =𝑥−3

𝑥−5. Result is 𝑓−1(𝑦) =

5𝑦−3

𝑦−1.

Once we have the formula for the function 𝑓−1, we can use any formula we want.


𝑥 𝑦

𝑓

𝑓−1

an element

of set A

an element

of set B

𝑥 𝑦

𝑓

𝑔

an element

of set A

an element

of set B

Day 2 Continued

X Continuing Day 2

Graphs of inverse functions

Observe that if point (𝑎, 𝑏) is on the graph of 𝑓, it means that 𝑓(𝑎) = 𝑏.

But that means that 𝑓−1(𝑏) = 1, which tells us that point (𝑏, 𝑎) will be on the graph of 𝑓−1.

So the graph of 𝑓−1 will look like the graph of 𝑓 but flipped across the line 𝑦 = 𝑥.

So in general, given a graph of 𝑓, one can build a graph of 𝑓−1 by simply flipping that given graph across

the line 𝑦 = 𝑥. Any point of the form (𝑥, 𝑦) = (𝑎, 𝑏) on the graph of 𝑓 will have a corresponding point

of the form (𝑥, 𝑦) = (𝑏, 𝑎) on the graph of 𝑓−1.

Do a graphical example

.

Day 3


Finish Leftover Topics from Day 2

Section 2.5 Increasing and Decreasing Functions (Book Section 2.5)

Return to idea of Analyzing functions. That is, making observations about them that will help us better

understand them.

In Section 2.5, we will consider two observations that are easy to make about a function when one has a

graph of the function available. (In coming days and weeks, we will consider how the question of how to

make the corresponding observations when one has only a formula for the function, not the graph.) The

two observations are the sign of the function and the increasing/decreasing behavior of the function.

The sign of a function

First, recall very basic stuff

• Terminology used to describe the sign of a number

• positive numbers are numbers that are greater than zero.

• non-negative numbers are numbers that are greater than or equal to zero.

• And of course, negative numbers are numbers that are less than zero.

Intervals

• an interval is a set of numbers of the form 𝐼 = (𝑝, 𝑞) or 𝐼 = [𝑝, 𝑞) or 𝐼 = (𝑝, 𝑞] or 𝐼 = [𝑝, 𝑞]. • Ask students to explain what these symbols mean.

• Ask them what symbols like form 𝐼 = (5, ∞) or 𝐼 = (−∞, 3] would mean.

• Which of these is a valid symbol? (−∞, 3) [−∞, 3) (∞, 3] [−∞, ∞]. • A symbol such as (5, ∞) Denotes the set of x such that 5 < 𝑥. Notice that I don’t write this

inequality as 5 < 𝑥 < ∞. The less than symbol The symbol ∞ does not represent a number. So a

symbol such as or 𝐼 = (5, ∞] or 𝐼 = [5, ∞] makes no sense. That is, the inequality 5 < 𝑥 ≤ ∞

makes no sense because x can never equal infinity. Infinity is not a number.. In other words, in

interval notation, the symbol ∞ must always be next to a parentheses, never a bracket.

Notation for the set of all real numbers.

The clearest thing one can say is the set of all real numbers.

A common symbol is the letter ℝ. This kind of font is called “double-struck”. It is also sometimes called

“blackboard bold”. The term “double-struck” comes from the fact that in the days of typewriters, one

would get a bold letter by typing a letter, backspacing, and typing it again in the same spot. The idea

behind the term “blackboard bold” is that one can’t really do bold letters on a blackboard, so one has to

just draw double lines to give emphasis.

A lot of mathematicians use the symbol (−∞, ∞) to denote the set of all real numbers. They would say

that this represents the set of all real numbers x such that −∞ < 𝑥 < ∞. I personally hate this symbol.

The less than sign < is used to compare two real numbers. For instance, 2 < 5 is true. 5 < 2 is false.

The symbol ∞ does not represent a real number. So a symbol like 𝑥 < ∞ is not really a valid

mathematical symbol. Ditto for the symbol −∞ < 𝑥 < ∞. Just say “all real numbers”.

We will often want to consider the sign behavior of a function 𝑓. In particular, we will consider the sign

of 𝑓 on an interval.

Definition of positive on an interval

• Words: 𝑓 is positive on the interval 𝐼.

• Meaning: at every 𝑥 value in the interval 𝐼, the corresponding 𝑦 value is positive.

• Graphical Description: at every 𝑥 value in the interval 𝐼, the graph of 𝑓 is above the 𝑥 axis.

Note that this is a trivial observation to make about a graph. But again, in coming weeks, we will have to

figure out how to make the same observation when one has only a formula for the function, not the graph


Day 3 Continued


Increasing/Decreasing Behavior of a function

Definition of increasing on an interval

• Words: 𝑓 is increasing on the interval 𝐼.

• Usage: 𝐼 is an interval of the form 𝐼 = (𝑝, 𝑞) or 𝐼 = [𝑝, 𝑞) or 𝐼 = (𝑝, 𝑞] or 𝐼 = [𝑝, 𝑞]. • Meaning: If 𝑝 ≤ 𝑎 < 𝑏 ≤ 𝑞 then 𝑓(𝑎) < 𝑓(𝑏).

• Graphical Description: As one moves from left to right in the interval I, the graph goes up.

This might seem like a simple observation to make. But there is a bit of subtlety here. And again, in

coming weeks, we will have to figure out how to make the same observation when one has only a

formula for the function, not the graph

Example: For the graph shown, answer the questions that follow:

On what intervals is 𝑓(𝑥) positive?

On what intervals is 𝑓(𝑥) negative?

On what intervals is 𝑓(𝑥) decreasing?

On what intervals if 𝑓(𝑥) increaseing?

What are the roots of 𝑓(𝑥)?.


𝑓(𝑥)

𝑥

(-5,0)

(-3,-1)

(-1,0)

(1,2)

(3,0)

x = 5

(6,0)

(8,1)

(10,3)

(12,0)

Day 3 Continued

X Section 2.6 Financial Mathematics

Project onscreen and discuss the reference page about Business Terminology

(You can find it on the main MATH 1350 web page in the calendar and in the list of Exercises)

.

.

Business Terminology

In our course, we will study hypothetical business examples in which a company make and sells some item. The

simplifying assumptions are

• The items are manufactured in batches.

• All of the items manufactured are sold, and they are all sold for the same price per item.

Here is the Business Terminology that we will be using.

Quantity, 𝑞 (small letter), is a variable that represents the number of items made. This sounds simple enough,

but there can be complications. For example, in some problems, 𝑞 represents the number of thousands of items

made. Sometimes the letter 𝑥 will be used instead of 𝑞.

Demand Price, 𝐷(𝑞) or 𝐷(𝑥),is a function. For a given input quantity 𝑞, the output 𝐷(𝑞) is the price that the

items will need to be sold for in order for consumers to be willing to buy 𝑞 of the items. The price is often

denoted by a small letter 𝑝, so we could write 𝑝 = 𝐷(𝑞) or 𝑝 = 𝐷(𝑥). Note that 𝐷(𝑞) will be a decreasing

function. If you don’t want to sell many items, the selling price will need to be high. If you want to sell a lot of

items, the selling price will need to be low. So the graph of 𝐷(𝑞) will go down as one moves from left to right.

Supply Price, 𝑆(𝑞) is a function. For a given input quantity 𝑞, the output 𝑆(𝑞) is the price that the items will

need to be sold for in order for producers to be willing to make 𝑞 of the items. The price is often denoted by a

small letter 𝑝, so we could write 𝑝 = 𝑆(𝑞) or 𝑝 = 𝑆(𝑥). Note that 𝑆(𝑞) will be an increasing function. If you

don’t want to producers to be willing to make many items, the selling price will need to be low. If you want

producers to be willing to make a lot of items, the selling price will need to be high. So the graph of 𝑃(𝑞) will

go up as one moves from left to right.

Since the Demand Price function 𝑝 = 𝐷(𝑞) is decreasing and the Supply Price function 𝑝 = 𝑆(𝑞) is increasing,

if we plot them on the same axes, they will cross at one point. This point is called the Equilibrium Point. Its

coordinates are given the special designation (𝑞0, 𝑝0). The symbols 𝑞0 and 𝑝0 are called the equilibrium

quantity and the equilibrium price. Note that because the Demand Price and Supply Price graphs cross at that

point, it must be true that 𝐷(𝑞0) = 𝑆(𝑞0) = 𝑝0.

Revenue, 𝑅(𝑞) is a function. It is the amount of money that comes into a company from the sale of 𝑞 items.

Because of our simplifying assumptions listed above, we can say that

𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = (𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠 𝑠𝑜𝑙𝑑) ⋅ (𝑠𝑒𝑙𝑙𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 𝑝𝑒𝑟 𝑖𝑡𝑒𝑚) 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 ⋅ 𝐷𝑒𝑚𝑎𝑛𝑑 𝑃𝑟𝑖𝑐𝑒

𝑅(𝑞) = 𝑞 ⋅ 𝐷(𝑞)

Cost, 𝐶(𝑞) (capital letter 𝐶), is a function that gives the cost of making the batch of 𝑞 items.

We say that a company Breaks Even when Revenue = Cost. That is, when 𝑅(𝑞) = 𝐶(𝑞).

Profit, 𝑃(𝑞) (capital letter 𝑃), is a function defined as follows

𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 − 𝐶𝑜𝑠𝑡 𝑃(𝑞) = 𝑅(𝑞) − 𝐶(𝑞)

The expression 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚, denoted by the symbol 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ , means 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦

𝑞.

That is, Average Revenue is �̅�(𝑞) =𝑅(𝑞)

𝑞, Average Cost is 𝐶̅(𝑞) =

𝐶(𝑞)

𝑞, and Average Profit is �̅�(𝑞) =

𝑃(𝑞)

𝑞.

The expression 𝑴𝒂𝒓𝒈𝒊𝒏𝒂𝒍 𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚 means 𝑻𝒉𝒆 𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 𝒐𝒇 𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚.

That is, Marginal Revenue is 𝑅′(𝑥), and Marginal Cost is 𝐶′(𝑥), and Marginal Profit is 𝑃′(𝑥).

The word Marginal can also be put in front of the Average Quantities. That is Marginal Average Revenue is

�̅�′(𝑥), and Marginal Average Cost is 𝐶̅′(𝑥), and Marginal Average Profit is �̅�′(𝑥).

Day 4

X Section 3 Families of functions

Section 3.2 Polynomial Functions

Definition of Polynomial

Function of the form 𝑓(𝑥) = 𝑎𝑛𝑥𝑛 + 𝑎𝑛−1𝑥𝑛−1 + ⋯ + 𝑎2𝑥2 + 𝑎1𝑥 + 𝑎0

where 𝑛 is a non-negative integer. That is, n is an integer greater and 𝑛 ≥ 0

and the symbols 𝑎𝑛, 𝑎𝑛−1, … 𝑎2, 𝑎1, 𝑎0 are all real number constants called coefficients.

Write examples of functions that are or are not polynomials. Include the zero polynomial.

The degree of a polynomial is the number n. That is, it is the largest exponent on the powers of x that

appear in the polynomial. Give examples of degrees

Facts about polynomials and their graphs

even degree poly of degree ≥ 2 with positive leading coefficient graph goes up on both side

etc for other combinationss

a polynomial of degree n that is not the zero polynomial can have up to n roots.

Consequently, the graph of a polynomial of degree n that is not the zero polynomial can have up to n x-

intercepts.

The natural domain of a polynomial is the set of all real numbers x. To understand why, consider the

formula for a polynomial. Given any value of x, nothing can go wrong when computing the value 𝑓(𝑥).

A consequence is that the graph of a polynomial has the property that every vertical line of the form 𝑥 =𝑐 touches the graph. The coordinates of the point, of course, will be (𝑥, 𝑦) = (𝑐, 𝑓(𝑐)).

End behavior of polynomial graphs. Project table showing the end behavior of polynomial graphs.

See book Section 3.2 for examples about this.

Section 3.3 Rational Functions

Definition of Function

a ratio of polynomials where the denominator polynomial is not the zero polynomial.

That is, 𝑓(𝑥) =𝑝(𝑥)

𝑞(𝑥) where 𝑝(𝑥) and 𝑞(𝑥) are polynomials and 𝑞(𝑥) is not the zero polynomial.

Write examples of functions that are or are not rational functions. Include the zero polynomial.

The natural domain of a polynomial is the set of all real numbers x except those that cause the

denominator to be zero. (the roots of the denominator). To understand why, consider the formula for a

polynomial. Given any value of x, nothing can go wrong when computing the value of the numerator,

𝑝(𝑥), and nothing can go wrong when computing the value of the denominator 𝑞(𝑥). But something can

go wrong when computing the value of the ratio 𝑓(𝑥) =𝑝(𝑥)

𝑞(𝑥). If the value of 𝑞(𝑥) is zero, then 𝑓(𝑥) is

undefined. A consequence is that the graph of a polynomial has the property that every vertical line of the

form 𝑥 = 𝑐, where x is a real number that is not a root of the denominator, touches the graph. The

coordinates of the point, of course, will be (𝑥, 𝑦) = (𝑐, 𝑓(𝑐)). But there is a large variety of kinds of

features that can appear in the graphs of rational functions.

See book section 3.3 for examples of graphs of rational functions.

Terminology that the book uses in its examples (and that we will use eventually, too)

Definition of “evaluated at”

• words: 𝑓(𝑥) evaluated at 𝑥 = 𝑐.

• symbol. 𝑓(𝑥)|𝑥=𝑐 or [𝑓(𝑥)]𝑥=𝑐

• meaning: 𝑓(𝑐) That is, the output value that results when the number 𝑥 = 𝑐 is used as input.


Day 4, continued

X Section 3 Families of functions, continued

Section 3.4 Exponential and Logarithmic Functions

Exponential Functions

Definition of Exponential function with base b:

A function of the form 𝑓(𝑥) = 𝑏(𝑥) where b is a number such that 𝑏 > 0 and 𝑏 ≠ 1.

The number b is called the base.

Special base: the number e

Called euler’s number

e is a real number, with 2 < 𝑒 < 3

A little more precisely, 𝑒 ≈ 2.718

But e is an irrational number

the value of e cannot be given exactly by a fraction, or a terminating decimal, or even a repeating decimal

The only way to write e exactly is to just write e.

The exponential function 𝑦 = 𝑒𝑥 is called the natural exponential function.

For the functions 𝑦 = 2(𝑥) and 𝑦 = 𝑒(𝑥) and 𝑦 = 3(𝑥), make a table of y values for x = -3,-2,0,1,2,3,

then graph. Put column for 𝑦 = 𝑒(𝑥) between the other two. We can only write y values for 𝑦 = 𝑒(𝑥) in

symbols, not decimals.

Observe common traits of Exponential function 𝑦 = 𝑏(𝑥) with base 𝑏 > 1:

• Domain is all real numbers x

• Range is all y > 0

• Function is one-to-one

• Graph goes through the point (𝑥, 𝑦) = (0,1) because 𝑏0 = 1.

• Graph goes through the point (𝑥, 𝑦) = (1, 𝑏) because 𝑏1 = 𝑏.

• Graph is increasing

• Graph has a horizontal asymptote on the left with line equation 𝑦 = 0

Do the same thing for 𝑦 = (

1

2)

(𝑥)

Observe common traits of Exponential function 𝑦 = 𝑏(𝑥) with base 0 < 𝑏 < 1:

Same traits as before except

• Graph is decreasing

• Graph has a horizontal asymptote on the right with line equation 𝑦 = 0


Day 4, continued

X Section 3 Families of functions, continued

Section 3.4 continued Logarithmic functions

Recall that for any base the exponential function 𝑦 = 𝑏(𝑥) is one-to-one. So we know how its inverse

function should look.

Make new sketch of 𝑦 = 2(𝑥), with (𝑥, 𝑦) coordinates of key points labeled.

Make new graph with keys point obtained by interchanging 𝑥, 𝑦 of those old key points.

The new graph is the graph of the inverse function for 𝑦 = 2(𝑥).

It is called 𝑦 = log2(𝑥)

Definition of Logarithmic function

• Words: The base b logarithm function

• Symbol: 𝑦 = log𝑏(𝑥)

• Meaning: the function that is the inverse function for the base b exponential function, 𝑦 = 𝑏(𝑥).

That is, to say that 𝑝 = log𝑏(𝑞) means that 𝑞 = 𝑏(𝑝). Graphically, this would mean that the graph

of 𝑦 = log𝑏(𝑥) contains point (𝑞, 𝑝) whenever the graph of 𝑦 = 𝑏(𝑥) contains point (𝑝, 𝑞)

• Additional terminology and notation: The base e logarithm is called the natural logarithm

function. Its full symbol would be 𝑦 = log𝑒(𝑥). A common shorter form is 𝑦 = ln(𝑥).

• Danger: The symbol 𝑦 = log(𝑥) is commonly used in two different ways.

• Some books use the symbol 𝑦 = log(𝑥) as an abbreviation for 𝑦 = log10(𝑥).

• Some books use the symbol 𝑦 = log(𝑥) as an abbreviation for 𝑦 = log𝑒(𝑥), the natural

logarithm.

• For that reason, I will avoid using this notation in class. But it also means that you will always

have to think carefully when you encounter the symbol 𝑦 = log(𝑥) in someone else’s work.

You will have to make sure that you know how they are using the symbol.

Observe common traits of Logarithmic function 𝑦 = log𝑏(𝑥) with base 𝑏 > 1:

• Domain is all positive real numbers. That is, all x > 0

• Range is all real numbers y

• Function is one-to-one

• Graph goes through the point (𝑥, 𝑦) = (1,0). This tells us that 𝑦 = log𝑏(1) = 0.

• Graph goes through the point (𝑥, 𝑦) = (1, 𝑏). This tells us that 𝑦 = log𝑏(𝑏) = 1.

• Graph is increasing

• Graph has a vertical asymptote with line equation 𝑥 = 0.

Examples

Note that the equations 𝑒(ln(𝑥)) = 𝑥 and ln(𝑒(𝑥)) = 𝑥 are true because 𝑦 = 𝑒(𝑥) and 𝑦 = ln(𝑥) are

inverses of one another. These equations are called the inverse relations.

Example of a computation involving logarithms

Find 𝑦 = log2(32).

Solution Two methods

Method 1: Realize that the equation 𝑦 = log2(32) means the same thing as the equation 2(𝑦) = 32. We

are looking for the number y that makes this equation true. It must be 𝑦 = 5. That is, log2(32) = 5.

Method 2: This is a trick. Notice that 32 = 25. So we can replace 32 by 25 in the expression. This will

allow us to then use the inverse relation. 𝑦 = log2(32) = log2(2(5)) = 5.

See book for more examples and for properties of exponents and logarithms

.

Day 5

X Today: Section 4.2 Limits: Graphical Approach

The Definition of limit

• symbol: lim𝑥→𝑎

𝑓(𝑥) = 𝐿

• spoken: “The limit, as x approaches a, of f(x) is L.”

• usage: 𝑥 is a variable, 𝑓 is a function, 𝑎 is a real number, and 𝐿 is a real number.

• meaning: as 𝑥 gets closer & closer to 𝑎, but not equal to 𝑎, the value of 𝑓(𝑥) gets closer & closer

to 𝐿 (may actually equal 𝐿).

• Leave blank line to be filled in later

today we will do graphical approach

Examples of graph ➔ description of limit behavior

Work on Printed Activity Worksheet about Limits.

(On next page of these notes. Can also project from MATH 1350 web page)

do row 𝑥 = 1.

Fill in the blank line in the definition of limit:

• graph behavior: The graph of 𝑓 appears to be heading for location (𝑥, 𝑦) = (𝑎, 𝐿).

Do row 𝑥 = 4

Point out the difference between the idea of the existence of a 𝑦-value at 𝑥 = 𝑎 and the existence of the

limit as 𝑥 → 𝑎.

do row 𝑥 = −1.

Define one-sided limits

The Definition of one-sided limit

• symbol: lim𝑥→𝑎−

𝑓(𝑥) = 𝐿

• spoken: “The limit, as 𝑥 approaches 𝑎 from the left, of 𝑓(𝑥) is 𝐿.”

• meaning: as 𝑥 gets closer & closer to 𝑎, but less than 𝑎, the value of 𝑓(𝑥) gets closer & closer to

𝐿 (may actually equal 𝐿).

• graph behavior: The graph of 𝑓 appears to be heading for location (𝑥, 𝑦) = (𝑎, 𝐿) from the left.

There is an analogous definition for the one-sided limit from the right.

Re-cast the definition of limit using 3-part test involving one-sided limits


𝑓(𝑥) = 𝐿


• meaning: 𝑓(𝑥) passes this three-part test:

• The one-sided limit from the left, lim𝑥→𝑎−

𝑓(𝑥) exists.

• The one-sided limit from the right, lim𝑥→𝑎+

𝑓(𝑥) exists.

• The left and right one-sided limits have the same value and that value is L. That is,

lim𝑥→𝑎−

𝑓(𝑥) = 𝐿 = lim𝑥→𝑎+

𝑓(𝑥)


finish row 𝑥 = −1.

do row 𝑥 = −3, and other rows if time permits

Now: Example of description of limit behavior ➔ graph

Example: Sketch a graph that satisfies all these conditions:

𝑓(1) = 3 lim

𝑥→1−𝑓(𝑥) = 2

lim𝑥→1+

𝑓(𝑥) = −4

. .

Activity for Day 5: Limits for a Function Given by a Graph (Section 4.2)

Use the graph to fill in the table. (Extra copies of the graph are on back.)

x-value limit from left limit from right limit y-value

−5 lim𝑥→−5−

𝑓(𝑥) = lim𝑥→−5+

𝑓(𝑥) = lim𝑥→−5

𝑓(𝑥) = 𝑓(−5) =

−3 lim𝑥→−3−



𝑓(𝑥) = 𝑓(−3) =

−1 lim𝑥→−1−



𝑓(𝑥) = 𝑓(−1) =

1 lim𝑥→1−

𝑓(𝑥) = lim𝑥→1+

𝑓(𝑥) = lim𝑥→1

𝑓(𝑥) = 𝑓(1) =

2 lim𝑥→2−



𝑓(𝑥) = 𝑓(2) =

4 lim𝑥→4−



𝑓(𝑥) = 𝑓(4) =

6 lim𝑥→6−



𝑓(𝑥) = 𝑓(6) =

𝑥

𝑓(𝑥)

.

𝑥

𝑓(𝑥)

𝑥

𝑓(𝑥)

𝑥

𝑓(𝑥)

Day 6

X Section 5 Limits of a Function Defined by a Formula

Recall Definition of Limit from Section 4.2

The Definition of limit


𝑓(𝑥) = 𝐿


• usage: 𝑥 is a variable, 𝑓 is a function, 𝑎 is a real number, and 𝐿 is a real number.




In the previous lecture, we considered limits for a function defined by a graph. In the examples from that

lecture, we used the graph behavior part of the definition of limit as our tool.

And in those examples, we also considered the values of the function. Remember from Lecture 1:

input 𝑎 ∈ 𝐴 produces an ouput 𝑓(𝑎) ↔ there is dot on the graph at location (𝑥, 𝑦) = (𝑎, 𝑓(𝑎)).

For example, consider the graph below:

• We say that lim

𝑥→2𝑓(𝑥) = 3 because the graph appears to be heading for the location (𝑥, 𝑦) = (2,3).

• We say that 𝑓(2) = 3 because there is a point on the graph at the location (𝑥, 𝑦) = (2,3).

• We say that lim𝑥→3

𝑓(𝑥) = 4 because the graph appears to be heading for the location (𝑥, 𝑦) = (3,4).

• We say that 𝑓(3)𝐷𝑁𝐸 because there is no point on the graph with 𝑥 = 3.

So finding the values of 𝑓(𝑎) and lim𝑥→𝑎

𝑓(𝑥) is easy when looking at the graph of a function.

But how do we find the values of 𝑓(𝑎) and lim𝑥→𝑎

𝑓(𝑥) when we only have a formula for 𝑓(𝑥), not a graph?

Of course, computing the value of 𝑓(𝑎) is easy if one has a formula for 𝑓(𝑥). Simply substitute 𝑥 = 𝑎

into the formula and compute the result. If something goes wrong and that computation cannot produce a

numerical result, then you say that 𝑓(𝑎) does not exist. That is, 𝑓(𝑎) 𝐷𝑁𝐸.

But given a formula for a function 𝑓(𝑥), determining that lim𝑥→𝑎

𝑓(𝑥) = 𝐿 using the definition of the limit,



is actually a very sophisticated process, one that is beyond the level of this course, even beyond the level

of the more difficult course MATH 2301.

How, then, are we supposed to find lim𝑥→𝑎

𝑓(𝑥) = 𝐿 in the case where we only have a formula for 𝑓(𝑥)??


𝑓(𝑥)

𝑥

Day 6, continued


Continuous Functions

It turns out for some special functions, the value of the limit lim𝑥→𝑎

𝑓(𝑥) will turn out to be the same as the

y-value of the function, 𝑦 = 𝑓(𝑎). In these special functions, we can compute the value of the limit by

simply computing the y-value of the function. These special functions are called continuous functions

Definition of a continuous function

• words: the function 𝑓 is continuous at 𝑥 = 𝑐.

• meaning: the function passes this three part test

1. the limit of the function 𝑓 exists at 𝑥 = 𝑐. That is, lim𝑥→𝑐

𝑓(𝑥) exists

2. the y-value of the function 𝑓 exists at 𝑥 = 𝑐. That is, 𝑓(𝑐)

3. The value of the limit equals the y-value. That is, lim𝑥→𝑐

𝑓(𝑥) = 𝑓(𝑐).

• graphical significance: The graph of 𝑓 has no breaks or jumps at 𝑥 = 𝑐.

• additional terminology:

o words: function 𝑓 is continuous on an interval (𝑎, 𝑏) at 𝑥 = 𝑐.

o meaning: 𝑓 is continuous at 𝑥 = 𝑐 for all 𝑎 < 𝑐 < 𝑏.

• For additional terminology of left continuity and right continuity, see the book Section 5.2

So if we know that a function 𝑓 is continuous at some 𝑥 = 𝑐, then computing the value of lim𝑥→𝑐

𝑓(𝑥) is

easy: we simply find the 𝑦-value 𝑓(𝑐), and then use the fact that 𝑓 is continuous to say lim𝑥→𝑐

𝑓(𝑥) = 𝑓(𝑐).

But how will we know if a function 𝑓 is continuous at 𝑥 = 𝑐? Proving that a function 𝑓 is continuous at

𝑥 = 𝑐 is a very sophisticated process, one that is beyond the level of this course. For this course, we

simply use theorems (that somebody else has proved) that state that certain famous functions are

continuous

Project onscreen from web page showing exercises: Theorems about Limits

Theorem 1: Certain basic famous functions that are continous everywhere on their domains

• Power functions of the form 𝑦 = 𝑥𝑛 where 𝑛 is a non-negative integer. (their domain is all real

numbers)

• Exponential Functions (their domain is all real numbers)

• Logarithmic Functions (their domain is all positive real numbers)

• Power functions of the form 𝑦 = 𝑥𝑛 where 𝑛 is a non-negative integer. (domain is all real numbers)

Theorem 1 continued: Certain Famous functions that are continous everywhere on their domains

More general Power Functions of the form 𝑦 = 𝑥𝑝/𝑞 are actually tricky.

• If the exponent is 𝑝

𝑞 is is a positive rational number in reduced form and the denominator 𝑞 is an odd

integer, then the function 𝑦 = 𝑥𝑝/𝑞 will have domain all real numbers. For example √𝑥3

means 𝑥1/3.

This has domain all real numbers. Observe: √83

= 2 and √03

= 0 and √−83

= −2.

• But if the exponent is 𝑝

𝑞 is a positive rational number in reduced form and the denominator 𝑞 is an

even integer, then the function 𝑦 = 𝑥𝑝/𝑞 will have domain all non-negative real numbers. For

example √𝑥 means 𝑥1/2. This has domain all non-negative real numbers. Observe: √4 = 2 and

√0 = 0 but √−4 𝐷𝑁𝐸.


𝑞 is is a negative rational number, then the domain of 𝑦 = 𝑥𝑝/𝑞 can’t include 𝑥 =

0. For example 𝑦 = 𝑥−1 means 𝑦 =1

𝑥. The domain of this function is all real numbers 𝑥 ≠ 0.

• For example 𝑦 = 𝑥−1/3 means 𝑦 =1

√𝑥3 . The domain of this function is all real numbers 𝑥 ≠ 0.


√𝑥. The domain of this function is all real numbers 𝑥 > 0.


Day 6, continued


Theorem 1 allows us to compute the values of limits of those basic functions. Give examples

Some functions are not as simple as the basic functions, so we can’t find their limits using Theorem 1.

But in many cases, a complicated function is built from simpler functions whose limits are known.

Project onscreen from web page showing exercises: Limit Laws

Theorem 2 Limit Laws

• Constant Multiple Rule: if 𝑘 is a constant, then lim𝑥→𝑐

𝑘𝑓(𝑥) = 𝑘 lim𝑥→𝑐

𝑓(𝑥)

• Sum Rule: lim𝑥→𝑐

[𝑓(𝑥) + 𝑔(𝑥)] = lim𝑥→𝑐

𝑓(𝑥) + lim𝑥→𝑐

𝑔(𝑥)

• Product Law: lim𝑥→𝑐

(𝑓(𝑥) ⋅ 𝑔(𝑥)) = (lim𝑥→𝑐

𝑓(𝑥)) ⋅ (lim𝑥→𝑐

𝑔(𝑥))

• Quotient Rule: lim𝑥→𝑐

𝑓(𝑥)

𝑔(𝑥)=

lim𝑥→𝑐

𝑓(𝑥)

lim𝑥→𝑐

𝑔(𝑥)as long as lim

𝑥→𝑐𝑔(𝑥) ≠ 0.

• Composition Limit Law: If lim𝑥→𝑐

𝑔(𝑥) = 𝐿 and 𝑓(𝑥) is continuous at 𝑥 = 𝐿,

then lim𝑥→𝑐

𝑓(𝑔(𝑥)) = 𝑓 (lim𝑥→𝑐

𝑔(𝑥)) = 𝑓(𝐿)

.

Special Cases that can be addressed using Theorem 1 and Theorem 2

• Polynomial Functions

• Rational Functions

• Nested Functions (Composition of Functions)

Examples Find the following limits using Theorem 1 and Theorem 2

(a) lim𝑥→−3

−5𝑥2 + 7𝑥 + 13

(b) lim𝑥→−3

𝑥

𝑥+5

(c) lim𝑥→3

√𝑥3 − 2

Examples that cannot be addressed using Theorem 1 and Theorem 2

Example 1: lim𝑥→3

𝑥2−2𝑥−3

𝑥−3 can’t be computed using those theorems because the limits of numerator and

denominator are both zero. However, observe that 𝑦 =𝑥2−2𝑥−3

𝑥−3=

(𝑥+1)(𝑥−3)

(𝑥−3) is the formula for the graph

shown at the beginning of the hour. From the graph, we can see that the graph appears to be heading for

the location (𝑥, 𝑦) = (3,4). This tells us that the value of the limit should be lim𝑥→3

𝑥2−2𝑥−3

𝑥−3= 4. But we are

not able to figure that out from Theorems 1 and 2.

Example 1: lim𝑥→3

𝑥+1

𝑥−3 can’t be computed using those theorems because the limit of the denominator is zero.

It turns out that this limit does not exist using the definition of limit that we have so far. But that fact is a

theorem:

Project on the board Theorem about certain limits that do not exist

Theorem 3: Certain limits that do not exist in the framework of Sections 4 and 5 of the textbook

If the lim𝑥→𝑐

𝑓(𝑥) = 𝑀 ≠ 0 and lim𝑥→𝑐

𝑔(𝑥) = 0, then the limit of the quotient lim𝑥→𝑐

𝑓(𝑥)

𝑔(𝑥) 𝐷𝑁𝐸.

That is, there is no number L such that the values of 𝑓(𝑥)

𝑔(𝑥) get closer and closer to L as x gets closer and

closer to c.

(Remark:In later sections of the book, we will expand the definition of limit. In those later section, some

limits of this form will exist.)

Content that will be projected onscreen on Day 6 is shown on the next 2 pages ➔

Theorems about Limits From Sections 5.2 and 5.3 of the Textbook

Theorem 1: Certain basic famous functions that are continous

everywhere on their domains

• Power functions of the form 𝑦 = 𝑥𝑛 where 𝑛 is a non-negative integer.

(their domain is all real numbers)

• Exponential Functions (their domain is all real numbers)

• Logarithmic Functions (their domain is all positive real numbers)

• Power functions of the form 𝑦 = 𝑥𝑛 where 𝑛 is a non-negative integer.

(domain is all real numbers)

More general Power Functions of the form 𝑦 = 𝑥𝑝/𝑞 are actually tricky.


𝑞 is is a positive rational number in reduced form and

the denominator 𝑞 is an odd integer, then the function 𝑦 = 𝑥𝑝/𝑞 will

have domain all real numbers. For example √𝑥3

means 𝑥1/3. This has

domain all real numbers. Observe: √83

= 2 and √03

= 0 and √−83

=−2.

• But if the exponent is 𝑝

𝑞 is a positive rational number in reduced form

and the denominator 𝑞 is an even integer, then the function 𝑦 = 𝑥𝑝/𝑞

will have domain all non-negative real numbers. For example √𝑥

means 𝑥1/2. This has domain all non-negative real numbers. Observe:

√4 = 2 and √0 = 0 but √−4 𝐷𝑁𝐸.


𝑞 is is a negative rational number, then the domain of

𝑦 = 𝑥𝑝/𝑞 can’t include 𝑥 = 0. For example 𝑦 = 𝑥−1 means 𝑦 =1

𝑥. The

domain of this function is all real numbers 𝑥 ≠ 0.


√𝑥3 . The domain of this function is

all real numbers 𝑥 ≠ 0. For example 𝑦 = 𝑥−1/2 means 𝑦 =1

√𝑥. The

domain of this function is all real numbers 𝑥 > 0.

Theorem 2: Limit Laws

• Constant Multiple Rule: if 𝑘 is a constant,

then lim𝑥→𝑐

𝑘𝑓(𝑥) = 𝑘 lim𝑥→𝑐

𝑓(𝑥)

• Sum Rule: lim𝑥→𝑐

[𝑓(𝑥) + 𝑔(𝑥)] = lim𝑥→𝑐

𝑓(𝑥) + lim𝑥→𝑐

𝑔(𝑥)

• Product Law: lim𝑥→𝑐

(𝑓(𝑥) ⋅ 𝑔(𝑥)) = (lim𝑥→𝑐

𝑓(𝑥)) ⋅ (lim𝑥→𝑐

𝑔(𝑥))

• Quotient Rule: lim𝑥→𝑐

𝑓(𝑥)

𝑔(𝑥)=

lim𝑥→𝑐

𝑓(𝑥)

lim𝑥→𝑐

𝑔(𝑥) as long as lim

𝑥→𝑐𝑔(𝑥) ≠ 0.

• Composition Limit Law:

If lim𝑥→𝑐

𝑔(𝑥) = 𝐿 and 𝑓(𝑥) is continuous at 𝑥 = 𝐿,

then lim𝑥→𝑐

𝑓(𝑔(𝑥)) = 𝑓 (lim𝑥→𝑐

𝑔(𝑥)) = 𝑓(𝐿)

Special Cases that can be addressed using Theorems 1 and 2:

• Polynomial Functions

• Rational Functions

• Nested Functions (compositions of functions)

Theorem about Certain Limits that Don’t Exist in the Framework of

Sections 5.2 and 5.3 of the Textbook (This theorem should be in the

book, but it is not.)

If the lim𝑥→𝑐

𝑓(𝑥) = 𝑀 ≠ 0 and lim𝑥→𝑐

𝑔(𝑥) = 0, then the limit of the

quotient lim𝑥→𝑐

𝑓(𝑥)

𝑔(𝑥) 𝐷𝑁𝐸.

That is, there is no number 𝐿 such that the values of 𝑓(𝑥)

𝑔(𝑥) get closer

and closer to 𝐿 as 𝑥 gets closer and closer to 𝑐.

(Remark:In later sections of the book, we will expand the definition

of limit. In those later section, some limits of this form will exist.)

Day 7

X Section 6.2 Limits of the form zero over zero

Recall the functions 𝑓(𝑥) =𝑥2−2𝑥−3

𝑥−3=

(𝑥+1)(𝑥−3)

(𝑥−3) and 𝑔(𝑥) = 𝑥 + 1 from lecture 1.

We discussed that they are not the same function. They don’t have the same domain.

Observe: that 𝑓(3)𝐷𝑁𝐸 because cannot divide 0

0. But 𝑔(3) = 4.

But when we made a table of values for the two functions, and then graphed them, we found that the

graphs looked very similar.

Graph of 𝑓 has a hole at (𝑥, 𝑦) = (3,4); graph of 𝑔 has a point at (𝑥, 𝑦) = (3,4);

What about the limit as x approaches 3?

Using our graphical interpretation of the limit, we would say:

The lim𝑥→3

𝑓(𝑥) = 4 because the graph of 𝑓 appears to be heading for the location (𝑥, 𝑦) = (3,4).

But can we find this limit analytically, using the formula for 𝑓(𝑥)?

Here is an incorrect computation. (Do this work in red, and when it is done, cross it out in red!)

lim𝑥→3


𝑥2 − 2𝑥 − 3

𝑥 − 3=

lim𝑥→3

𝑥2 − 2𝑥 − 3

lim𝑥→3

𝑥 − 3=

0

0 𝐷𝑁𝐸

This says lim𝑥→3

𝑓(𝑥) does not exists.

Is this correct?

It is certainly true that 0

0 DNE.

But remember that we already know that lim𝑥→3

𝑓(𝑥) = 4, so the answer DNE is definitely wrong!

What went wrong?

The problem is in the second equal sign. The rule for the limit of a quotient says that that step is allowed

only if the limit of the denominator is not zero. So in the next step, when we realized that the limit of the

denominator is zero, what we have to conclude is not that the limit does not exist, but rather that we can’t

do the limit this way. We have to find some other way.


𝑓(𝑥)

𝑥

𝑔(𝑥)

𝑥

Day 7, continued

X Section 6.2 Limits of the form zero over zero

Definition: of indeterminate form

A limit of the form lim𝑥→𝑎

𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑥)

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑥) where lim

𝑥→𝑎𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑥) = 0 and lim

𝑥→𝑎𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑥) = 0 is

called an indeterminate form 0

0.

A limit of this form cannot be determined by using the rule for the limit of quotients.

That is, one cannot say lim𝑥→𝑎


𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑥)=

lim𝑥→𝑎


lim𝑥→𝑎

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑥) because the rule for quotients does not

apply in any situation where lim𝑥→𝑎

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑥) = 0.

One must find some other way to find the limit of an indeterminate form.

How are we supposed to find the limit of an indeterminate form?

Example of calculation of a limit for an indeterminate form.

We will find lim𝑥→3


𝑥2−2𝑥−3

𝑥−3

First of all, one must use the factored form of the function, rather than the standard form. So the right

hand side should be replaced: lim𝑥→3

𝑥2−2𝑥−3

𝑥−3= lim

𝑥→3

(𝑥+1)(𝑥−3)

(𝑥−3)

Next, observe that the symbol 𝑥 → 3 tells us that 𝑥 is close to 3 but 𝑥 ≠ 3.

This will mean that 𝑥 − 3 ≠ 0, so (𝑥−3)

(𝑥−3) is not

0

0, so we can can cancel

(𝑥−3)

(𝑥−3) inside the limit.

That is, the right hand side can be replaced: lim𝑥→3

(𝑥+1)(𝑥−3)

(𝑥−3)= lim

𝑥→3(𝑥 − 1)

It is important to read and write the above equation correctly.

The limit equation does not say that (𝑥+1)(𝑥−3)

(𝑥−3)= (𝑥 + 1). Indeed that equation without the limit symbols

is not true! The left side of that equal sign is a function with domain all 𝑥 ≠ 3, while the right side is a

function with domain all real numbers. They are not the same function, so they cannot be said to be

equal. The equation without the limit symbol is incorrect. The correct equation has the limit symbol.

Now observe that the right had side of the equation is a the limit of a polynomial. We know from

Theorem 2 (project it) that because polynomials are continuous, lim𝑥→𝑎

𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙(𝑥) = 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙(𝑎).

In our case, this means lim𝑥→3

(𝑥 − 1) = ((3) + 1) = 4

Put the whole calculation together with explanations written off to the side:

lim𝑥→3


𝑥2 − 2𝑥 − 3

𝑥 − 3

= lim𝑥→3

(𝑥 + 1)(𝑥 − 3)

(𝑥 − 3) (𝑢𝑠𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑓𝑜𝑟𝑚)

= lim𝑥→3

(𝑥 − 1) (𝑠𝑖𝑛𝑐𝑒 𝑥 → 3, 𝑤𝑒 𝑘𝑛𝑜𝑤 𝑥 ≠ 3, 𝑠𝑜 𝑥 − 3 ≠ 0, 𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑐𝑎𝑛𝑐𝑒𝑙)

= ((3) + 1) (𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙: 𝑇ℎ𝑚 2 𝑠𝑎𝑦𝑠 𝑡ℎ𝑎𝑡 𝑤𝑒 𝑐𝑎𝑛 𝑠𝑢𝑏 𝑖𝑛 𝑥 = 3)

= 4

.

Observation: The limit expression appears in front of each expression until the step where we substitute

in 𝑥 = 3. At that step, the limit has been done. Everything after that step is just arithmetic.

common mistake: lim𝑥→3


(𝑥−3)(𝑥+1)

(𝑥−3)=

(3−3)(3+1)

(3−3)=

0

0𝐷𝑁𝐸 (one mistake on this line.)

another: lim𝑥→3


(𝑥−3)(𝑥+1)

(𝑥−3)=

(3−3)(3+1)

(3−3)= (3 + 1) = 4 (two mistakes on this line!)

Another observation: Can’t cancel 0/0 ever. That is why 𝑓(3)𝐷𝑁𝐸. But when computing the limit, we

can cancel factors (𝑥−3)

(𝑥−3) because they are NOT 0/0. That’s how limit can exist while y value doesn’t.


Day 7, continued

X Section 6.2 Continued

Example of a similar-looking limit that is NOT an indeterminate form.

Let 𝑓(𝑥) =𝑥2+4𝑥+3

𝑥−3=

(𝑥+1)(𝑥+3)

(𝑥−3). What is the lim

𝑥→3𝑓(𝑥) = lim

𝑥→3

𝑥2+4𝑥+3

𝑥−3= lim

𝑥→3

(𝑥+1)(𝑥+3)

(𝑥−3)?

Observe:

• The limit of the numerator is lim𝑥→3

𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑥) = lim𝑥→3

𝑥2 + 4𝑥 + 3 = (3)2 + 4(3) + 3 = 24.

• The limit of the denominator is lim𝑥→3

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑥) = lim𝑥→3

𝑥 − 3 = (3) − 3 = 0.

Since lim𝑥→3

𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑥) = 24 ≠ 0 and lim𝑥→3

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑥) = 0, this is not an indeterminate form.

In fact, Theorem 4 (project onscreen) tells us that this limit does not exist!

But what does that mean?

• It means that there is not a number 𝐿 such that lim𝑥→3

𝑓(𝑥) = 𝐿.

• For the graph, it means that there is not a location (𝑥, 𝑦) = (3, 𝐿) that the graph is heading for.

.

Another example of calculating a limit of an indeterminate form. Find the following limit:

limℎ→0

15 + ℎ

−15

ℎ

Observe this is an indeterminate form, because limℎ→0

𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(ℎ) = 0 and limℎ→0

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(ℎ) = 0.

That is, we cannot simply substitute in ℎ = 0. Must do something else.

Here are the steps.

limℎ→0

15 + ℎ

−15

ℎ= lim

ℎ→0

1

ℎ(

1

5 + ℎ−

1

5) (𝑟𝑒𝑤𝑟𝑖𝑡𝑒 𝑤𝑖𝑡ℎ 𝑠𝑖𝑛𝑔𝑙𝑒 𝑏𝑎𝑠𝑒𝑙𝑖𝑛𝑒 𝑓𝑜𝑟 𝑐𝑙𝑎𝑟𝑖𝑡𝑦)

= limℎ→0

1

ℎ(

1

(5 + ℎ)

(5)

(5)−

1

5

(5 + ℎ)

(5 + ℎ)) (𝑔𝑒𝑡 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟)

= limℎ→0

1

ℎ(

5 − (5 + ℎ)

(5 + ℎ)(5)) (𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦)

= limℎ→0

1

ℎ(

−ℎ

25 + 5ℎ) (𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦 𝑚𝑜𝑟𝑒)

= limℎ→0

ℎ

ℎ(

−1

25 + 5ℎ) (𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦 𝑚𝑜𝑟𝑒)

= limℎ→0

(−1

25 + 5ℎ) (𝑠𝑖𝑛𝑐𝑒 ℎ → 0, 𝑤𝑒 𝑘𝑛𝑜𝑤 ℎ ≠ 3, 𝑠𝑜 𝑤𝑒 𝑐𝑎𝑛 𝑐𝑎𝑛𝑐𝑒𝑙)

=−1

25 + 5(0) (𝑛𝑜𝑡 𝑎𝑛 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑡𝑒 𝑓𝑜𝑟𝑚. 𝑇ℎ𝑚 2 𝑠𝑎𝑦𝑠 𝑤𝑒 𝑐𝑎𝑛 𝑠𝑢𝑏 ℎ = 0)

=−1

25 (𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦)

Observation: The limit expression appears in front of each expression until the step where we substitute

in ℎ = 0. At that step, the limit has been done. Everything after that step is just arithmetic.

Conclusion of the day: The limit of an indeterminate form is not 0

0𝐷𝑁𝐸. Must use some other method!

.

Day 8

X Section 6.3 Limits of the form nonzero over zero

Review of last few days’ discussion of limits

In Section 4, we introduced the definition of limits

In Section 5, we discussed continuous functions, For those functions, the value of the limit is just the

same as the value of the y-coordinate. lim𝑥→𝑐

𝑓(𝑥) = 𝑓(𝑐). The Limit Laws are basically about limits of

continuous functions.

In Section 6.2, we discussed limits of the form zero over zero, the so-called indeterminate forms.

These are limits lim𝑥→𝑐

𝑓(𝑥) of functions that are not continuous at 𝑥 = 𝑐, but rather

𝑓(𝑥) =𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑥)

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑥) 𝑤ℎ𝑒𝑟𝑒 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑐) = 0 𝑎𝑛𝑑 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑐) = 0

For these, if we (incorrectly) tried to compute lim𝑥→𝑐

𝑓(𝑥) by using the rule for quotients

lim𝑥→𝑎



lim𝑥→𝑎


lim𝑥→𝑎


𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟(𝑐)

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟(𝑐)=

0

0𝐷𝑁𝐸

But the rule for quotients does not apply in the case, so the entire calculation is invalid.

So our Limit Laws from Section 5 (which are designed for continuous functions) don’t tell us anything

about what the value of the limit will be. That’s why the limit form is called “indeterminate form”.

In Section 6.2, we developed other techniques for finding the value of limits that are indeterminate forms.

But if a limit had the form 𝑛𝑜𝑛𝑧𝑒𝑟𝑜

0, the limit was just declared to not exist by Special Theorem.

In Section 6.3 Limits of the form nonzero over zero, we will reexamine these limits of the form 𝑛𝑜𝑛𝑧𝑒𝑟𝑜

0.

We will expand our definition of limit, using the terminology and notation of infinity.

Introduce Limits Involving Infinity

Define infinite limits: symbol: lim𝑥→𝑐

𝑓(𝑥) = ∞

spoken: The limit, as 𝑥 approaches 𝑐, of 𝑓(𝑥), is infinity.

Meaning: as 𝑥 gets closer & closer to 𝑐 but not equal to 𝑐, the 𝑦-values get more & more positive

without bound.

Graphical Significance: Graph of 𝑓 has a vertical asymptote at 𝑥 = 𝑐 and the graph goes up on both

sides of the asymptote. Note that the line equation for the asymptote is 𝑥 = 𝑐.

Obvious variations: lim𝑥→𝑐+

𝑓(𝑥) = ∞ or lim𝑥→𝑐

𝑓(𝑥) = −∞, etc. Ask the students to explain.

See Activity for Day 8: Graph of a function with infinite limits → description of limit behavior

(available at a link on the page of exercises, and is also included at the end of this day’s lecture notes.)


Day 8, continued

X Section 6.3 continued

Limits Involving Infinity for a function given by a formula

Example 1 involving 𝑓(𝑥) =1

𝑥−6

make table of values for 𝑓(𝑥) =1

𝑥−6 for integer x values to the right of 6, including some numbers getting

very close to 6.

make table of values for 𝑓(𝑥) =1

𝑥−6 for integer x values to the left of 6, including some numbers getting

very close to 6.

Observe that dividing the number 1 by a positive number close to 0 results in a very large positive y

value. Observe that when x gets closer & closer to 6 from the right, the y values get more & more

positive without bound. Abbreviate with limit notation: lim𝑥→6+


1

𝑥−6= ∞.

Similarly Observe that dividing the number 1 by a negative number close to 0 results in a very large

negative y value. Observe that when x gets closer & closer to 6 from the negative, the y values get more

& more negative without bound. Abbreviate with limit notation: lim𝑥→6−

𝑓(𝑥) = lim𝑥→6−

1

𝑥−6= −∞.

Since the left and right limits don’t match, we say lim𝑥→6

𝑓(𝑥) 𝐷𝑁𝐸.

Observe what we don’t do:

We don’t do this:

• Right limit: lim𝑥→6+


1

𝑥−6=

1

6−6=

1

0 𝐷𝑁𝐸

• Left limit: lim𝑥→6−


1

𝑥−6=

1

6−6=

1

0 𝐷𝑁𝐸

• Limit: lim𝑥→6


1

𝑥−6=

1

6−6=

1

0 𝐷𝑁𝐸

And we don’t do this:

• Right limit: lim𝑥→6+


1

𝑥−6=

1

6−6=

1

0= ∞

• Left limit: lim𝑥→6−


1

𝑥−6=

1

6−6=

1

0= ∞



1

𝑥−6=

1

6−6=

1

0= ∞

.

Example 2 involving 𝑓(𝑥) =𝑥−2

𝑥−6. Find the following:

Study behavior at 𝑥 = 2:

• Function value: 𝑓(2) =


𝑓(𝑥) =

What does that tell us about the graph of 𝑓 in the vicinity of 𝑥 = 2?

Study behavior at 𝑥 = 6:

• Function value: 𝑓(6) =

• Limit: lim𝑥→6−

𝑓(𝑥) =

• Limit: lim𝑥→6+

𝑓(𝑥) =


𝑓(𝑥) =

What does that tell us about the graph of 𝑓 in the vicinity of 𝑥 = 6?

Compare to computer graph.

Example 3 involving 𝑓(𝑥) =(𝑥−2)(𝑥−4)

(𝑥−6)(𝑥−4). Similar questions.:


Activity for Day 8: Limits for a Function Given by a Graph (Section 6.3)

Use the graph to fill in the table. (Extra copies of the graph are on back.)

(𝐴) lim𝑥→−3

𝑓(𝑥) =

(𝐵) 𝑓(−3) =

(𝐶) lim

𝑥→1−𝑓(𝑥) =

(𝐷) lim𝑥→1+

𝑓(𝑥) =

(𝐸) lim𝑥→1

𝑓(𝑥) =

(𝐹) 𝑓(1) =

(𝐺) lim

𝑥→4−𝑓(𝑥) =

(𝐻) lim𝑥→4+

𝑓(𝑥) =

(𝐼) lim𝑥→4

𝑓(𝑥) =

(𝐽) 𝑓(4) =

𝑥

𝑓(𝑥)

.

𝑥

𝑓(𝑥)

𝑥

𝑓(𝑥)

𝑥

𝑓(𝑥)

Day 12 Lecture on Section 8

X Section 8 Rates of Change

We will do a series of examples involving 𝑓(𝑥) = −𝑥2 + 10𝑥 − 16 = −(𝑥 − 2)(𝑥 − 8).

(A) Draw the graph on paper.

(B) Draw the secant line that passes through points (3, 𝑓(3)) and (5, 𝑓(5)).

(C) Find slope of that secant line. Result: using 𝑚 =Δ𝑦

Δ𝑥, we get 𝑚 = 2.

Introduce Average Rate of Change

Definition of Average Rate of Change

• words: the average rate of change of 𝑓 as the input changes from 𝑎 to 𝑏

• usage: 𝑓 is a function that is continuous on the interval [𝑎, 𝑏].

• meaning: the number 𝑚 =𝑓(𝑏)−𝑓(𝑎)

𝑏−𝑎

• graphical interpretation: The number 𝑚 is the slope of the secant line that touches the graph of

𝑓 at the points (𝑎, 𝑓(𝑎)) and (𝑏, 𝑓(𝑏)).

• remark: The average rate of change 𝑚 is a number.

Discuss the definition of Average Rate of Change, and the fact that the secant line slope calculation that

we just did was an example of a calculation of an average rate of change. That is, the average rate of

change of 𝑓(𝑥) = −𝑥2 + 10𝑥 − 16 from 𝑥 = 3 to 𝑥 = 5 is number 𝑚 = 2.

(D) Find avg rate of change of 𝑓(𝑥) = −𝑥2 + 10𝑥 − 16 from 𝑥 = 3 to 𝑥 = 3 + ℎ where ℎ ≠ 0.

Result: 𝑚 =𝑓(3+ℎ)−𝑓(3)

ℎ= 4 − ℎ. (Very important can cancel h/h because ℎ ≠ 0)

Explore resulting formula by plugging in some numbers for ℎ. Draw corresponding secant lines.

(E) Make a new graph of 𝑓 and Illustrate this quantity on that new graph.


Day 12 Continued

X Section 8 Rates of Change

Introduce the Tangent Line

(F) Draw a new graph and Draw the line tangent to the graph of 𝑓 at 𝑥 = 3.

Informal definition of the tangent line (not precise)

The line tangent to the graph of 𝑓 at 𝑥 = 𝑎 is defined to be the line that has these two properties:

• The line touches the graph of 𝑓 at 𝑥 = 𝑎. This means that the point (𝑎, 𝑓(𝑎)) is on the line. This

point is called the point of tangency. (Note that this requires that 𝑓(𝑎) must exist.)

• The line looks like it is going the same direction as the graph of 𝑓 at that point.

Goal: Find the slope of that tangent line. That is, find the slope 𝑚 of the line tangent to the graph of

𝑓(𝑥) = −𝑥2 + 10𝑥 − 16 at 𝑥 = 3.

Discuss the fact that we don’t have two known points on the tangent line, so we can’t use our slope

formula 𝑚 =Δ𝑦

Δ𝑥 to find the slope of the tangent line.

Observe that by bringing right intersection point of Secant Line closer to left intersection point (closer to

𝑥 = 3), the secant line changes to look more & more like tangent line.

And observe that the slope of secant line seems to be getting closer and closer to the number 𝑚 = 4, and

it seems that this number 𝑚 = 4 is probably the slope of the tangent line.

But when second point gets moved to 𝑥 = 3, the slope is not defined. Need two points!

Analytically, pulling second point closer to the first corresponds to finding the limit 𝑚 = limℎ→0

𝑓(3+ℎ)−𝑓(3)

ℎ

Do that limit for our computation. We find 𝑚 = limℎ→0

𝑓(3+ℎ)−𝑓(3)

ℎ= lim

ℎ→04 − 0 = 4.

On computer graph, check box to show limit for Δ𝑥 = 0. Now see what happens when second point gets

moved on top of the first point at x = 3.

Official, precise definition of the tangent line

The line tangent to the graph of 𝑓 at 𝑥 = 𝑎 is defined to be the line that has these two properties:

• The line touches the graph of 𝑓 at 𝑥 = 𝑎. This means that the point (𝑎, 𝑓(𝑎)) is on the line. This

point is called the point of tangency. (Note that this requires that 𝑓(𝑎) must exist.)

• The line has slope 𝑚 = limℎ→0

𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ if this limit exists and is a real number.

Graphical interpretation: The tangent line (if it exists) is the line that passes through the point of tangency

(𝑎, 𝑓(𝑎)) and looks like it is going the same direction as the graph of 𝑓 at that point.

Introduce Instantaneous Rate of Change

Definition of Instantaneous Rate of Change

• words: the instantaneous rate of change of 𝑓 at 𝑎

• alternate words: the derivative of 𝑓 at 𝑎

• symbol: 𝑓′(𝑎)

• meaning: the number 𝑚 = limℎ→0

𝑓(𝑎+ℎ)−𝑓(𝑎)

ℎ

• graphical interpretation: The number 𝑚 is the slope of the line tangent to the graph of 𝑓 at the

point (𝑥, 𝑦) = (𝑎, 𝑓(𝑎)).

• remark: The instantaneous rate of change 𝑓′(𝑎) is a number.

Class Activity #1 for Section 8 Representations of Slopes (The Class Activity is printed on next page

of this lecture outline and is available on the course web page of Exercises.)


Class Activity #1 for Section 8: Representations of Slopes

In textbook Section 8, you learned about average rate of change and instantaneous rate of change:

Definition of Average Rate of Change

• words: the average rate of change of 𝑓 as the input changes from 𝑎 to 𝑏

• usage: 𝑓 is a function that is continuous on the interval [𝑎, 𝑏].

• meaning: the number 𝑚 =𝑓(𝑏)−𝑓(𝑎)

𝑏−𝑎

• graphical interpretation: The number 𝑚 is the slope of the secant line that touches the graph of 𝑓 at

the points (𝑎, 𝑓(𝑎)) and (𝑏, 𝑓(𝑏)).

• remark: The average rate of change 𝑚 is a number.

Definition of Instantaneous Rate of Change

• words: the instantaneous rate of change of 𝑓 at 𝑎

• alternate words: the derivative of 𝑓 at 𝑎

• symbol: 𝑓′(𝑎)

• meaning: the number 𝑚 = limℎ→0


ℎ

• graphical interpretation: The number 𝑚 is the slope of the line tangent to the graph of 𝑓 at the point

(𝑥, 𝑦) = (𝑎, 𝑓(𝑎)).

• remark: The instantaneous rate of change 𝑓′(𝑎) is a number.

Each expression in the left column represents a number 𝑚 that can be interpreted as the slope of a line on the

graph of 𝑓. In each example, draw the line on the graph of 𝑓, or write the missing expression based on the line

shown in the graph, and then give the value of the number 𝑚 represented by the expression.

Example Expression representing 𝑚 Line whose slope is 𝑚 Value of 𝑚

(1)

the average rate of

change of 𝑓 as the input

changes from 1 to 5

𝑚 =

(2) the derivative of 𝑓 at 𝑥 = 1

𝑚 =

𝑥

𝑓

𝑥

𝑓

Example Expression representing 𝑚 Line whose slope is 𝑚 Value of 𝑚

(3) the instantaneous rate of

change of 𝑓 at 𝑥 = 4

𝑚 =

(4) limℎ→0

𝑓(3 + ℎ) − 𝑓(3)

ℎ

𝑚 =

(5) 𝑓(4) − 𝑓(2)

4 − 2

𝑚 =

(6) 𝑓′(2)

𝑚 =

(7)

𝑚 =

𝑥

𝑓

𝑥

𝑓

𝑥

𝑓

𝑥

𝑓

𝑓

𝑥

Day 13 Lecture on Sections 8 and 9

X Today: Sections 8 and 9 Introduction to the Derivative.

Introduce the idea of the derivative function, 𝑓′. Imagine a machine:

input: number “𝑎”

Instructions inside machine: find slope of the line tangent to graph of 𝑓 at 𝑥 = 𝑎

output obtained graphically: number 𝑚 that is slope of line tangent to the graph of 𝑓 at 𝑥 = 𝑎.

output obtained analytically: the number 𝑚 = 𝑓′(𝑎) = limℎ→0


ℎ.

diagram illustrating the machine:

This machine is really called the derivative machine. Discuss using 𝑥 instead of 𝑎.

Example of computing the derivative graphically:

Class Activity #2 for Section 8: Finding derivatives graphically using a ruler. (The Class Activity is

printed on next page of this lecture outline and is available on the course web page of Exercises.)

Examples of computing 𝑓′(𝑥) using the Analytic Definition of the Derivative:

𝑓′(𝑥) = limℎ→0

𝑓(𝑥 + ℎ) − 𝑓(𝑥)

ℎ

[Example] Let 𝑓(𝑥) = 𝑥2 − 2𝑥 − 3. Find 𝑓′(𝑥). using the Analytic Definition of the Derivative

Write explanation for most important step: Cancelling ℎ

ℎ. Result: 𝑓′(𝑥) = 2𝑥 − 2.

Observe: 𝑓(𝑥) = 𝑥2 − 2𝑥 − 3 and 𝑓′(𝑥) = 2𝑥 − 2 match the graphs from class activity that we just did.

Related questions

(A) Find the slope of the line that is tangent to the graph of 𝑓 at 𝑥 = 2.

(B) Find the slope of the line that is tangent to the graph of 𝑓 at 𝑥 = 0.

(C) Find the 𝑥-coordinates of all points on the graph of 𝑓 that have horizontal tangent lines.

Nonexistence of the Derivative

Draw graph with common pathologies:

• Missing point

• Jump in Graph

• Cusp

• Vertical tangent

Discuss why the derivative does not exist at each point.

Remark that both Cusp Point and Vertical Tangent are locations where the function exists, the limit

exists, and the function is continuous. Only with the concept of derivative are we finally able to articulate

what is “bad” about these locations.

.


Find the slope

𝑚 of the line

tangent to the

graph of 𝑓 at

𝑥 = 𝑎. input

the number 𝑎

output

the number 𝑚

Class Activity #2 for Section 8: Finding Derivatives Graphically Using a Ruler

The goal: Given the graph of 𝑓 on the top axes on the next page, make a graph of 𝑓′ on the

bottom axes.

On the graph of 𝑓′, the input will be 𝑥 and the output will be 𝑓′(𝑥). Remember the graphical

interpretation of 𝑓′(𝑥):

Definition of the Derivative

• symbol: 𝑓′(𝑎) • graphical interpretation: 𝑓′(𝑎) is the number that is the slope of the line tangent to the

graph of 𝑓 at the point where 𝑥 = 𝑎.

Part 1: Prepare the data for your graph of 𝑓′ by filling out the following table.

𝑥 what to do on the graph of 𝑓 𝑓′(𝑥)

−2 Draw the line tangent to the graph of 𝑓 at the point where 𝑥 = −2

and find its slope 𝑚. This slope 𝑚 will be the value of 𝑓′(−2).

−1 Draw the line tangent to the graph of 𝑓 at the point where 𝑥 = −1

and find its slope 𝑚. This slope 𝑚 will be the value of 𝑓′(−1).

0 Draw the line tangent to the graph of 𝑓 at the point where 𝑥 = 0

and find its slope 𝑚. This slope 𝑚 will be the value of 𝑓′(0).









Part 2 is on the next page.

.

𝑥

𝑓(𝑥)

𝑥

Part 2: Using the (𝑥, 𝑓′(𝑥)) data from your table, make a graph of 𝑓′.

𝑓′(𝑥)

Extra Graphs for Class Activity #2 for Section 8: Finding Derivatives Graphically Using a Ruler

.

.

𝑥

𝑓(𝑥)

𝑥

𝑓(𝑥)

𝑥

𝑓(𝑥)

𝑥

𝑓(𝑥)

Days 14 and 15 Lectures on Sections 8 and 9

X Today: Sections 8 and 9 Introduction to the Derivative.

More Examples of computing 𝑓′(𝑥) using the Analytic Definition of the Derivative

Remember the Analytic Definition of the Derivative:

𝑓′(𝑥) = limℎ→0

𝑓(𝑥 + ℎ) − 𝑓(𝑥)

ℎ

[Example 1] Let 𝑓(𝑥) = −3𝑥2 + 5𝑥 − 7 Find 𝑓′(𝑥). using the Analytic Definition of the Derivative

[Example 2] 𝑓(𝑥) =1

𝑥 Find 𝑓′(𝑥). using the Analytic Definition of the Derivative

[Example 3] Harder example involving a 1

𝑥 type function:

Let 𝑓(𝑥) = 3 −7

𝑥. Find 𝑓′(𝑥). using the Analytic Definition of the Derivative

[Example 4] 𝑓(𝑥) = √𝑥. Find 𝑓′(𝑥). using the Analytic Definition of the Derivative

[Example 5] harder example involving a √𝑥 type function:

Let 𝑓(𝑥) = 3 − 7√𝑥. Find 𝑓′(𝑥). using the Analytic Definition of the Derivative

Important: For this course, students need to know how to use the Analytic Definition of the Derivative

to find derivatives of three kinds of functions:

1) Polynomial functions of degree 2 or less

2) 1

𝑥 type functions.

3) √𝑥 type functions.

There are Exercises of this type in the Exercise set online.

.

Day 16 Lecture on Section 10 Rules of Differentiation

X Section 10 Rules of Differentiation

Review Analytic Definition of the Derivative: 𝑓′(𝑥) = limℎ→0

𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ

For this course, students need to know how to use the Analytic Definition of the Derivative to find

derivatives of three kinds of functions:

1) Polynomial functions of degree 2 or less

2) 1

𝑥 type functions.

3) √𝑥 type functions.

There are Exercises of this type in the Exercise set online. We did examples in class last week.

In Section 10 Rules of Differentiation, we will learn rules for taking derivatives. We won’t use def.

Notation for Derivatives

• Using the familiar prime symbol

o With variable: 𝑓′(𝑥) or 𝑓′(𝑡) or whatever the variable is.

o Without variable: The empty version 𝑓′( ) or even just 𝑓′.

• New symbol: 𝑑𝑓(𝑥)

𝑑𝑥 or

𝑑𝑓(𝑡)

𝑑𝑡 or whatever the variable is.

o The symbol 𝑑

𝑑𝑥𝐵𝐿𝐴𝐻 means the derivative of BLAH. so

𝑑

𝑑𝑥𝑓(𝑥) means

𝑑𝑓(𝑥)

𝑑𝑥.

The Constant Function Rule

Present the Constant Function Rule:

• Two equation form: If 𝑓(𝑥) = 𝑐, then 𝑓′(𝑥) = 0.

• Single Equation form: 𝑑

𝑑𝑥𝑐 = 0.

[Example 1]: If 𝑓(𝑥) = 7 then 𝑓′(𝑥) = 0. That is, 𝑑

𝑑𝑥7 = 0.

Does this make sense graphically?

• Draw graphs of 𝑓(𝑥) = 7 and 𝑓′(𝑥) = 0.

• Observe that at all points on the graph of 𝑓(𝑥), the tangent lines are horizontal, with slope 𝑚 = 0.

• So it makes sense that all of the points on the graph of 𝑓′(𝑥) = 0 have 𝑦 = 0.

The Power Rule

Present the Power Rule:

• Two equation form: If 𝑓(𝑥) = 𝑥𝑛, then 𝑓′(𝑥) = 𝑛𝑥𝑛−1.

• Single Equation form: 𝑑

𝑑𝑥𝑥𝑛 = 𝑛𝑥𝑛−1.

[Example 2] 𝑓(𝑥) = 𝑥4.

[Example 3] 𝑓(𝑥) =1

𝑥. Compare to result of Day 14 Example using definition of derivative.

[Example 4] 𝑓(𝑥) = √𝑥. Compare to result of Day 15 Example using definition of derivative.

Discuss importance of step 1: rewriting function then step 2: take derivative.

Discuss common incorrect notation.

[Example 5] 𝑓(𝑥) = 𝑥. Use Power Rule. Then check to see if it makes sense graphically.

[Example 6] 𝑓(𝑥) = 1. Compare to result obtained using Constant Function Rule.

.


Day 16, continued

X Section 10 Rules of Differentiation

The Tangent Line

Remember what we know about the line tangent to graph of 𝑓 at 𝑥 = 𝑎.

Tangent line is the line that has these two properties:

1) Touches graph at 𝑥 = 𝑎. That point is called “point of tangency”. The 𝑦 coord is 𝑓(𝑎). So point of

tangency has coordinates (𝑥, 𝑦) = (𝑎, 𝑓(𝑎)).

2) Tangent line has slope 𝑚 = 𝑓′(𝑎). This is the KNOWN SLOPE of the tangent line.

Example involving Slope of the Tangent Line

Example: Let 𝑓(𝑥) = √𝑥. Find slope of the line tangent to the graph of 𝑓 at 𝑥 = 9. Illustrate on graph.

The Equation for the Tangent Line

Review point-slope form (𝑦 − 𝑏) = 𝑚(𝑥 − 𝑎).

Remember what we know about the line tangent to graph of 𝑓 at 𝑥 = 𝑎.

Analytical description: tangent line is the line that has these two properties:

1) Touches graph at 𝑥. That point is called “point of tangency”. The y-coord is 𝑓(𝑎). So point of

tangency has coordinates (𝑥, 𝑦) = (𝑎, 𝑓(𝑎)).

2) Tangent line has slope 𝑚 = 𝑓′(𝑎). This is the KNOWN SLOPE of the tangent line.

So the Point Slope form of equation for tangent line will be: (𝑦 − 𝑓(𝑎)) = 𝑓′(𝑎)(𝑥 − 𝑎).

Example: Let 𝑓(𝑥) = √𝑥. Find equation of the line tangent to the graph of 𝑓 at 𝑥 = 9.

Present the line equation in slope intercept form.

Illustrate on graph with important points labeled with their (𝑥, 𝑦) coordinates.

.

.

Day 17 Lecture on Section 10 Rules of Differentiation

X continuing Section 10 Today: The Sum and Constant Multiple Rule

The Sum and Constant Multiple Rule

Present the Sum and Constant Multiple rule: 𝑑

𝑑𝑥(𝑎𝑓(𝑥) + 𝑏𝑔(𝑥)) = 𝑎

𝑑

𝑑𝑥𝑓(𝑥) + 𝑏

𝑑

𝑑𝑥𝑔(𝑥)

[Example 1]

Let 𝑓(𝑥) = −3𝑥2 + 5𝑥 − 7. Find 𝑓′(𝑥). (using the techniques of Section 10, not the Definition of the

Derivative)

Compare to result of Day 14 (Wed Feb 5) Example 2 (page 5,6) using definition of derivative.

[Example 2] Find derivative of 𝑓(𝑥) = 2 −3

𝑥. (using the techniques of Section 10)

Present method that I want them to be able to do

Step 1: rewrite 𝑓 as sum of terms of form constant*power function.

Step 2: Only after f is rewritten do they start work on the derivative. The first step should be to identify

multiplicative constants and use sum and constant multiple rule. Simplify the result to eliminate negative

exponents.

Compare to result of Day 14 (Wed Feb 5) Example 4 (page 10-12) using definition of derivative.

[Example 3] Find derivative of 𝑓(𝑥) = 3 − 7√𝑥. (using the techniques of Section 10)

Compare to result of Day 15 (Fri Feb 7) Example (page 5,6) using definition of derivative.

Class Activities are included on the next two pages of these lecture notes and are also available on

the MATH 1350 Exercises web page

[Class Activity #1] Rewriting Functions in Different Form

[Class Activity #2] Finding a Derivative Using Sum, Constant Multiple, and Power Rules

Find deriv of 𝑓(𝑥) =7 √𝑥

3

5+

3

11𝑥2/5.

[Example 4] Trick problem: Let 𝑓(𝑥) =2𝑥5−4𝑥3+2𝑥

𝑥3 . Find 𝑓′(𝑥).

[Example 5] tangent line problem: Let 𝑓(𝑥) = 𝑥3 − 9𝑥2 + 15𝑥 + 25 = (𝑥 + 1)(𝑥 − 5)2.

(a) Find 𝑓′(𝑥) (using the techniques of Section 10, not the Definition of the Derivative)

(b) Find the slope of the line that is tangent to the graph of 𝑓 at 𝑥 = 2.

(c) Find the slope of the line that is tangent to the graph of 𝑓 at 𝑥 = 0.

(d) Find the 𝑥 coordinates of all points on the graph of 𝑓 that have horizontal tangent lines.

(e) Find equation of the line tangent to graph at 𝑥 = 6.

..

Class Activity #1 for Section 10: Rewriting Functions in Different Forms

Su

m o

f te

rms

of

the

form

con

sta

nt

× p

ow

er f

un

ctio

n

Th

at

is,

𝒂𝒙

𝒑+

𝒃𝒙

𝒒

5𝑥

−2

+9

𝑥−

1

1.2

𝑥−

1/2

−0

.6𝑥

−2

/3

−1

0𝑥

−3

−9

𝑥−

2

7 15

𝑥−

2/3

−6 55

𝑥−

7/5

=

=

=

=

=

=

=

Sep

ara

te c

on

sta

nts

5(

1 𝑥2

)+

9(

1 𝑥)

1.2

(1 √𝑥

)−

0.6

(1

√𝑥

23

)

=

=

=

=

=

=

=

Sim

pli

fied

fo

rm

𝑓( 𝑥

)=

5 𝑥2

+9 𝑥

𝑓( 𝑥

)=

1.2

√𝑥

−0

.6√

𝑥2

3

𝑓( 𝑥

)=

5 √𝑥

3−

6

𝑥1

/2

𝑓( 𝑥

)=

7√

𝑥3 5

+3

11

𝑥2

/5

.

Class Activity #2 for Section 10: Rewriting a Function Before Finding its Derivative

𝑓(𝑥) =7√𝑥

3

5+

3

11𝑥2/5

(A) Rewrite 𝑓(𝑥) as a sum of terms of the form constant × power function. That is, 𝑎𝑥𝑝 + 𝑏𝑥𝑞.

(B) Find 𝑓′(𝑥).

• Use the Derivative Rules (That is, DO NOT use the Definition of the Derivative.)

• Show all details clearly and use correct notation.

• Simplify your final answer, and rewrite it so that it does not have any negative exponents.

.

Day 18 Lecture on Section 11 The Product Rule and Quotient Rule

X Section 11: The Product Rule and Quotient Rule

Present the product rule

• Obvious thing: 𝑑

𝑑𝑥(𝑓(𝑥)𝑔(𝑥)) = 𝑓′(𝑥)𝑔′(𝑥) is WRONG!

• Present the correct rule: 𝑑

𝑑𝑥(𝑓(𝑥)𝑔(𝑥)) = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)

[Example 1] 𝑓(𝑥) = (−3𝑥2 + 5𝑥 − 7)(3𝑥 − 2). Find 𝑓′(𝑥) using the product rule. Simplify answer.

[Example 2] 𝑓(𝑥) = 5(−3𝑥2 + 5𝑥 − 7). Find 𝑓′(𝑥) using product rule. Then use constant multiple rule.

Present the quotient rule

• Obvious thing: 𝑑

𝑑𝑥(

𝑡𝑜𝑝(𝑥)

𝑏𝑜𝑡𝑡𝑜𝑚(𝑥)) =

𝑡𝑜𝑝′(𝑥)

𝑏𝑜𝑡𝑡𝑜𝑚′(𝑥) is WRONG!

• Present the correct rule: 𝑑

𝑑𝑥(

𝑡𝑜𝑝(𝑥)

𝑏𝑜𝑡𝑡𝑜𝑚(𝑥)) =

𝑡𝑜𝑝′(𝑥)𝑏𝑜𝑡𝑡𝑜𝑚(𝑥)−𝑡𝑜𝑝(𝑥)𝑏𝑜𝑡𝑡𝑜𝑚′(𝑥)

(𝑏𝑜𝑡𝑡𝑜𝑚(𝑥))2

[Example 3] Let 𝑓(𝑥) =(3𝑥+5)

(𝑥2−3). Find 𝑓′(𝑥) and simplify answer.

[Example 4] Trick problem: Let 𝑓(𝑥) =2𝑥5−4𝑥3+2𝑥

𝑥3. Find 𝑓′(𝑥) using the quotient Rule, and then point

out that we did this same problem in Section 17 by first rewriting the function and then using the simpler

derivative rules. The simpler derivative rules are always far better, when they will do the job.

List of other derivative problems that look like they might be quotient rule problems, but that are

far better solved by first rewriting the function and then using the simpler derivative rules

Let 𝑓(𝑥) =625−𝑥2

√𝑥 Find 𝑓′(𝑥). Point out that in Book Section 11.2, they do this derivative two ways: the

quotient rule, and then the produt rule. Both of those are dumb ways to do this derivative.

Let 𝑓(𝑥) =3𝑥5−2𝑥7

√𝑥35 Find 𝑓′(𝑥). (Similar to previous. Would be incredibly hard with Quotient Rule.)

Let 𝑓(𝑥) =7 √𝑥

3

5+

3

11𝑥2/5 Find 𝑓′(𝑥). (Problem from Class Activity that we did yesterday.)

Tangent Line Problem

[Example 5] Let 𝑓(𝑥) =𝑥

𝑥2+25.

(A) find 𝑓′(𝑥).

(B) find 𝑓′(0)

(C) find the 𝑥-coordinates of all the points on the graph of 𝑓 that have horizontal tangent lines.

(D) Illustrate your solutions to (B),(C) on the given graph of 𝑓(𝑥).

.

𝑓(𝑥)

𝑥

Day 19 Lecture on Section 11 Quotient Rule

X Section 11 The Product Rule and Quotient Rule

Finish Section 11 Two problems involving Derivatives of Quotients

Do Tangent Line Problem from Day 18 if not already done

Rate of Change Problem

[Example] Sales of a game are described by the function 𝑆(𝑡) =7000𝑡

𝑡+6 where 𝑡 is the time (in months)

since the game was introduced and 𝑆(𝑡) is the total number of games that have been sold at time 𝑡.

(A) Find 𝑆(4).

(B) Find 𝑆′(𝑡). Show all details clearly, use correct notation, and simplify your answer.

(C) Find 𝑆′(4).

(D) Write a brief interpretation of the answers from (A) and (C). That is, explain what the answers tell us

about sales of the game. Include the correct units in your explanation.

(E) Illustrate the answers to A,C on a given graph of 𝑆(𝑡).

Quiz 4

.

𝑓′(𝑎) the slope 𝑚 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.

Diagram illustrating important relationships:

𝑆(𝑡)

𝑡 0 1 2 3 4 5 6 7 8 9 10

Day 20 Lecture on Section 12 The Chain Rule

X Section 12 The Chain Rule

introduce the chain rule

used for taking derivatives of “nested functions” Functions of form 𝑜𝑢𝑡𝑒𝑟(𝑖𝑛𝑛𝑒𝑟(𝑥))

The Chain Rule: 𝑑

𝑑𝑥𝑜𝑢𝑡𝑒𝑟(𝑖𝑛𝑛𝑒𝑟(𝑥)) = 𝑜𝑢𝑡𝑒𝑟′(𝑖𝑛𝑛𝑒𝑟(𝑥)) ⋅ 𝑖𝑛𝑛𝑒𝑟′(𝑥)

Basic Examples

[Example 1] For 𝑓(𝑥) = 2(3𝑥4 + 5𝑥2 + 6)7 find 𝑓′(𝑥).

[Example 2] For 𝑓(𝑥) =2

(3𝑥4+5𝑥2+6)7 find 𝑓′(𝑥).

[Example 3] For 𝑓(𝑥) = 3√𝑥2 − 3𝑥 + 21 Find 𝑓′(𝑥).

Harder Chain Rule Problems:

[Example 4]: (Tangent line problem) For 𝑓(𝑥) = 3√𝑥2 − 3𝑥 + 21

(A) Find the equation of the line tangent to the graph of 𝑓 at 𝑥 = 4.

(B) Find 𝑥 coordinate of all points on the graph of 𝑓 that have a horizontal tangent line.

[Example 5] (Product Rule AND Chain Rule) For 𝑓(𝑥) = 𝑥3(𝑥 − 7)4

Find 𝑓′(𝑥) and find all values of 𝑥 where the tangent line is horizontal. Discuss common mistake.

.

Day 21 Lecture on Section 13 Derivatives of Exponential Functions

X Section 13 Derivatives of Exponential Functions

New Rules

Exponential Function Rule #1 𝑑

𝑑𝑥𝑒(𝑥) = 𝑒(𝑥)

See book for discussion of why this is true. Does it make sense graphically?

Exponential Function Rule #2 𝑑

𝑑𝑥𝑏(𝑥) = 𝑏(𝑥) ⋅ ln(𝑏). Again, see book for discussion.

Use Chain Rule and Rule #1 to prove Exponential Function Rule #3 𝑑

𝑑𝑥𝑒(𝑘𝑥) = 𝑐𝑒(𝑘𝑥)

The book does not present this in its list of derivative rules.

That is a shame, because it is one of the most-used derivative rules.

[Example 1] Derivatives of Basic Functions Involving Exponents

(A) 𝑓(𝑥) = 5𝑒(𝑥), (B) 𝑓(𝑥) = 5𝑒(7𝑥), (C) 𝑓(𝑥) = 5 ⋅ 7(𝑥), (D) 𝑓(𝑥) = 5𝑒(7), (E) 𝑓(𝑥) = 5𝑥𝑒

More difficult example involving exponential function

Exponential Function Rule and Product Rule

[Example 2] Let 𝑓(𝑥) = (−3𝑥2 + 5𝑥 + 7)𝑒(𝑥). (A) Find 𝑓′(𝑥) and simplify.(B) 𝑓′(0). (C) 𝑓′(1).

Exponential Function Rule and Quotient Rule

[Example 3] Let 𝑓(𝑥) =

𝑒(𝑥)

(−3𝑥2+5𝑥+7). (A) Find 𝑓′(𝑥) and simplify.(B) 𝑓′(0). (C) 𝑓′(1).

Tangent Line Example

[Example 4] For the function: 𝑓(𝑥) = 11𝑒(𝑥) + 23𝑥.

Find equation of line tangent to graph of 𝑓 at 𝑥 = 0.

Find equation of line tangent to graph of 𝑓 at 𝑥 = 1.

Rate of Change Example

Given: Formula for Continuously Compounded Interest: 𝐴 = 𝑃𝑒(𝑟𝑡)

𝑃 is the amount of the initial deposit.

𝑡 is the time in years since the initial deposit.

𝑟 is the interest rate, expressed as a decimal.

𝐴 is the amount in the account (the account balance) at time 𝑡.

Question: What is the instantaneous rate of change of the balance at time 𝑡 years?

Solution: Think of 𝐴 as a function of the variable 𝑡. That is, 𝐴(𝑡) = 𝑃𝑒(𝑟𝑡)

Find 𝐴′(𝑡) using the chain rule. Go through the steps. Result 𝐴′(𝑡) = 𝑟𝑃𝑒(𝑟𝑡).

[Example 5] Investment of $500 earns interest at an annual rate of 2% compounded continuously.

(A) What is the balance at time 𝑡 = 0 years?

(B) What is the balance at time 𝑡 = 6 years?

(C) What is the average rate of change of the balance from time 𝑡 = 0 to 𝑡 = 6 years?

(D) What is the instantaneous rate of change of the balance at time 𝑡 = 0 years?

(E) What is the instantaneous rate of change of the balance at time 𝑡 = 6 years?

(F) Illustrate the quantities from Parts (A)-(E) on the given graph.

..

.

Balance A

time t 0 1 2 3 4 5 6 7

Day 22,23 Lectures on Section 14 Derivatives of Logarithmic Functions

X Section 14 Derivatives of Logarithmic Functions

Review Logarithmic Functions and their graphs

Review graph of 𝑦 = 2(𝑥), including discussing its domain and range and discussing famous points:

• We know 2(0) = 1, so the point (0,1) is on graph.

• We know 2(1) = 2, so the point (1,2) is on graph.

.

Discuss graph obtained by interchanging 𝑥 ↔ 𝑦. It is the function 𝑦 = log2(𝑥).

Discuss domain, range and discuss famous points:

• The point (1,0) is on graph. This tells us that log2(1) = 0.

• The point (2,1) is on graph. This tells us that log2(2) = 1.

.

Similarly, review graph of 𝑦 = 𝑒(𝑥). Discussing its domain and range and discuss famous points:

• We know 𝑒(0) = 1, so the point (0,1) is on graph.

• We know 𝑒(1) = 𝑒, so the point (1, 𝑒) is on graph.

.

Discuss graph obtained by interchanging 𝑥 ↔ 𝑦. It is the function 𝑦 = ln(𝑥).

Discuss domain, range and discuss famous points:

• The point (1,0) is on graph. This tells us that ln(1) = 0.

• The point (𝑒, 1) is on graph. This tells us that ln(𝑒) = 1.

.

Derivatives of Logarithmic Functions

Logarithmic Function Rule #1 𝑑

𝑑𝑥ln(𝑥) =

1

𝑥.

See book for discussion of why this is true, with discussion involving formulas.

Here, just discuss why it makes sense in terms of the graph.

Logarithmic Function Rule #2 𝑑

𝑑𝑥log𝑏(𝑥) =

1

𝑥 ln(𝑏).

Examples

[Example #1] Find the derivatives of the following functions:

(A) 𝑓(𝑥) = 5 ln(𝑥).

(B) 𝑓(𝑥) = 5 log7(𝑥).

(C) 𝑓(𝑥) = 5 log(𝑥).

(D) 𝑓(𝑥) = 5 ln(7).

(E) 𝑓(𝑥) = 5 ln(7𝑥).

(F) 𝑓(𝑥) = 5 ln (7

𝑥).

(G) 𝑓(𝑥) = 5 ln(𝑥7).

(H) 𝑓(𝑥) = 5𝑥 ln(7).

[Example #2] (Logarithmic Function Rule and Product Rule)

Let 𝑓(𝑥) = 5𝑥7 ln(𝑥) (A) find 𝑓′(𝑥) and simplify. (B) Find 𝑓′(1), simplify. (C) Find 𝑓′(𝑒), simplify.

[Example #3] (Logarithmic Function Rule and Chain Rule)

Let 𝑓(𝑥) = 7 ln(5𝑥2 − 30𝑥 + 65) Find 𝑓′(𝑥).

[Example #4] (Tangent Line Problem)

Find equation of the line tangent to 𝑓(𝑥) = 7 + ln(𝑥5) at 𝑥 = 𝑒3.

.

.

Day 24 was Exam 2

Day 25 Lecture on Section 18 Tangent Lines and Linear Approximation; Differentials

X Section 18 Linearizations and Approximating y values

The linearization

Given function f, we know that the equation for the line tangent to graph of f at x=a is

(𝑦 − 𝑓(𝑎)) = 𝑓′(𝑎)(𝑥 − 𝑎). In this equation, 𝑎, 𝑓(𝑎), 𝑓′(𝑎) are all constants. This equation expresses

relationship between x&y for all points on the tangent line.

Solve this equation for y in terms of x. (just add f(a) to both sides): 𝑦 = 𝑓(𝑎) + 𝑓′(𝑎)(𝑥 − 𝑎).

Since this equation is solved for y in terms of x, the equation expresses y as a function of x.

The function gives us the y-value of a point on the tangent line if you know the x-value.

Invent a symbol & name: 𝐿(𝑥) = 𝑓(𝑎) + 𝑓′(𝑎)(𝑥 − 𝑎) called the Linearization of 𝑓 at 𝑎.

Example 2.8#1 𝑓(𝑥) = 𝑥4 + 3𝑥2. Find linearization of 𝑓 at 𝑎 = −1.

Solution: 𝐿(𝑥) = 4 − 10(𝑥 + 1) = −10𝑥 − 6. Surprisingly, the y=mx+b form is less useful.

Example 𝑓(𝑥) = 𝑥1/3. Find linearization of 𝑓 at 𝑎 = 1000.

Result: 𝐿(𝑥) = 10 + (1

300) (𝑥 − 1000).

Graph 𝑓(𝑥) and 𝐿(𝑥) from previous example. Observe at 𝑥 = 1000, the values 𝑓(1000) and 𝐿(1000)

agree (because that is the point of tangency).

And for 𝑥 close to 1000, it looks like 𝑓(𝑥) ≈ 𝐿(𝑥). Close, but not exact.

Compare values of 𝑓(𝑥) and 𝐿(𝑥) when 𝑥 = 1001. Which do you think will be bigger?

Result: 𝐿(1001) = 10.033̅̅̅̅ while 𝑓(1001) = 10.3332222829 …

We can use the idea that for x valuses near x=a, the value of 𝑓(𝑥) will be very close to the value of 𝐿(𝑥),

so 𝐿(𝑥) can give us an estimate of 𝑓(𝑥).

Method:

Identify the function

Identify a convenient nearby x-value that will play the role of the number a

Build the Linearization

Use the linearization to estimate the value of 𝑓(𝑥). That is, 𝑓(𝑥) ≈ 𝐿(𝑥)

Example Use a linear approximation to estimate the number √10013

.

Solution: We need to figure out the function. In this case, 𝑓(𝑥) = 𝑥1/3.

We have to identify a convenient nearby x-value that we can use for the number a. In this case, we use

a=1000 because we know that the value of 𝑓(1000) = 10001/3 = 10.

Build the linearization of 𝑓(𝑥) = 𝑥1/3 at a=1000. Result: 𝐿(𝑥) = 10 + (1

300) (𝑥 − 1000)

Use Linearization to estimate the number: Result: √10013

= 𝑓(1001) ≈ 𝐿(1001) = 10.033̅̅̅̅ .

Example Use linear approximation to estimate the value of (1.999)4.

Solution: We need to figure out the function. In this case, 𝑓(𝑥) = 𝑥4.

convenient nearby 𝑥 value that we will use for the number a, We will use a=2 (convenient because it is

easy to compute 𝑓(2) = 24 = 16.)

Build the linearization of 𝑓(𝑥) = 𝑥2 at 𝑎 = 2 Result: 𝐿(𝑥) = 16 + 32(𝑥 − 2).

Use Linearization to estimate the number: Result:

1.9994 = 𝑓(1.999) ≈ 𝐿(1.999) = 15.968

Compare to calculator result: 𝑓(1.999) = 1.9994 = 15.968023992001 …

Notice that in this case our estimate is an underestimate, which makes sense because tangent line is below

the graph of f(x)

Day 26 Lecture on Section 18 Tangent Lines and Linear Approximation; Differentials

X Section 18 Differentials and Approximating Changes in y values

Yesterday, we used the Linearization of a function to estimate 𝑦 values.

Now we will turn our attention to estimating change in 𝑦 values.

Consider a function 𝑓 and two known 𝑥 values 𝑥1 and 𝑥2. How does the 𝑦 value on graph of 𝑓 change

when 𝑥 changes from 𝑥1 to 𝑥2? Draw picture of increasing, concave down chunk of graph.

Introduce terminology:

• change in 𝑥 is denoted Δ𝑥 = 𝑥2 − 𝑥1

• corresponding change in 𝑦 is denoted Δ𝑦 = 𝑦2 − 𝑦1 = 𝑓(𝑥2) − 𝑓(𝑥1)

Example: 𝑓(𝑥) = √𝑥, values 𝑥1 = 4 and 𝑥2 = 9. find Δ𝑥 and Δ𝑦. Draw picture.

notice that Δ𝑦 = change in 𝑦 = actual change in 𝑦 values on graph.

Also show how Δ𝑦 can be illustrated as a vertical length on the graph of 𝑓. (above 𝑥2). Draw in green.

There is a different way to get a vertical length. That is to find the change in height on the tangent line to

the graph of 𝑓 at 𝑥 = 𝑥1, when 𝑥 changes from 𝑥1 to 𝑥2. Draw in red.

Examine triangle: determine that ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑚 ⋅ Δ𝑥 = 𝑓′(𝑥1) ⋅ Δ𝑥.

Compute for our example. Determine that red height = 5/4 = 1.25.

Call this quantity 𝑓′(𝑥1) ⋅ Δ𝑥 the differential of 𝑓 at 𝑥1 = 4 when Δ𝑥 = 5.

Abbreviate it by the symbol 𝑑𝑦.

Draw picture showing Δ𝑦 and 𝑑𝑦 together.

The usefulness of the differential 𝑑𝑦 is that

• when Δ𝑥 is small, 𝑑𝑦 is very close to Δ𝑦.

• we often want to know Δ𝑦, but it can be hard to calculate, while 𝑑𝑦 is usually much easier to

calculate.

So 𝑑𝑦 is an easy way to get an approximation for Δ𝑦

[Example #1] The total cost of producing a batch of 𝑥 guitars is 𝐶(𝑥) = 1000 + 100𝑥 − 0.25𝑥2 dollars

(A) What is the cost of producing a batch of 50 guitars?

(B) What is the cost of producing a batch of 51 guitars?

(C) If batch size changes from 𝑥 = 50 guitars to 𝑥 = 51 guitars, what will be change in the cost of

producing a batch of guitars? That is, if 𝑥 = 51 and Δ𝑥 = 1, what is the exact change in cost, Δ𝐶? (exact

value) Use Wolfram, result is Δ𝐶 = $74.75.

(d) Find the differential 𝑑𝐶 to find the approximate change in cost

Result: 𝑑𝐶 = 𝐶′(𝑥1) ⋅ Δ𝑥 = 75 ⋅ 1 = 75

Compare the exact and approximate results:

Exact change in cost is Δ𝐶 = $74.75 Approximate change in cost is 𝑑𝐶 = $75. Observe that Δ𝐶 ≈ 𝑑𝐶

Relationship between Differentials and Marginal Quantities

Recall that for the cost function 𝐶(𝑥), the function 𝐶′(𝑥) is called the marginal cost function.

The quantity 𝐶′(75) is called the marginal cost at a production level of 75.

We see the usefulness of this quantity: It gives us an estimate of how much cost will change if quantity

changes by Δ𝑥 = 1. That is, Δ𝐶 = 𝑒𝑥𝑎𝑐𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑐𝑜𝑠𝑡 ≈ 𝑎𝑝𝑝𝑟𝑜𝑥 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑐𝑜𝑠𝑡 = 𝑑𝐶 = 𝐶′(75). [Example #2] A company makes and sells 𝑥 cell phone cases per week. The revenue from the sale of 𝑥

cases is 𝑅(𝑥) = 𝑥𝑝 = 𝑥 (−𝑥

20+ 50) = −

𝑥2

20+ 50𝑥. Domain is 0 ≤ 𝑥 ≤ 1000.

(C) Find the marginal revenue at a production level of 400 cell phone cases and interpret the results.

Result: 𝑅′(𝑥) = −𝑥

10+ 50. So 𝑅′(400) = −

400

10+ 50 = −40 + 50 = 10. This tells us that if the

company changes from batch size of 400 to batch size of 401, revenue will increase approximately $10.

(D) Find the marginal revenue at a production level of 650 cases and interpret the results.

Result: 𝑅′(650) = −650

10+ 50 = −65 + 50 = −15. Interpretation: This tells us that if the company

changes from batch size of 650 to batch size of 651, revenue will decrease by approximately $15.

.

Day 27 Lecture on Section 19 Increasing and Decreasing Functions

X Section 19 Increasing and Decreasing Functions

Note that there is a correspondence at each 𝑥 = 𝑐:

Sign of 𝑓′(𝑥) at 𝑥 = 𝑐 ↔ Behavior of 𝑓(𝑥) at 𝑥 = 𝑐

• 𝑓′(𝑥) is positive at 𝑥 = 𝑐 ↔ the line tangent to graph of 𝑓(𝑥) at 𝑥 = 𝑐 tilts upward.

• 𝑓′(𝑥) is negative at 𝑥 = 𝑐 ↔ the line tangent to graph of 𝑓(𝑥) at 𝑥 = 𝑐 tilts downward.

• 𝑓′(𝑥) is zero at 𝑥 = 𝑐 ↔ the line tangent to graph of 𝑓(𝑥) at 𝑥 = 𝑐 is horizontal.

Recall Definition of Increasing Function from Day 3 of the semester:

Definition of increasing on an interval

• Words: 𝑓 is increasing on the interval 𝐼.

• Usage: 𝐼 is an interval of the form 𝐼 = (𝑝, 𝑞) or 𝐼 = [𝑝, 𝑞) or 𝐼 = (𝑝, 𝑞] or 𝐼 = [𝑝, 𝑞]. • Meaning: If 𝑝 ≤ 𝑎 < 𝑏 ≤ 𝑞 then 𝑓(𝑎) < 𝑓(𝑏).

• Graphical Description: As one moves from left to right in the interval I, the graph goes up.

Of course, increasing/decreasing behavior of a function 𝑓(𝑥) is easy to determine if one has a graph of

𝑓(𝑥) to look at. But what if one only has a formula for 𝑓(𝑥), not a graph. Is there a way to determine the

increasing/decreasing behavior of 𝑓(𝑥) by analyzing the formula for 𝑓(𝑥)?

Note that there is a correspondence about behavior on an interval:

Sign of 𝑓′(𝑥) on an interval (𝑎, 𝑏) → Behavior of 𝑓(𝑥) on the interval (𝑎, 𝑏).

• 𝑓′(𝑥) is positive on the whole interval (𝑎, 𝑏). → 𝑓(𝑥) is increasing on the whole interval (𝑎, 𝑏).

• 𝑓′(𝑥) is negative on the whole interval (𝑎, 𝑏). → 𝑓(𝑥) is decreasing on the whole interval (𝑎, 𝑏).

• 𝑓′(𝑥) is zero on the whole interval (𝑎, 𝑏). ↔ 𝑓(𝑥) is constant on the whole interval (𝑎, 𝑏).

It is actually a subtle point that some of these arrowheads only go one way. It is possible for a function to

be increasing on a whole interval even though the derivative 𝑓′(𝑥) is not positive on the whole interval.

Draw an example.

So if we want to determine the increasing/decreasing behavior of 𝑓(𝑥), we can do it by studying the sign

behavior of 𝑓′(𝑥) and then using the correspondence above to draw conclusions about the

increasing/decreasing behavior of 𝑓(𝑥).

That leads us to discuss methods for determining sign behavior of functions (Concepts from section 7.3)

Note that a function 𝑔(𝑥) can only change sign at 𝑥 values 𝑥 = 𝑐 where 𝑔(𝑐) = 0 or where 𝑔(𝑥) is

discontinuous. (Draw sample graph illustrating this behavior.)

Definition of Partition Number: The partition numbers for a function 𝑔(𝑥) are the 𝑥 values 𝑥 = 𝑐

where 𝑔(𝑐) = 0 or 𝑔(𝑥) is discontinuous

So a function 𝑔(𝑥) can only change sign at 𝑥 values that are partition numbers for 𝑔. In between those

partition numbers, 𝑔 does not change sign.

We need a methodical way of determining the sign behavior of a function. That method is the sign chart.

To make a sign chart for a function 𝒈(𝒙)

• Find the partition numbers for 𝑔(𝑥). (𝑥 values 𝑥 = 𝑐 where 𝑔(𝑐) = 0 or 𝑔(𝑥) is discontinuous.)

• Make a number line.

• Put the title “Sign chart for 𝑔(𝑥) = 𝑓𝑜𝑟𝑚𝑢𝑙𝑎” above the number line.

• Put the partition numbers for 𝑔(𝑥) on the number line, in order from left to right

• Put the sign behavior of 𝑔(𝑥) above the number line at these partition numbers

• Choose one sample number (one x value) in each of the intervals created by the sample numbers.

• Determine the sign of 𝑔(𝑥) at these sample numbers by substituting the sample number into the

formula for 𝑔(𝑥) and finding the sign of the resulting 𝑦 value.

• The sign of 𝑔(𝑥) at each sample number determines the sign of 𝑔(𝑥) on the whole interval

containing that sample number. Write this sign above the number line in that interval.

Example: Let 𝑔(𝑥) = 6𝑥2 − 6𝑥 − 12. Find the intervals where 𝑔(𝑥) is positive and the intervals where

𝑔(𝑥) is negative.

.

Day 28 Lecture on Section 19 Increasing and Decreasing Functions


Example about determining increasing/decreasing behavior of 𝑓(𝑥) by studying sign behavior of 𝑓′(𝑥):

Let 𝑓(𝑥) = 2𝑥3 − 3𝑥2 − 12𝑥 + 5

(a) Find the 𝑥 values where 𝑓(𝑥) has a horizontal tangent line.

(b) Find the intervals where 𝑓(𝑥) is increasing and the intervals where 𝑓(𝑥) is decreasing.

Local Extrema (Local Max, Local Min)

Definition of Local Max

• words: f has a local max at (𝑐, 𝑓(𝑐))

• meaning:

• The y value 𝑓(𝑐) exists

• for x values near c, 𝑓(𝑥) ≤ 𝑓(𝑐). In other words, the point (𝑐, 𝑓(𝑐)) is the highest point nearby.

• Examples:

Of course, local max & min are easy to see on a graph of 𝑓(𝑥).

Question: When you have a function given by a formula, not by a graph, is there some kind of test that

can be performed on the formula that will find where the max & mins occur?

Answer: Yes, the test is called the First Derivative Test for Local Max/Min

The presentation of the test will be clearer if we first introduce some more terminology

The Terminology of Critical Number, Critical Value, Critical Point

Definition of Critical Number

• Words: 𝑥 = 𝑐 is a critical number for the function 𝑓(𝑥).

• Meaning: Both of these things are true:

o 𝑓(𝑐) exists. (That is, 𝑥 = 𝑐 is in the Domain of 𝑓(𝑥).)

o 𝑓′(𝑐) = 0 or 𝑓′(𝑐) does not exist (That is, 𝑥 = 𝑐 is a Partition Number for 𝑓′(𝑥)

Notice that a critical number for a function 𝑓(𝑥) is an 𝑥 value. The corresponding 𝑦 value and

corresponding point also get special names.

Definition of Critical Value and Critical Point

• A critical value of a function 𝑓(𝑥) is the 𝑦 value 𝑦 = 𝑓(𝑐) corresponding to a critical number 𝑥 = 𝑐.

• A critical point of a function 𝑓(𝑥) is the point (𝑥, 𝑦) = (𝑐, 𝑓(𝑐)) corresponding to a critical number

𝑥 = 𝑐.

Here is the test that can be performed on the formula for a function 𝑓(𝑥) to determine where 𝑓(𝑥) will

have a local max or local min

The First Derivative Test for Local Max & Local Min

If 𝑥 = 𝑐 passes all three of these tests:

1. 𝑓(𝑥) is continuous on some interval containing 𝑥 = 𝑐.

2. 𝑓′(𝑐) = 0 or 𝑓′(𝑐) does not exist.

3. 𝑓′(𝑥) changes sign at 𝑥 = 𝑐.

then 𝑓(𝑥) is guaranteed to have either a local max or local min at 𝑥 = 𝑐.

The coordinates of the max or min will be (𝑥, 𝑦) = (𝑐, 𝑓(𝑐)).

max max no max no max

Day 29 Lecture on Section 19 Finding Local Extrema


Examples about

(a) Determining increasing/decreasing behavior of 𝑓(𝑥) by studying sign behavior of 𝑓′(𝑥).

(b) Finding the local extrema of 𝑓(𝑥) by using the First Derivative Test.

(c) Sketching

Example #1 Let 𝑓(𝑥) = −𝑥3 + 3𝑥2 + 24𝑥 − 17

(a) Find the intervals where 𝑓(𝑥) is increasing and the intervals where 𝑓 is decreasing.

(b) Find the local extrema of 𝑓(𝑥).

(c) Sketch the graph of 𝑓(𝑥). Label all important stuff.

Result:

• local min at (-2,-45)

• local max at (4,63)

Example #2 Let 𝑓(𝑥) = 𝑥4 − 4𝑥3 + 13 Same questions as in previous example.

Result:

• local min at (3,-14)

• horizontal tangent but no extremum at (0,13)

Example #3 Let 𝑓(𝑥) = −𝑥4 + 50𝑥2 + 7. Same questions as in previous example.

Result: 𝑓′(𝑥) = −4𝑥3 + 100𝑥 = −4𝑥(𝑥2 − 25) = −4𝑥(𝑥 + 5)(𝑥 − 5)

• local max at (−5,632).

• local max at (5,632).

• Local min at (0,7).

Day 30 Lecture on Section 17 Higher Derivatives and Section 20 Concavity

X Section 17 Higher Derivatives and Section 20 Concavity

Section 17 Higher Derivatives

Definition of higher order Derivatives

symbols: 𝑓′′′′′(𝑥) or 𝑓(𝑛)(𝑥) or 𝑑𝑛

𝑑𝑥𝑛 𝑓(𝑥)

spoken: the nth derivative of 𝑓(𝑥).

meaning: 𝑑𝑛

𝑑𝑥𝑛 𝑓(𝑥) =𝑑

𝑑𝑥(… (

𝑑

𝑑𝑥(

𝑑

𝑑𝑥𝑓(𝑥)))) Tke the derivative 𝑛 times.

Example #1 Let 𝑓(𝑥) = 𝑥5 − 3𝑥4 + 7𝑥3 − 11𝑥2 + 13𝑥 + 17. Find 𝑓(3)(𝑥). That is, find 𝑓′′′(𝑥).

Observe that when finding higher order derivatives of a polynomial, each successive derivative is one

degree lower than the previous one. So each successive derivative is simpler.

But for more general functions, higher order derivatives can get messy quickly.

Example #2 Let 𝑓(𝑥) = 𝑒(𝑥2+5𝑥+7). Find 𝑓′′(𝑥).

Result:

• We found 𝑓′(𝑥) = 𝑒(𝑥2+5𝑥+7) ⋅ (2𝑥 + 5) using Chain Rule.

• We found 𝑓′′(𝑥)𝑒(𝑥2+5𝑥+7) ⋅ (4𝑥2 + 20𝑥 + 27) using both Product Rule and Chain Rule, but the

Chain Rule part was just the same computation that was just done in computing 𝑓′(𝑥).

Simplification was many steps.

Example #3 Let 𝑓(𝑥) = ln(𝑥2 + 5𝑥 + 7). Find 𝑓′′(𝑥).

Result:

• We found 𝑓′(𝑥) =2𝑥+5

𝑥2+5𝑥+7 using Chain Rule.

• We found 𝑓′′(𝑥) = −2𝑥2+10𝑥+11

(𝑥2+5𝑥+7)2 using Quotient Rule.

Section 20 Concavity

Definition of concave up at a particular 𝑥 = 𝑐

words: 𝑓 is concave up at 𝑥 = 𝑐.

meaning: The graph of 𝑓 has a tangent line at 𝑥 = 𝑐, and for 𝑥-values near 𝑐, the graph of 𝑓 stays above

the tangent line.

pictures:

Associated Terminology:

We say that 𝑓 is concave up on a whole interval (𝑎, 𝑏) if 𝑓 is concave up at all 𝑥 = 𝑐 where 𝑎 < 𝑐 < 𝑏.

We say that 𝑓 has an inflection point at (𝑐, 𝑓(𝑐)) if

• The point (𝑐, 𝑓(𝑐)) exists.

• The concavity of 𝑓 changes (from down to up or from up to down) at that point.

[Class Drill] Identifying Three Kinds of Graph Behavior See Next Page


f is concave up at x=3

x=3

f is concave up at x=3

x=3

f is not concave up or

concave down at x=3

x=3

m=0

m=0

Class Drill: Identifying Three Kinds of Graph Behavior

The graph of a function 𝑓 is shown below.

(1) At which 𝑥 values is 𝑓 zero?

(2) On what intervals is 𝑓 positive?

(3) On what intervals is 𝑓 positive

(4) At which 𝑥 values the line tangent to the graph of 𝑓 horizontal?

(5) On what intervals is 𝑓 increasing?

(6) On what intervals is 𝑓 decreasing?

(7) On what intervals is 𝑓 concave up?

(8) On what intervals is 𝑓 concave down?

(9) At which 𝑥 values is 𝑓 not concave?

(10) At which 𝑥 values does 𝑓 have a point of inflection?

x

f (x)

Lecture 31 Final lecture on Concavity (Book Section 20)

X Lecture 31 Final lecture on Concavity (Book Section 20)

We will start by discussing the relationship between the

behavior of derivatives of 𝑓(𝑥) and the concavity of 𝑓(𝑥)

Draw three examples of functions 𝑓(𝑥) that are concave up.

Draw a few tangent lines on the graph of each function.

Describe the tangent line slopes (pos or neg? large, medium, small?

And use that info about the tangent line slopes to draw a graph of 𝑓′(𝑥) below the graph of 𝑓(𝑥).

Observe common behavior: all graphs of 𝑓′(𝑥) are increasing. We recognize that there is a

correspondence:

increasing/decreasing behavior of 𝑓′(𝑥) ➔ concavity behavior of of 𝑓(𝑥)

• 𝑓′(𝑥) increasing on whole interval (𝑎, 𝑏) ➔ 𝑓(𝑥) concave up on the whole interval (𝑎, 𝑏).

• 𝑓′(𝑥) decreasing on whole interval (𝑎, 𝑏) ➔ 𝑓(𝑥) concave down on the whole interval (𝑎, 𝑏).

How would 𝑓′′ fit into this? Remember that 𝑓′′ is the derivative of 𝑓′, so we there are correspondences

between behavior of 𝑓′′(𝑥) and behavior of of 𝑓′(𝑥). We can combine the correspondences.

sign behavior of 𝑓′′(𝑥) ➔ increasing/decreasing behavior of 𝑓′(𝑥) ➔ concavity behavior of of 𝑓(𝑥)

• 𝑓′′(𝑥) positive on interval (𝑎, 𝑏) ➔ 𝑓′(𝑥) increasing on (𝑎, 𝑏) ➔ 𝑓(𝑥) concave up on (𝑎, 𝑏).

• 𝑓′′(𝑥) negative on interval (𝑎, 𝑏) ➔ 𝑓′(𝑥) decreasing on (𝑎, 𝑏) ➔ 𝑓(𝑥) concave down on (𝑎, 𝑏).

Examples involving Given graph of 𝑓 ➔ what are the signs of 𝑓, 𝑓′, 𝑓′′?

Example (a)

• 𝑓 is + because graph is above axis.

• 𝑓′ is – because 𝑓 is decreasing (all tangent lines on graph of 𝑓 have negative slope).

• 𝑓′′ is + because 𝑓 is concave up.

Example (b)

• 𝑓 is - because graph is below axis.

• 𝑓′ is + because 𝑓 is increasing.

• 𝑓′′ is - because 𝑓 is concave down.

Example: Given descriptions of the signs of 𝒇, 𝒇′, 𝒇′′, sketch 𝒇.

Suppose 𝑓 is pos, 𝑓′ is pos, 𝑓′′ is neg, sketch a possible graph of 𝑓.

Solution: Start by translating given info about signs of 𝑓, 𝑓′, 𝑓′′ into

info about the graph of 𝑓,

• 𝑓 is + ➔ graph of 𝑓 should be above axis.

• 𝑓′ is + ➔ graph of 𝑓 should be increasing.

• 𝑓′′ is - ➔ graph of 𝑓 should be concave down.

Then sketch graph of 𝑓. Possible result is shown at right


Example (b) Example (a)

Lecture 31, continued

X Lecture 31, continued

Example: Given sign charts for 𝒇′, 𝒇′′, sketch 𝒇.

Suppose 𝑓 is a continuous function and that 𝑓(𝑥) positive.

And suppose the 𝑓′ and 𝑓′′ have these sign charts (present these sign charts on number lines)

• Sign chart for 𝑓′ (𝑥 (−∞, 5) 5 (5,15) 15 (15, ∞)

𝑓′ − − − 0 + + + 0 − − −).

• Sign chart for 𝑓′′ (𝑥 (−∞, 10) 10 (10, ∞)

𝑓′′ + + + 0 − − −).

Sketch a graph of 𝑓(𝑥). Solution:

The sign charts for 𝑓′, 𝑓′′ have partition numbers 𝑥 = 5,10,15.

We can put these numbers on one number line and gather up the info that we have about the signs of

𝑓, 𝑓′, 𝑓′′ in each of the intervals.

From this info, we can make conclusions about the behavior of the behavior of the graph of 𝑓 in each of

the intervals.

That will allow us to envision the shape of the graph of 𝑓 in each interval.

We can stitch these pieces together to build a graph of 𝑓(𝑥).

.

Ohio University moved all of its 2019 – 2020 Spring Semester courses to an online format during Spring Break,

in March 2020.

Material from this point onward in this outline will be organized by Day XX, rather than by Lecture XX.

(This begins with Day 32. Past Lecture 1 through Lecture 31 will continue to be referred to as Lectures.)

The outline could be used for producing videos or reading material to be posted online.

Day 32 Analyzing Functions to Determine Concavity; Second Derivative Test (Section 20)

X Day 32 Analyzing Functions to Determine Concavity; Second Derivative Test (Section 20)

Analyzing Functions to determine Concavity

In Lecture 31 (Book Section 20), we discussed the following correspondence:

sign behavior of 𝑓′′(𝑥) ➔ increasing/decreasing behavior of 𝑓′(𝑥) ➔ concavity behavior of of 𝑓(𝑥)

• 𝑓′′(𝑥) positive on interval (𝑎, 𝑏) ➔ 𝑓′(𝑥) increasing on (𝑎, 𝑏) ➔ 𝑓(𝑥) concave up on (𝑎, 𝑏).

• 𝑓′′(𝑥) negative on interval (𝑎, 𝑏) ➔ 𝑓′(𝑥) decreasing on (𝑎, 𝑏) ➔ 𝑓(𝑥) concave down on (𝑎, 𝑏).

In that lecture (and in the accompanying exercises), we explored this from a graphical perspective.

Today, we will explore this correspondence analytically.

That is, we will investigate the concavity of functions 𝑓(𝑥) that are given by formulas, not by graphs.

[Example 1 (easy)] Let 𝑓(𝑥) = 𝑥3 − 3𝑥2 − 9𝑥 − 5.

(a) Find the intervals where 𝑓 is concave up, and the intervals where 𝑓 is concave down. Present your

answers in three ways: inequality notation, interval notation, and pictures involving the number line

Be sure to present the answers in sentences.

(b) Find all inflection points in the graph of 𝑓. Be sure to present your answer in a sentence that explains

how you know that those are inflection points.

Figures for the solution of [Example 1]


𝑓′′(1) = 0

Sign Chart for 𝑓′′(𝑥) = 6(𝑥 − 1)

𝑥 = 1

partition

𝑓′′ 𝑛𝑒𝑔 − − − − − −

𝑥 = 0

sample

𝑓′′ 𝑝𝑜𝑠 + + + + + +

𝑥 = 2

sample

𝑥 = 1

𝑥 = 1

𝑓(𝑥) is concave down on the red set.

𝑓(𝑥) is concave up on the red set.


X Day 32, continued

[Example 2 (moderate)] Let 𝑓(𝑥) = ln(𝑥2 + 9). Answer the same questions as in [Example 1].

Figures for the solution of [Example 1]


𝑓′′(−3) = 0

Sign Chart for 𝑓′′(𝑥) = −2(𝑥+3)(𝑥−3)

(𝑥2+9)2

𝑥 = −3

partition

𝑓′′ 𝑛𝑒𝑔 − − − − − −

𝑥 = −4

sample

𝑓′′ 𝑝𝑜𝑠 + + + + + +

𝑥 = 0

sample

𝑥 = −3

𝑓(𝑥) is concave down on the red set.

𝑓′′(3) = 0

𝑥 = 3

partition

𝑓′′ 𝑛𝑒𝑔 − − − − − −

𝑥 = 4

sample

𝑥 = 3

𝑥 = −3

𝑓(𝑥) is concave up on the red set.

𝑥 = 3


X Day 32 Analyzing Functions to Determine Concavity; Second Derivative Test (Section 20)

Second Derivative Test

In Lecture 28 (Book Section 19), we used the 1st Derivative Test to locate local extrema.

The concept of concavity can be used to formulate a 2nd Derivative Test for local extrema

Second Derivative Test for Local Extrema

• If a function 𝑓(𝑥) has 𝑓′(𝑐) = 0 and 𝑓′′(𝑐) > 0, then 𝑓 has a local min at the point (𝑐, 𝑓(𝑐)).

• If a function 𝑓(𝑥) has 𝑓′(𝑐) = 0 and 𝑓′′(𝑐) < 0, then 𝑓 has a local max at the point (𝑐, 𝑓(𝑐)).

• If a function 𝑓(𝑥) has 𝑓′(𝑐) = 0 and 𝑓′′(𝑐) = 0, then this is not enough information to determine

whether 𝑓 has a local max, local min, or neither at the point (𝑐, 𝑓(𝑐)).

[Example 3] Consider two functions that illustrate the cases when the Second Derivative Test says there

is a min or max

Let 𝑓(𝑥) = 𝑥2 + 1.

• Then 𝑓′(𝑥) = 2𝑥, and 𝑓′′(𝑥) = 2

• Then 𝑓′(0) = 2(0) = 0, and 𝑓′′(0) = 2 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒

• The second derivative test tells us that 𝑓 has a local min at the point (0, 𝑓(0)) = (0,1). The graph

below confirms this.

Let 𝑔(𝑥) = −𝑥2 + 1.

• Then 𝑔′(𝑥) = −2𝑥, and 𝑔′′(𝑥) = −2

• Then 𝑔′(0) = −2(0) = 0, and 𝑔′′(0) = −2 = 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒

• The second derivative test tells us that 𝑔 has a local max at the point (0, 𝑓(0)) = (0,1). The graph

below confirms this.


𝑓(𝑥) 𝑔(𝑥)

Graphs for Day 32 [Example 3]

Day 32, continued

X Day 32, continued

[Example 4] Now consider three functions that illustrate the case when the Second Derivative Test does

not say whether there is a min or max

Let 𝑓(𝑥) = 𝑥4 + 1.

• Then 𝑓′(𝑥) = 4𝑥3, and 𝑓′′(𝑥) = 12𝑥2

• So 𝑓′(0) = 4(0)3 = 0, and 𝑓′′(0) = 12(0)2 = 0

• The second derivative test does not tell us whether 𝑓 has a local min or max or neither at 𝑥 = 0.

• From the graph below, we can see that 𝑓 has a local min.

Let 𝑔(𝑥) = −𝑥4 + 1.

• Then 𝑔′(𝑥) = −4𝑥3, and 𝑔′′(𝑥) = −12𝑥2

• So 𝑔′(0) = −4(0)3 = 0, and 𝑔′′(0) = −12(0)2 = 0

• The second derivative test does not tell us whether 𝑔 has a local min or max or neither at 𝑥 = 0.

• From the graph below, we can see that 𝑔 has a local max.

Let ℎ(𝑥) = 𝑥3 + 1.

• Then ℎ′(𝑥) = 3𝑥2, and ℎ′′(𝑥) = 6𝑥

• So ℎ′(0) = 3(0)2 = 0, and ℎ′′(0) = 6(0) = 0

• The second derivative test does not tell us whether ℎ has a local min or max or neither at 𝑥 = 0.

• From the graph below, we can see that ℎ has neither a local min nor max there.

Now we will try out the 2nd Derivative test on more sophisticated functions. We will revisit the functions

from our earlier examples.

[Example 5] Return to the function 𝑓(𝑥) = 𝑥3 − 3𝑥2 − 9𝑥 − 5 from [Example 1].

Use the second derivative test to find all local extrema.

Example 6] Return to the function 𝑓(𝑥) = ln(𝑥2 + 9) from [Example 2].

Use the second derivative test to find all local extrema.

𝑓(𝑥) 𝑔(𝑥) ℎ(𝑥)

Graphs for Day 32 [Example 4]

Day 33 Global Extrema on Closed Intervals (Section 21)

X Day 33 Global Extrema on Closed Intervals (Section 21)

Introduction: On Days 28 and 29, we discussed the idea of Local Extrema. That is, Local Max and Local

Min. These are points that are the highest (or lowest) of all points nearby. (content from Section 19 of the

online Ximera Business Calculus textbook.) Today, we will discuss Global Extrema. This content is

discussed in Section 21 of the online textbook.

Definition of Absolute Max Value

• Words: The absolute max value of 𝑓(𝑥) on the domain 𝐷.

• Meaning: the 𝑦 value that has these two properties:

o 𝑦 = 𝑓(𝑐) for some 𝑥 = 𝑐 in the domain 𝐷.

o 𝑓(𝑐) ≥ 𝑓(𝑥) for all 𝑥 in the domain 𝐷.

Related Definition of Absolute Min Value

• Words: The absolute min value of 𝑓(𝑥) on the domain 𝐷.

• Meaning: the 𝑦 value that has these two properties:

o 𝑦 = 𝑓(𝑐) for some 𝑥 = 𝑐 in the domain 𝐷.

o 𝑓(𝑐) ≤ 𝑓(𝑥) for all 𝑥 in the domain 𝐷.

Remarks:

• The absolute max (or min) value is a 𝑦 value.

• For a particular 𝑓(𝑥) and domain 𝐷, there may or may not be an absolute max (or min) value.

[Example 1] Graphical Examples using the graph shown.

[Example 1a] Using the domain 𝐷 = [11,15]. • The global max value on the domain 𝐷 = [11,15]

is 𝑦 = 7. It occurs at 𝑥 = 15.

• The global min value on the domain 𝐷 = [11,15] is 𝑦 = 4. It occurs at 𝑥 = 13.

Observations gleaned from this example:

• A global extremum can occur at an endpoint

[Example 1b] Using the domain 𝐷 = (11,15).

• The global min value on the domain 𝐷 = (11,15)

is 𝑦 = 4. It occurs at 𝑥 = 13.

• There is no global max value on domain 𝐷 =(11,15). (Note that the global max value cannot be 𝑦 = 7, because there is no 𝑥 = 𝑐 in the

domain 𝐷 = (11,15) that causes 𝑓(𝑐) = 7. And the global max value cannot be 𝑦 = 6, because

there are some 𝑥 values in the domain 𝐷 = (11,15) that have y values larger than 𝑦 = 6.)


• The global max value and global min value depend both on the function 𝑓(𝑥) and on the domain

𝐷: If you change the domain, the global max & min may change (or even not exist)

• For a particular 𝑓(𝑥) and domain 𝐷, there might not be a global max value or a global min value.

[Example 1c] Using the domain 𝐷 = [9,15]. • The global max value on the domain 𝐷 = [9,15] is 𝑦 = 7. It occurs at 𝑥 = 15.

• The global min value on the domain 𝐷 = [9,15] is 𝑦 = 4. It occurs at 𝑥 = 9 and at 𝑥 = 13.


• A global extremum can occur at more than one 𝑥 value. But that does not mean that there is more

than one global extremum. Remember, a global max or min is a 𝑦 value. In this example, there is

only one global min value: the 𝑦 value 𝑦 = 4. It happens to occur at two 𝑥 values. But there is

only one global min value.


𝑓(𝑥)

Graph for Day 33 [Example 1]

(15,7)

(13,4)

(11,6)

(9,4)

Day 33 Global Extrema on Closed Intervals (Section 21), continued

X Day 33 Global Extrema on Closed Intervals (Section 21), continued

Global max and min values for a function 𝑓 are easy to determine if one has a graph of the function. Like

so many of our concepts in Calculus, we are interested in figuring out a way to analyze the formula for a

function 𝑓 to determine this behavior. The key facts that will enable us to do this are two theorems that

describe behavior that we have actually observed happen in [Example 1].

Theorem 1 (about where Global Extrema can occur)

For a function 𝑓(𝑥) and domain 𝐷, global max and global min values can only occur at 𝑥 values that are

critical numbers of 𝑓 or that are endpoints of the domain 𝐷. (Note that if domain 𝐷 does not have

endpoints, then what this theorem tells us is that the global max and min values can only occur at 𝑥

values that are critical numbers.)

Theorem 2 (Extreme Value Theorem)

If the domain 𝐷 is a closed interval [𝑎, 𝑏] and the function 𝑓(𝑥) is continuous on that closed interval,

then there will be a global max value and a global min value for 𝑓(𝑥) on the domain 𝐷.

These two theorems give us enough information to determine the global max and min values in the

particular situation where we have a domain 𝐷 that is a closed interval [𝑎, 𝑏] and a function 𝑓(𝑥) that is

known to be continuous on that closed interval. That is,

• We know (by Theorem 2) that there will be a global max value and a global min value for 𝑓(𝑥)

on the domain 𝐷.

• And we know (by Theorem 1) where to look for those global extrema: We need only look at 𝑥

values that are critical numbers of 𝑓 or that are endpoints of the domain 𝐷.

That is the essence of the Closed Interval Method for finding Global Extrema.

The Closed Interval Method

(used for Finding Global Extrema of a Continuous Function on a Closed Interval)

1. Verify that you are allowed to use the Closed Interval Method

• Confirm that the domain 𝐷 is a closed interval [𝑎, 𝑏]. Identify the endpoints 𝑥 = 𝑎, 𝑥 = 𝑏.

• Confirm that the function 𝑓(𝑥) is continuous on that closed interval. (Say how you know.)

2. Find the critical numbers for the function 𝑓(𝑥).

3. Make a list of the important 𝑥 values and find the corresponding 𝑦 values. 𝐼𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑡 𝑥 𝑣𝑎𝑙𝑢𝑒𝑠 (𝑎𝑛𝑑 𝑤ℎ𝑦 𝑎𝑟𝑒 𝑡ℎ𝑒𝑦 𝑖𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑡? ) 𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑦 𝑣𝑎𝑙𝑢𝑒𝑠

𝑥 = 𝑎 (𝑒𝑛𝑑𝑝𝑜𝑖𝑛𝑡) 𝑓(𝑎)

𝑥 = 𝑐 (𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙) 𝑓(𝑎)⋮ ⋮ ⋮

𝑥 = 𝑐 (𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙) 𝑓(𝑏)

𝑥 = 𝑏 (𝑒𝑛𝑑𝑝𝑜𝑖𝑛𝑡) 𝑓(𝑏)

4. Identify the greatest and least of the y values. These are the global max value and global min

value. Use them to write the conclusion clearly in sentences like this:

• The global max value is 𝑦 = 𝑠𝑜𝑚𝑒 𝑣𝑎𝑙𝑢𝑒 and it occurs at 𝑥 = 𝑠𝑜𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟.

• The global min value is 𝑦 = 𝑠𝑜𝑚𝑒 𝑣𝑎𝑙𝑢𝑒 and it occurs at 𝑥 = 𝑠𝑜𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟.

[Example 2] Find global extrema of 𝑓(𝑥) = 𝑥3 + 3𝑥2 − 9𝑥 + 7 on the domain 𝐷 = [−4,4]. Solution: Following the steps in the Closed Interval Method, we find

• Global max value of 𝑓 on domain 𝐷 = [−4,4] is 𝑦 = 83, which occurs at endpoint 𝑥 = 4.

Global min value of 𝑓 on domain 𝐷 = [−4,4] is 𝑦 = 2, which occurs at critical number 𝑥 = 1.

By now, you have noticed that the Closed Interval Method involves a lot of computing: One must find the

critical numbers, and then one must compute all of those 𝑦 values. But it is important to notice what one

does not do in the Closed Interval Method. One does not make a sign chart for 𝑓′(𝑥) and use it to

determine where 𝑓(𝑥) is increasing and decreasing to determine the location of maxs and mins (All the

stuff we did in Section 19 when finding local maxs and local mins.) In the Closed Interval Method, we

just make the list of important 𝑥 values, compare their corresponding 𝑦 values, and write a conclusion.


Day 33 Global Extrema on Closed Intervals (Section 21), continued

X Day 33 Global Extrema on Closed Intervals (Section 21), continued

[Example 2] Find global extrema of 𝑓(𝑥) = 𝑥3 + 3𝑥2 − 9𝑥 + 7 on the domain 𝐷 = [−4,4]. Solution: Following the steps in the Closed Interval Method, we find

• Global max value of 𝑓 on domain 𝐷 = [−4,4] is 𝑦 = 83, which occurs at endpoint 𝑥 = 4.

• Global min value of 𝑓 on domain 𝐷 = [−4,4] is 𝑦 = 2, which occurs at critical number 𝑥 = 1.

[Example 3] Consider the job of finding global extrema of 𝑓(𝑥) = 𝑥3 − 6𝑥 + 5 on domain 𝐷 = [−3,2]. What about Alice’s solution involving just plugging in all the 𝑥 values?

𝐿𝑖𝑠𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑥 𝑣𝑎𝑙𝑢𝑒𝑠 𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑦 𝑣𝑎𝑙𝑢𝑒𝑠

𝑥 = −3 𝑓(−3) = −4

𝑥 = −2 𝑓(−2) = 9

𝑥 = −1 𝑓(−1) = 10

𝑥 = 0 𝑓(0) = 5

𝑥 = 1 𝑓(1) = 0

𝑥 = 2 𝑓(2) = 1

Based on this list, Alice claims that

• The global max value is 𝑦 = 10 and it occurs at 𝑥 = −1.

• The global min value is 𝑦 = 0 and it occurs at 𝑥 = 1.

Is Alice’s solution valid? Observe that Alice has not actually checked all of the 𝑥 values. She has only

checked the integer 𝑥 values. She has not found the critical numbers.

It turns out that the critical numbers are 𝑥 = −√2 ≈ −1.414 and 𝑥 = √2 ≈ 1.414. Using this

information, the Closed Interval Method would proceed as follows. 𝐿𝑖𝑠𝑡 𝑜𝑓 𝑖𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑡 𝑥 𝑣𝑎𝑙𝑢𝑒𝑠 𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑦 𝑣𝑎𝑙𝑢𝑒𝑠

𝑥 = −3 𝑓(−3) = −4

𝑥 = −√2 𝑓(−√2) = 5 + 4√2 ≈ 10.657

𝑥 = √2 𝑓(√2) = 5 − 4√2 ≈ −0.657

𝑥 = 2 𝑓(2) = 1

Based on this list, the correct conclusion is

• the global max value is 𝑦 = 5 + 4√2 ≈ 10.657, which occurs at 𝑥 = −√2.

• the global min value is 𝑦 = 5 − 4√2 ≈ −0.657, which occurs at 𝑥 = √2.

So we see that Alice did not actually find the global min value or global max value, because they did not

occur at integer 𝑥 values.

And what if the global min value and global max value did occur at integer 𝑥 values? For instance, what

if Alice did her method of just trying all the integer 𝑥 values for the function 𝑓 and domain 𝐷 from

[Example 2]. Well, in that case, she would have stumbled up the correct answers for the global min value

and the global max value, because they occur at 𝑥 = 1 and 𝑥 = 4, which are integers (and so they would

be on Alice’s list of 𝑥 values). But because she did not find the critical numbers, Alice would not really

have proven that those were actually the correct global max value and global min value.

In conclusion: To find the global extreme values of a continuous function 𝑓 on a closed interval 𝐷, one

must find the critical numbers, make the list of all important 𝑥 values, and compare the corresponding 𝑦

values. It is not a valid solution to just make a list of the integer 𝑥 values in the interval, and find the 𝑦

values at those 𝑥 values.

End of Lecture Notes for Day 33 Global Extrema on Closed Intervals (Section 21).

.

Day 34 Limits and Asymptotes (Section 22)

X Day 34 Limits and Asymptotes (Section 22)

Part 1: Limits and Vertical Asymptotes (content from Section 22.2)

Most of this part is a repeat of discussion from Class Days 9,10 (Jan 28, 29) (book section 6.3)

Definition of Infinite Limits

symbol: lim𝑥→𝑐

𝑓(𝑥) = ∞

spoken: The limit, as 𝑥 approaches 𝑐, of 𝑓(𝑥), is infinity.

Meaning: as 𝑥 gets closer & closer to 𝑐 but not equal to 𝑐, the 𝑦-values get more & more positive

without bound.

Graphical Significance: Graph of 𝑓 has a vertical asymptote at 𝑥 = 𝑐 and the graph goes up on

both sides of the asymptote. Note that the line equation for the asymptote is 𝑥 = 𝑐.

Obvious variations: lim𝑥→𝑐+

𝑓(𝑥) = ∞ or lim𝑥→𝑐

𝑓(𝑥) = −∞, etc. Ask the students to explain.

We did an Activity for Day 8: Infinite Limits for a Function Given by a Graph

Then investigated limits for a function given by a formula.

[Example 1] involved 𝑓(𝑥) =1

𝑥−6 investigating behavior near 𝑥 = 6 by making tables of 𝑥, 𝑦 values.

Compared our results to a computer graph.

[Example 2] involving 𝑓(𝑥) =𝑥−2

𝑥−6

Investigate the behavior near 𝑥 = 2.

• found the value of 𝑓(2) = 0. Tells us graph has point at (𝑥, 𝑦) = (2,0).

• found lim𝑥→2

𝑓(𝑥) = 0 using standard limit techniques from Sections 4 and 5 of the textbook. Tells

us graph is heading for the location (𝑥, 𝑦) = (2,0).

Now investigate the behavior near 𝑥 = 6.

We find 𝑓(6) does not exist. So there is no point on the graph at 𝑥 = 6.

Investigate the behavior near 𝑥 = 6 by considering the limits as 𝑥 approaches 6.

First, we consider that using the concept of limit from Sections 4, 5, all three limits lim𝑥→6+

𝑥−2

𝑥−6 and lim

𝑥→6−

𝑥−2

𝑥−6

and lim𝑥→6

𝑥−2

𝑥−6 do not exist, because they are limits of the form

𝑛𝑜𝑛𝑧𝑒𝑟𝑜

0.

Now, reconsider the limits as 𝑥 approaches 6 allowing for infinite behavior. This time, we do not

investigate by making tables of 𝑥, 𝑦 values. Rather, this time, we consider the size of the numerator, the

size of the denominator, and the size of the ratio.

Result: lim𝑥→6+

𝑥−2

𝑥−6= ∞ and lim

𝑥→6−

𝑥−2

𝑥−6= −∞ and lim

𝑥→6

𝑥−2

𝑥−6 does not exist.

Compared to computer graph.

Part 2 Limits at Infinity, and and their relation to Horizontal Asymptotes (Section 22.3 Concepts)

Definition of Limits at Infinity:

• symbol: lim𝑥→∞

𝑓(𝑥) = 𝑏

• spoken: The limit, as 𝑥 goes to infinity, of 𝑓(𝑥), is 𝑏.

• Meaning: as 𝑥 gets more & more positive without bound, the 𝑦-values get closer & closer to 𝑏.

• Graphical Significance: Graph of 𝑓 has a horizontal asymptote on the right at 𝑦 = 𝑏. Note that

the line equation for the asymptote is 𝑦 = 𝑏.

• Draw picture: Graph with horizontal asymptote

• Obvious variations:

• lim𝑥→−∞

𝑓(𝑥) = 𝑏 Graph of 𝑓 has a horizontal asymptote on the left at 𝑦 = 𝑏.

• lim𝑥→∞

𝑓(𝑥) = ∞ Graph of 𝑓 goes up on the right (and does not have a horizontal asymptote).

• lim𝑥→∞

𝑓(𝑥) = 𝐷𝑁𝐸 Graph of 𝑓 does not have a simple behavior on the right.


Day 34 Limits and Asymptotes (Section 22)continued

X Day 34 Limits and Asymptotes (Section 22), continued

[Example 3] again involving 𝑓(𝑥) =𝑥−2

𝑥−6, whose graph we saw above in [Example 2]. Observe that its

graph has horizontal asymptote on the right. That would mean that lim𝑥→∞

𝑓(𝑥) = 𝑏, where 𝑦 = 𝑏 is the

equation of the line that is the horizontal asymptote.

An obvious question is, what is the value of 𝑏? That is, what is the value of lim𝑥→∞

𝑥−2

𝑥−6?

Study by making a table of 𝑥, 𝑦 values.

Result: We see that as 𝑥 gets more & more positive without bound, the 𝑦-values get closer & closer to 1.

Using limit notation to abbreviate this, we write lim𝑥→∞

𝑥−2

𝑥−6= 1. That tells us that 𝑏 = 1.

So the line equation for the horizontal asymptote is 𝑦 =

1. Another obvious question is, can we find lim

𝑥→∞

𝑥−2

𝑥−6 analytically, not using a table of 𝑥, 𝑦 values?

The answer is yes, but it involves a new technique.

The technique for finding the lim𝑥→∞

𝑓(𝑥) of a rational function 𝑓(𝑥) =𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙

𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑝𝑜𝑙𝑦𝑛𝑜𝑚𝑖𝑎𝑙 is based on

the key observation about two very simple forms of limits.

Important Simple Limit #1: If 𝑘 is a constant, then lim𝑥→∞

𝑘 = 𝑘. Discuss why.

Important Simple Limit #2:

If 𝑘 is a constant and 𝑝 is a constant such that 𝑝 > 0, then lim𝑥→∞

𝑘

𝑥𝑝 = 0. Discuss why.

The Divide by 𝒙𝒑 Method for finding the 𝐥𝐢𝐦𝒙→∞

𝒇(𝒙) of a rational function 𝒇(𝒙).

• Identify the highest power of 𝑥 that appears in the denominator Call it 𝑥𝑝. • Divide every term in the rational function by 𝑥𝑝. This does not change the value of the function,

but it does change the way the function 𝑓(𝑥) looks: it will be made up of some constant terms and

a bunch of terms of the form 𝑘

𝑥𝑝.

• Take the lim𝑥→∞

𝑓(𝑥) by taking the limit of each of the terms in the expression for 𝑓(𝑥). (Those

constant terms and those terms of the form 𝑘

𝑥𝑝.) Using the results about Simple Limit #1 and

Simple Limit #2 above, we see that the limits of each of the terms in the expression for 𝑓(𝑥) will

be 𝑘 or 0.

• Make a conclusion about the value of lim𝑥→∞

𝑓(𝑥).

our examples involve using the Divide by 𝒙𝒑 Method.

[Example 4] Find lim𝑥→∞

𝑥−2

𝑥−6 analytically using the Divide by 𝒙𝒑 Method.


7𝑥2−42𝑥+35

2𝑥2−16𝑥+30 analytically using the Divide by 𝒙𝒑 Method.


7𝑥2−42𝑥+35

2𝑥3−16𝑥2+30𝑥 analytically using the Divide by 𝒙𝒑 Method.


7𝑥3−42𝑥2+35𝑥

2𝑥2−16𝑥+30 analytically using the Divide by 𝒙𝒑 Method. (skip in class)


Day 34 Limits and Asymptotes (Section 22)continued

X Day 34 Limits and Asymptotes (Section 22), continued

Conclusion of Discussion of Limits at Infinity and Horizontal Asymptotes

We have observed that

• The standard form of a rational function is most useful for determining behavior as 𝑥 → ∞.

• If lim𝑥→∞

𝑓(𝑥) = 𝑏, then the graph of 𝑓 has a horizontal asymptote on the right at 𝑦 = 𝑏. That is, the

line equation of the asymptote is 𝑦 = 𝑏.

• If lim𝑥→∞

𝑓(𝑥) = ∞, then graph goes up on the right. There is no horizontal asymptote on the right.

Furthermore, we observed the following behavior in [Examples 3,4,5,6,7]

Summary of the 𝐥𝐢𝐦𝒙→∞

𝒇(𝒙) of a Rational Function 𝒇(𝒙).

Suppose a rational function is of the form 𝑓(𝑥) =𝑎𝑥𝑝+𝑙𝑜𝑤𝑒𝑟 𝑝𝑜𝑤𝑒𝑟 𝑡𝑒𝑟𝑚𝑠

𝑏𝑥𝑞+𝑙𝑜𝑤𝑒𝑟 𝑝𝑜𝑤𝑒𝑟 𝑡𝑒𝑟𝑚𝑠

• If 𝑝 = 𝑞, then lim𝑥→∞

𝑓(𝑥) =𝑎

𝑏 and the graph of 𝑓(𝑥) has horiz asymptote on right with equation 𝑦 =

𝑎

𝑏

• If 𝑝 < 𝑞, then lim𝑥→∞

𝑓(𝑥) = 0 and the graph of 𝑓(𝑥) has horiz asymptote on right with equation 𝑦 = 0

• If 𝑝 > 𝑞, then lim𝑥→∞

𝑓(𝑥) = ∞ and the graph of 𝑓(𝑥) goes up on the right.

[Example 8] For each rational function 𝑓(𝑥), do the following two things:

(i) Find lim𝑥→∞

𝑓(𝑥) using the Summary of the 𝐥𝐢𝐦𝒙→∞

𝒇(𝒙) of a Rational Function 𝒇(𝒙) presented

above. (Don’t do the the Divide by 𝒙𝒑 Method.)

(ii) Does the graph of 𝑓(𝑥) have any horizontal asymptotes? If so, give their line equations.

(a) Let 𝑓(𝑥) =5𝑥7−8𝑥+43

3𝑥4+𝑥2+23

(b) Let 𝑓(𝑥) =5𝑥7−8𝑥+43

3𝑥7+𝑥2+23

(c) Let 𝑓(𝑥) =5𝑥4−8𝑥+43

3𝑥7+𝑥2+23

End of Lecture Outline for Day 34 Limits and Asymptotes (Section 22)

Lecture Outline for Day 35 Graphing (Section 23, 24)

X Day 35 Graphing (Section 23, 24)

Motivation for learning how to graph by hand:

• Leads to better understanding of how functions work.

• Even when we do use a computer to make a graph, we will understand why it looks the way it

does.

Vague description of the graphing strategy: Given a function 𝑓(𝑥),

• Start by finding answers to the simplest questions about 𝑓(𝑥) that will give info about the graph

of 𝑓(𝑥).

• Then analyze 𝑓′(𝑥) to get information about the graph of 𝑓(𝑥).

• Then analyze 𝑓′′(𝑥) to get information about the graph of 𝑓(𝑥).

• Finally, assemble all the known info about the graph of 𝑓(𝑥) and sketch the graph.

Project More Detailed Version of the Graphing Strategy from MATH 1350 Web page. (see next page)

[Example 1] (Most of the analysis of 𝑓, 𝑓′, 𝑓′′ has been done.)

Given the following information about 𝑓, 𝑓′, 𝑓′′, use the Graphing Strategy to sketch a possible graph of

𝑓(𝑥).

Be sure to label all important points with their (𝑥, 𝑦) coordinates.

𝑥 10 20 30

𝑓(𝑥) 7 14 21

[Example 2] (Analysis of 𝑓, 𝑓′, 𝑓′′ has not been done.)

Use the Graphing Strategy to graph the function 𝑓(𝑥) = 𝑥3 − 3𝑥2 − 9𝑥 − 5 = (𝑥 + 1)2(𝑥 − 5).

[Example 23 Use the Graphing Strategy to graph the function 𝑓(𝑥) = 𝑥4 − 4𝑥3.

[Example 4] The graph of a function 𝑓(𝑥) is shown at right. Notice that the coordinate axes are not

shown.

The formulas for 𝑓(𝑥) and its derivatives

are

𝑓(𝑥) = 3 + ln(𝑥2 + 1)

𝑓′(𝑥) =2𝑥

𝑥2 + 1

𝑓′′(𝑥) =−2(𝑥2 − 1)

(𝑥2 + 1)2=

−2(𝑥 + 1)(𝑥 − 1)

(𝑥2 + 1)2

(A) It looks like there is a local min on the graph. What are the (𝑥, 𝑦) coordinates of that point?

(B) It looks like there are two inflection points on the graph. What are their (𝑥, 𝑦) coordinates?


x

𝑓′ = 0

𝑥 = 10

𝑓′ − − − − −

𝑓′ + + + + +

𝑓′ = 0

𝑥 = 30

𝑓′

+ + + + +

x

𝑓′′ = 0

𝑥 = 20

𝑓′′ + + + + +

𝑓′′ − − − − −

𝑓′′ = 0

𝑥 = 30

𝑓′′

+ + + + +

Lecture Outline for Day 35 Graphing (Section 23, 24), continued

The Graphing Strategy (Concepts from Sections 23, 24)

Given a function 𝑓(𝑥), do any of these steps that are easy. Otherwise skip.

• Analyze 𝑓(𝑥) to get info about the graph of 𝑓(𝑥).

o Find the 𝑦 intercept by setting 𝑥 = 0 and finding 𝑦.

o Find all 𝑥 intercepts by setting 𝑦 = 0 and finding 𝑥.

o Make a sign chart for 𝑓 to determine where 𝑓 lies above or below the 𝑥 axis. (Label your sign

chart!)

o Will there be any vertical asymptotes in the graph of 𝑓(𝑥)? If so, what are their line equations?

o If there are vertical asymptotes, does the graph go up or down alongside each asymptote? (This

question can be answered most simply by considering the sign behavior of 𝑓.)

o Are there any holes in the graph of 𝑓(𝑥)? If so, what are their (𝑥, 𝑦) coordinates?

o What is the end behavior of the graph? (Does the right end of the graph go up, or down, or level

off at a horizontal asymptote? If there is a horizontal asymptote on the right, what is its line

equation? Same questions for the left end of the graph.)

• Find 𝑓′(𝑥) and analyze it to get information about the graph of 𝑓(𝑥).

o Find the partition numbers for 𝑓′(𝑥).

o Make a sign chart for 𝑓′(𝑥). (Label your sign chart!)

o Use the sign chart for 𝑓′(𝑥) to determine where 𝑓(𝑥) is increasing/decreasing and where it has

local extrema. Find the 𝑦 coordinates of the local extrema.

• Then analyze 𝑓′′(𝑥) to get information about the graph of 𝑓(𝑥).

o Find the partition numbers for 𝑓′′(𝑥).

o Make a sign chart for 𝑓′′(𝑥). (Label your sign chart!)

o Use the sign chart for 𝑓′′(𝑥) to determine where 𝑓(𝑥) is concave up/down and where it has

inflection points. Find the 𝑦 coordinates of the inflection points.

• Finally, assemble all the known info about the graph of 𝑓(𝑥) and sketch the graph.

End of Lecture Outline for Day 35

Day 36 Optimization I: Single Variable Applied Optimization (Section 26)

X Day 36 Optimization I: Single Variable Applied Optimization (Section 26)

Discuss general issues:

Optimization problems are simply Global Max/Min problems, but they may have complications

• May be presented as word problems. Problems that involve applying math to solve some real-

world problem are called Application Problems. Math problems that are not about some real

world problem, but rather are just about abstract math, could be called Abstract Problems.

• You will probably have to figure out the function and its domain

• May have domains that are not closed intervals

• May involve more than one variable

Today: Single Variable I Optimization Problems: Maximizing Revenue and Profit

Project onscreen and discuss the reference page about Business Terminology

(You can find it on the main MATH 1350 web page in the calendar and in the list of Exercises. It is also

reprinted on the next page of this lecture outline.)

[Example 1] A company manufactures and sells 𝑥 cameras per week.

The weekly price-demand equation is 𝑝 = 300 − 𝑥/30

The weekly cost equation is 𝐶(𝑥) = 90,000 + 30𝑥

(A) If the goal is to maximize the weekly revenue, what price should the company charge for the

cameras, and how many cameras should be produced per week?

Set up Revenue Function 𝑅(𝑥) = −𝑥2

30+ 300𝑥.

Discuss the shape, that we know 𝑅(𝑥) will have a max at its only critical number.

So our strategy is: Find 𝑅′(𝑥). Set 𝑅′(𝑥) = 0. Solve for 𝑥.

Result: 𝑥 = 4500 cameras per week at 𝑝 = $150 per camera.

(B) What is the maximum possible weekly revenue? Result: 𝑅(4500) = 675,000

(C) If the goal is to maximize the weekly profit, what price should the company charge for the cameras,

and how many cameras should be produced per week? Use calculus methods.

Set up Profit Function 𝑃(𝑥) = −𝑥2

30+ 270𝑥 − 90,000.

Discuss the shape and solution strategy as we did with Revenue function.

Result 𝑥 = 4050 cameras per week at 𝑝 = $165 per camera.

(D) What is the maximum possible weekly profit? Result: 𝑃(4050) = 456,750

(E) Confirm with graph on Desmos.

[Example 2]

(A) Coffee shop sells 1600 cups of coffee per day when price is $2.40 per cup. What is the daily revenue?

Solution: 𝑅 = 3840.

(B) Market survey predicts that for every $0.05 price reduction, 50 more cups of coffee will be sold.

How much should the coffee shop charge per cup in order to maximize revenue?

How many cups will be sold? What will be the resulting Revenue?

Solution: To avoid confusion with similar example in Section 26.2 of the book, I will use use different

variables. Let 𝑞 denote Revenue, let 𝑝 denote Price, and ntroduce variable 𝑘 = number of $0.05 price

reductions.

then selling price 𝑝 = 2.40 − 0.05𝑘 and demand 𝑞 = 1600 + 50𝑘.

𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑑𝑒𝑚𝑎𝑛𝑑 ⋅ 𝑝𝑟𝑖𝑐𝑒 = 𝑞 ⋅ 𝑝 = (1600 + 50𝑘) ⋅ (2.40 − 0.05𝑘) = −2.5𝑘2 + 40𝑘 + 3840

Discuss the shape, that we know 𝑅(𝑘) will have a max at its only critical number.

So our strategy is: Find 𝑅′(𝑘). Set 𝑅′(𝑘) = 0. Solve for 𝑘. Result: 𝑘 = 8, 𝑝 = 2, 𝑅 = 4000.

End of Lecture Outline for Day 36 Optimization I: Single Variable Applied Optimization

Business Terminology

In our course, we will study hypothetical business examples in which a company make and sells some item. The

simplifying assumptions are

• The items are manufactured in batches.

• All of the items manufactured are sold, and they are all sold for the same price per item.

Here is the Business Terminology that we will be using.

Quantity, 𝑞 (small letter), is a variable that represents the number of items made. This sounds simple enough,

but there can be complications. For example, in some problems, 𝑞 represents the number of thousands of items

made. Sometimes the letter 𝑥 will be used instead of 𝑞.

Demand Price, 𝐷(𝑞) is a function. For a given input quantity 𝑞, the output 𝐷(𝑞) is the price that the items will

need to be sold for in order for consumers to be willing to buy 𝑞 of the items. Note that 𝐷(𝑞) will be a

decreasing function. If you don’t want to sell many items, the selling price will need to be high. If you want to

sell a lot of items, the selling price will need to be low. So the graph of 𝐷(𝑞) will go down as one moves from

left to right.

Supply Price, 𝑆(𝑞) is a function. For a given input quantity 𝑞, the output 𝑆(𝑞) is the price that the items will

need to be sold for in order for producers to be willing to make 𝑞 of the items. Note that 𝑆(𝑞) will be an

increasing function. If you don’t want to producers to be willing to make many items, the selling price will need

to be low. If you want producers to be willing to make a lot of items, the selling price will need to be high. So

the graph of 𝑃(𝑞) will go up as one moves from left to right.

Since the Demand Price function 𝐷(𝑞) is decreasing and the Supply Price function 𝐷(𝑞) is increasing, if we

plot them on the same axes, they will cross at one point. This point is called the Equilibrium Point. Its

coordinates are given the special designation (𝑞0, 𝑝0). The symbols 𝑞0 and 𝑝0 are called the equilibrium

quantity and the equilibrium price. Note that because the Demand Price and Supply Price graphs cross at that

point, it must be true that 𝐷(𝑞0) = 𝑆(𝑞0) = 𝑝0.

Revenue, 𝑅(𝑞) is a function. It is the amount of money that comes into a company from the sale of 𝑞 items.

Because of our simplifying assumptions listed above, we can say that

𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = (𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠 𝑠𝑜𝑙𝑑) ⋅ (𝑠𝑒𝑙𝑙𝑖𝑛𝑔 𝑝𝑟𝑖𝑐𝑒 𝑝𝑒𝑟 𝑖𝑡𝑒𝑚) 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 ⋅ 𝐷𝑒𝑚𝑎𝑛𝑑 𝑃𝑟𝑖𝑐𝑒

𝑅(𝑞) = 𝑞 ⋅ 𝐷(𝑞)

Cost, 𝐶(𝑞) (capital letter 𝐶), is a function that gives the cost of making the batch of 𝑞 items.

We say that a company Breaks Even when Revenue = Cost. That is, when 𝑅(𝑞) = 𝐶(𝑞).

Profit, 𝑃(𝑞) (capital letter 𝑃), is a function defined as follows

𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 − 𝐶𝑜𝑠𝑡 𝑃(𝑞) = 𝑅(𝑞) − 𝐶(𝑞)

The expression 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚, denoted by the symbol 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ̅̅ , means 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦

𝑞.

That is, Average Revenue is �̅�(𝑞) =𝑅(𝑞)

𝑞, Average Cost is 𝐶̅(𝑞) =

𝐶(𝑞)

𝑞, and Average Profit is �̅�(𝑞) =

𝑃(𝑞)

𝑞.

The expression 𝑴𝒂𝒓𝒈𝒊𝒏𝒂𝒍 𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚 means 𝑻𝒉𝒆 𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒗𝒆 𝒐𝒇 𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚.

That is, Marginal Revenue is 𝑅′(𝑥), and Marginal Cost is 𝐶′(𝑥), and Marginal Profit is 𝑃′(𝑥).

The word Marginal can also be put in front of the Average Quantities. That is Marginal Average Revenue is

�̅�′(𝑥), and Marginal Average Cost is 𝐶̅′(𝑥), and Marginal Average Profit is �̅�′(𝑥).

Day 37 Optimization II: Two variable Abstract Optimization (Section 25)

X Day 37 Optimization II: Two variable Abstract Optimization (Section 25)

In the last meeting, we discussed Single Variable Applied Optimization problems. (Word problems that

involve applying math to solve some real-world problem.) Today, we will discuss Two Variable Abstract

Optimization Problems: They are not about applying math to solve a real world problem. Our general

strategy will be a four-step process:

Strategy for Two Variable Optimization Problems

Given a description of an abstract problem to be solved,

Step I Set up the two-variable problem

• Identify equations that are either directly mentioned in the problem statement, or that are

somehow implicated in the problem statement. (This may involve introducing variables.)

• Identify the quantity 𝑄 (or some other name) that is to be maximized or minimized.

• The result should be two equations involving the three letters 𝑥, 𝑦, 𝑄 (or some other names)

• Identify the domains for the variables 𝑥, 𝑦.

Step 2 Eliminate one of the variables

• Solve one of the two equations for one of the variables, and substituting into the other equation to

eliminate that variable.

• The goal is to end up with a single equation involving the letters 𝑥, 𝑄 (or some other names).

Solve this equation for 𝑄 in terms of 𝑥, and then write it in function notation as a function 𝑄(𝑥).

• Identify the domain 𝐷 for the variable 𝑥 in this new equation.

• The goal now is to find the value of 𝑥 that optimizes (that is, maximizes or minimizes) 𝑄. (So we

have turned the two-variable optimization problem into a single-variable optimization problem.)

Step 3 Single Variable Optimization

• Using calculus, find the value of 𝑥 that optimizes (that is, maximizes or minimizes) 𝑄 on the

domain 𝐷.

• Find the corresponding values of 𝑦 and 𝑄.

Step 4 Write a ClearConclusion

• The max (or min) value of 𝑄 is 𝑄 = 𝑣𝑎𝑙𝑢𝑒 and it occurs when 𝑥 = 𝑛𝑢𝑚𝑏𝑒𝑟 and 𝑦 = 𝑛𝑢𝑚𝑏𝑒𝑟.

We will study two examples using this strategy.

[Example 1]

Find positive numbers 𝑥, 𝑦 such that

• The sum 2𝑥 + 𝑦 = 900. • The product maximized.

Result: The maximum product possible is 𝑃 = 101,250 and it occurs when 𝑥 = 225 and 𝑦 = 450.

[Example 2]

Find positive numbers 𝑥, 𝑦 such that

• The product is 9000.

• The sum 10𝑥 + 25𝑦 is minimized

Result: The minimum possible value of the sum 𝑆 = 10𝑥 + 25𝑦 for positive numbers 𝑥, 𝑦 is 𝑆 = 3000

and it occurs when 𝑥 = 150 and 𝑦 = 60.

End of Lecture Notes for Day 37 Optimization II: Two variable Abstract Optimization

Day 38 Optimization III: Two variable Applied Optimization (Section 26)

X Day 38 Optimization II: Two variable Applied Optimization (Section 26)

On Day 36, we discussed Single Variable Applied Optimization problems. (Word problems that involve

applying math to solve some real-world problem.)

On Day 37, we discussed Two Variable Abstract Optimization Problems: (not about applying math to

solve a real world problem. Our general strategy was a four-step process.

Today, we will discuss Two Variable Applied Optimization Optimization problems. Remember that these

are word problems, because they are about the application of math so some real world problem. And they

will involve two variables, although in the original statement of the problem there might not be any

mention of variables or equations. Our job will be to figure out the variables and equations that we need

to solve the problems.

Our general approach will be to

• Draw pictures that help us visualize the situation being described in the word problem.

• Introduce variables to respresent the quantities mentioned in the problem, with domains.

• Formulate equations that express the relationships among the quantities that are described in the

problem

• Identify the quantity to be maximized or minimized.

• At this point, we will have turned the two-variable applied optimization problem into a two

variable abstract optimization problem. We use the method described in yesterday’s meeting, the

Strategy for Two Variable Optimization Problems, to solve the problem.

The examples that we will study have to do with finding the best dimensions of a rectangular enclosure

on a farm. In the first example, the goal is to maximize area; in the second, to minimize cost. These

examples may seem rather hokey, but they are chosen for three reasons:

1. They are easy to visualize.

2. Their underlying math appears in lots of business settings.

3. Their problem statements are very similar, but the underlying math is very different.

[Example 1] A farmer needs to build a fence to make a rectangular corral next to an adjacent pasture. He

needs only needs to fence three sides, because the fourth side has already been fenced. He has 900 feet of

fencing. What are the dimensions of the corral that will enclose the largest possible area?

Result: Once the problem is set up, we realize that the underlying abstract optimization problem is the

same as the problem from [Example 1] on Day 37. The result of that example can be used here.

The maximum area possible is 𝐴 = 101,250 and it occurs when the dimensions are as follows:

• The side perpendicular to the pasture fence has length 𝑥 = 225.

• The side parallel to the pasture fence has length 𝑦 = 450.

[Example 2] A farmer needs to build a rectangular fenced chicken yard. He wants an area of 9000ft2. The

fence on the west side of the yard must be wood, which costs $20/foot. The fence on the east, north, and

south sides can be wire that costs $5/foot. What are the dimensions of the cheapest fence? How much

does it cost?

Result: Once the problem is set up, we realize that the underlying abstract optimization problem is the

same as the problem from [Example 2] on Day 37. The result of that example can be used here.

End of Notes for Day 38 Optimization III: Two variable Applied Optimization

Day 39 was Exam 3 on Unit 3: Applications of the Derivative.

.

Day 40 Antiderivatives and Indefinite Integrals (Section 27)

X Day 40 Antiderivatives and Indefinite Integrals (Section 27)

Definition of Antiderivative

Words: 𝐹 is an antiderivative of 𝑓.

Meaning: 𝑓 is the derivative of 𝐹. That is, 𝑓 = 𝐹′.

Arrow diagram:

[Example 1] 𝐹(𝑥) = 𝑥3/3 is an antiderivative of 𝑓(𝑥) = 𝑥2. Check

[Example 2] 𝐺(𝑥) =𝑥3

3+ 17 is an antiderivative of 𝑓(𝑥) = 𝑥2. Check

Notice that there is more than one antiderivative of 𝑓(𝑥). That is, given one function 𝐹(𝑥) that is known

to be an antiderivative of 𝑓(𝑥), we can make lots of other antiderivatives. Any function of the form

𝐹(𝑥) + 𝐶 (where 𝐶 is a constant that can be any real number) will also be an antiderivative of 𝑓(𝑥).

An obvious question is: Are there any other antiderivatives of 𝑓(𝑥)?

The answer is provided by the following Theorem:

Theorem: If one 𝐹(𝑥) is an antiderivative of 𝑓(𝑥), then all functions of the form 𝐹(𝑥) + 𝐶 are

antiderivatives of 𝑓(𝑥). Furthermore, these are the only antiderivatives of 𝑓(𝑥). That is if some function

𝐺(𝑥) is known to also be an antiderivative of 𝑓(𝑥), then it must be that 𝐺(𝑥) = 𝐹(𝑥) + 𝐶 for some

constant 𝐶.

Draw family of all antiderivatives of 𝑓(𝑥) = 𝑥2, showing them as vertical translations of 𝐹(𝑥). By

contrast, draw family of horizontal translations of 𝐹(𝑥). They are not antiderivatives of 𝑓(𝑥).

Remark on wording: say “an antiderivative”, not “the antiderivative”

But even though we don’t know how to find antiderivatives yet, we are equipped to answer a particular

common form of question:

Common Form of Question: Is One Given Function an Antiderivative of Another Given Function?

That is, given a function 𝐹 and a function 𝑓, is 𝐹 an antiderivative of 𝑓?

To answer this question, simply find the derivative of 𝐹.

• If 𝐹′ = 𝑓, then 𝐹 is an antiderivative of 𝑓.

• If 𝐹′ ≠ 𝑓, then 𝐹 is not an antiderivative of 𝑓.

Write a clear conclusion.

[Example 3] Is 𝐹(𝑥) =(5𝑥+7)3

3 an antiderivative of 𝑓(𝑥) = (5𝑥 + 7)2? NO

[Example 4] Is 𝐹(𝑥) = 𝑥 ln(𝑥) − 𝑥 an antiderivative of 𝑓(𝑥) = ln(𝑥)? YES

[Example 5] Is 𝐹(𝑥) = 𝑒(𝑥2) an antiderivative of 𝑓(𝑥) = 𝑒(𝑥2)? NO

[Example 6] Is 𝐹(𝑥) = 𝑒(

𝑥3

3) an antiderivative of 𝑓(𝑥) = 𝑒(𝑥2)? NO

Discuss fact that 𝑓(𝑥) = 𝑒(𝑥2) does not have an antiderivative that can be written using basic functions.

Important fact from higher math: The function 𝑓(𝑥) = 𝑒(𝑥2) does have an antiderivative, but not one

that can be written using functions from our usual library of functions.

Do [Class Activity] Is One Given Function an Antiderivative of Another Given Function?

[1] The constant function 𝑓(𝑥) = 𝜋 is an antiderivative of the constant function 𝑘(𝑥) = 0. T F

[2] The constant function 𝑘(𝑥) = 0 is an antiderivative of the constant function 𝑓(𝑥) = 𝜋. T F

[3] The constant function 𝑘(𝑥) = 0 is an antiderivative of itself. T F

[4] The function 𝑔(𝑥) = 5𝑒(𝑥) is an antiderivative of itself. T F

[5] The function is ℎ(𝑥) = 5𝑒𝜋 an antiderivative of itself. T F

[6] If 𝑛 is a real number and 𝑛 ≠ −1, then 𝑥𝑛+1

𝑛+1 is an antiderivative of 𝑥𝑛. T F

(There is a basic exercise of this type in the exercises, but not in the reading.)

Meeting Outline for Day 40 Continues on the Next Page ➔

𝐹 𝑓

take derivative

Day 40 Antiderivatives and Indefinite Integrals (Section 27) continued

X Day 40 Antiderivatives and Indefinite Integrals (Section 27) continued

We saw earlier that for a given function 𝑓(𝑥), there is an infinite collection of antiderivatives of 𝑓(𝑥): the

set of all functions of the form 𝐹(𝑥) + 𝐶, where 𝐹(𝑥) is any one known antiderivative of 𝑓(𝑥). It is

helpful to have a name and a symbol for this collection of all antiderivatives of 𝑓(𝑥). The name of the

collection is the indefinite integral of 𝑓. Here is the definition:

Definition of Indefinite Integral

words: the indefinite integral of 𝑓(𝑥)

symbol: ∫ 𝑓(𝑥)𝑑𝑥

Additional Terminology: The function 𝑓(𝑥) inside the integral symbol is called the integrand.

meaning: the collection of all antiderivatives of 𝑓(𝑥).

Remark: We know that given one function 𝐹(𝑥) that is known to be an antiderivative of 𝑓(𝑥), we can

get all other antiderivatives by adding constants to 𝐹(𝑥). We can represent the whole collection of

antiderivatives of 𝑓(𝑥), by writing 𝐹(𝑥) + 𝐶, where 𝐶 is a constant that can be any real number. That

is,

If 𝐹′(𝑥) = 𝑓(𝑥) then ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶

Additional Terminology: The unknown constant 𝐶 is called the constant of integration.

Diagrams: The statement above can be illustrated by the diagrams below.

Additional Terminology: general antiderivative, particular antiderivative

Another Remark on Wording:

Notice the above definition says “the definite integral of 𝑓(𝑥)”. Contrast this with the earlier definition

that said “an antiderivative of 𝑓”. There are many antiderivatives of 𝑓(𝑥), but there is only one definite

integral of 𝑓(𝑥).

Revisit our examples of antiderivatives and rewrite each result using the new terminology and notation of

indefinite integrals.

[Example 7] In [Example 1] and [Example 2] above, we found that 𝐹(𝑥) =𝑥3

3 and 𝐺(𝑥) =

𝑥3

3+ 17 were

both antiderivatives of 𝑓(𝑥) = 𝑥2. Using the notation and terminology of indefinite integrals, we can

write.

∫ 𝑥2𝑑𝑥 =𝑥3

3+ 𝐶

[Example 8] In [Example 4] above, we found that 𝐹(𝑥) = 𝑥 ln(𝑥) − 𝑥 is an antiderivative of 𝑓(𝑥) =ln(𝑥). Using the notation and terminology of indefinite integrals, we can write.

∫ ln(𝑥) 𝑑𝑥 = 𝑥 ln(𝑥) − 𝑥 + 𝐶

End of outline for Day 40 Antiderivatives and Indefinite Integrals (Section 27)

𝑓(𝑥)

take derivative

∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶 find indefinite integral

the collection of all

antiderivatives of 𝑓(𝑥)

Then:

𝐹(𝑥) 𝑓(𝑥) take derivative Suppose:

Day 41 Indefinite Integral Rules (Section 27)

X Day 41 Indefinite Integrals (Section 27)

Indefinite integrals of Basic functions

Rule 1: The power Rule. If 𝑛 is a real number and 𝑛 ≠ 1, then ∫ 𝑥𝑛𝑑𝑥 =𝑥𝑛+1

𝑛+1+ 𝐶 Check it!

[Example 1] Find ∫ 𝑥8 𝑑𝑥. Check this result by finding the corresponding derivative.

[Example 2] Find ∫ 𝑥 𝑑𝑥. Check this result by finding the corresponding derivative.

[Example 3] Find ∫ 1 𝑑𝑥. Check this result by finding the corresponding derivative.

[Example 4] Find ∫1

𝑥8 𝑑𝑥. Check this result by finding the corresponding derivative.

Finding ∫1

𝑥𝑑𝑥

Consider the indefinite integral of 1/𝑥. Try using power rule. Gives undefined result. What went wrong?

Remember that the Power Rule for Indefinite Integrals is only allowed when 𝑛 is not equal to −1.

So when 𝑛 = −1, then 𝑥𝑛+1

𝑛+1 +C is NOT the indefinite integral of 𝑥𝑛.

That is, 𝑥𝑛+1

𝑛+1 +C is NOT the indefinite integral of 𝑥𝑛.

Remark: Recall Day 40 [Class Activity] question (6) says 𝑛 ≠ 1. In most cases, 𝑥𝑛+1

𝑛+1 is an antiderivative

of 𝑥𝑛. But ther is this one case when it is not.

Then how DO we find the indefinite integral of 1/x?

Recall that the derivative of ln(𝑥) is 1/𝑥. (domain restricted to 𝑥 > 0)

The corresponding integral equation is ∫1

𝑥𝑑𝑥 = ln(𝑥) + 𝐶. (domain restricted to 𝑥 > 0)

It looks as though we have found the indefinite integral rule for ∫1

𝑥𝑑𝑥. But the indefinite integral

equation that we have found is not quite good enough. The problem is that the domain is restricted to 𝑥 >

0. The domain of the integrand 𝑓(𝑥) =1

𝑥 is the set of all 𝑥 ≠ 0. We would like to find an indefinite

integral rule for ∫1

𝑥𝑑𝑥 that is not restricted to a smaller domain

It turns out that there is another derivative equation, less well-known, that has a larger domain. (The

Derivative Equation can be proven using the chain rule and the definition of the absolute value function,

and it can be visualized with graphs of 𝑦 = ln(|𝑥|) and 𝑦 =1

𝑥, but I will not do that here.) Here is that

equation, along with the corresponding indefinite integral equation:

• Derivative Equation: 𝑑

𝑑𝑥ln(|𝑥|) =

1

𝑥 (true for 𝑥 ≠ 0)

• Corresponding Indefinite Integral Equation: ∫1

𝑥𝑑𝑥 = ln(|𝑥|) + 𝐶 (true for 𝑥 ≠ 0)

So we have a new indefinite integral rule: The 1

𝑥 Rule for Indefinite Integrals ∫

1

𝑥𝑑𝑥 = ln(|𝑥|) + 𝐶

Meeting Outline for Day 41 Continues on the Next Page ➔

Day 41 Indefinite Integral Rules (Section 27)

X Day 41 Indefinite Integral Rules (Section 27), Continued

More Indefinite Integral Rules Recycled from earlier results

Recall Day 40 [Example 4]: We found that 𝐹(𝑥) = 𝑥 ln(𝑥) − 𝑥 is an antiderivative of 𝑓(𝑥) = ln(𝑥)

Here is the corresponding indefinite integral rule: ∫ ln(𝑥) 𝑑𝑥 = 𝑥 ln(𝑥) − 𝑥 + 𝐶

Exponential Function Rule #1

• For Derivatives: 𝑑

𝑑𝑥𝑒(𝑥) = 𝑒(𝑥)

• For Indefinite Integrals: ∫ 𝑒(𝑥)𝑑𝑥 = 𝑒(𝑥) + 𝐶



𝑑𝑥𝑏(𝑥) = 𝑏(𝑥) ⋅ ln(𝑏)

• For Indefinite Integrals: ∫ 𝑏(𝑥)𝑑𝑥 =𝑏(𝑥)

ln(𝑏)+ 𝐶



𝑑𝑥𝑒(𝑘𝑥) = 𝑒(𝑘𝑥)

• For Indefinite Integrals: ∫ 𝑒(𝑘𝑥)𝑑𝑥 =𝑒(𝑥)

𝑘+ 𝐶

Sum and Constant Multiple Rule


𝑑𝑥(𝑎𝑓(𝑥) + 𝑏𝑔(𝑥)) = 𝑎

𝑑

𝑑𝑥(𝑓(𝑥)) + 𝑏

𝑑

𝑑𝑥(𝑔(𝑥))

• For Indefinite Integrals: ∫ 𝑎𝑓(𝑥) + 𝑏𝑔(𝑥)𝑑𝑥 = 𝑎 ∫ 𝑓(𝑥)𝑑𝑥 + 𝑏 ∫ 𝑔(𝑥)𝑑𝑥 + 𝐶

This last rule looks simple enough, but there is some subtlety in the handling of the constant of

integration.

[Example 5] The Integral of a Constant Function

Find ∫ 𝑘 𝑑𝑥, where 𝑘 is a constant. Check this result by finding the corresponding derivative.

Show students Reference page on the course website: Derivative and Indefinite Integral Rules

[Example 6] The Integral of a Polynomial Function

Find ∫ 3𝑥5 − 8𝑥3 + 2𝑥 − 6 𝑑𝑥.

[Example 7] The Indefinite Integral of a Sum of Fractions Built from Constants and Powers of 𝑥.

Find ∫3𝑥7

5−

13

𝑥3𝑑𝑥

[Example 8] Valid and Invalid Solutions to an Idefinite Integral Problem

See the Supplemental Reading for Day 41 for more examples

End of Outline for Day 41 Indefinite Integral Rules (Section 27)

.

Day 42 Integrating with Initial Conditions (Section 28)

Recall that if some function 𝐹(𝑥) is an antiderivative of a function 𝑓(𝑥), then all functions of the form 𝐹(𝑥) +

𝐶, where 𝐶 is a constant, will also be antiderivative of 𝑓(𝑥). The function 𝐹(𝑥) is called a particular

antiderivative of 𝑓(𝑥). The expression 𝐹(𝑥) + 𝐶 is called the general antiderivative of 𝑓(𝑥). It is not a single

function, but rather the function form that all antiderivatives of 𝑓(𝑥) must have. Another name for the

expression 𝐹(𝑥) + 𝐶 is the indefinite integral of 𝑓(𝑥). So the following four expressions mean the same thing:

• the general antiderivative of 𝑓(𝑥).

• the indefinite integral of 𝑓(𝑥).

• ∫ 𝑓(𝑥)𝑑𝑥

• 𝐹(𝑥) + 𝐶

And recall that the collection of all of the 𝐹(𝑥) + 𝐶, that is, all of the antiderivatives of of 𝑓(𝑥), will have

graphs that are vertical translations of the graph of 𝐹(𝑥).

For instance, our first examples of antiderivatives, were on Day 40 (Monday) in [Examples 1,2]. In those

examples, we saw that the functions 𝐹(𝑥) =𝑥3

3 and 𝐺(𝑥) =

𝑥3

3+ 17 were both antiderivative of 𝑓(𝑥) = 𝑥2.

Functions 𝐹(𝑥) and 𝐺(𝑥) are both particular antiderivative of 𝑓(𝑥) = 𝑥2. The general antiderivative of 𝑓(𝑥),

that is, the indefinite integral of 𝑓(𝑥), is ∫ 𝑥2𝑑𝑥 =𝑥3

3+ 𝐶

Finding Antiderivatives that satisfy an Extra Condition

It often happens that one wants to find the particular antiderivative that has some desired 𝑦 value for a chosen 𝑥

value. That is, one wants to find the particular antiderivative whose graph goes through some desired point. This

amounts to an extra requirement, an extra condition, being imposed on the antiderivative.

[Example 1] What is the particular antiderivative of 𝑓(𝑥) = 𝑥2 that goes through the point (𝑥, 𝑦) = (−3,5).

Solution: The function 𝐻(𝑥) =𝑥3

3+ 14 is particular antiderivative of 𝑓(𝑥) = 𝑥2 that goes through the point

(𝑥, 𝑦) = (−3,5).

End of [Example 1]

So far, this discussion has used the language of being given a function and looking for a particular

antiderivative of that function that satisfies an extra condition. A more common type of problem statement is

one that states that the goal is to find a function 𝑓(𝑥) that satisfies two requirements:

derivative equation: 𝑓′(𝑥) = 𝑠𝑜𝑚𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑖𝑛𝑣𝑜𝑙𝑣𝑖𝑛𝑔 𝑥

extra condition: 𝑓(𝑎) = 𝑏

Here is the method that we will use to solve these kinds of problems:

Solving a Derivative Equation with Extra Condition

To find a function 𝑓(𝑥) that satisfies two requirements:

derivative equation: 𝑓′(𝑥) = 𝑠𝑜𝑚𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑖𝑛𝑣𝑜𝑙𝑣𝑖𝑛𝑔 𝑥

extra condition: 𝑓(𝑎) = 𝑏

Step 1: Integrate to find the General Solution

Find the general solution of the derivative equation by integrating. The result will be

𝑓(𝑥) = ∫ 𝑠𝑜𝑚𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑖𝑛𝑣𝑜𝑙𝑣𝑖𝑛𝑔 𝑥 𝑑𝑥 = 𝑠𝑜𝑚𝑒 𝑛𝑒𝑤 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑖𝑛𝑣𝑜𝑙𝑣𝑖𝑛𝑔 𝑥 𝑒𝑛𝑑𝑖𝑛𝑔 𝑖𝑛 + 𝐶

Notice that the general solution of the derivative equation will be a function form containing +𝐶.

Step 2: Determine the value of the Constant

Find the value of 𝐶 by requiring 𝑓(𝑎) = 𝑏. This will give an equation involving 𝐶. Solve the equation to

find the value of 𝐶.

Step 3: Write the Particular Solution and check it

Using the known value of 𝐶, write the function 𝑓(𝑥). This is the particular solution.

Check to see if 𝑓(𝑥) satisfies the derivative equation.

Check to see if 𝑓(𝑥) satisfies the extra condition.

Step 4: Write a Conclusion presenting the Particular Solution

Write a sentence explaining that the function 𝑓(𝑥)is the particular solution of the derivative equation

that satisfies the extra condition.

[Example 2] Find the function 𝑓(𝑥) that satisfies both of these equations:

derivative equation: 𝑓′(𝑥) = 6𝑥2 − 10𝑥

extra condition: 𝑓(10) = 4000

[Example 3] Find the function 𝑓(𝑥) that satisfies both of these equations:

derivative equation: 𝑓′(𝑥) = 3𝑒𝑥 − 4

extra condition: 𝑓(0) = 8

Remark on Initial Conditions: Notice that in [Example 4], the extra condition 𝑓(0) = 8 involved 𝑥 = 0. This

kind of extra condition is called an initial condition. Notice that the name of this section of the reading is called

Integrating with Initial Conditions.

Initial Conditions involving Revenue

Recall that in our class, when using math to model situations involving business

• The letter 𝑥 or 𝑞 is used to denote quantity, which is the number of items sold. (I will use 𝑞 in this

discussion, because that is what the online Ximera textbook uses.)

• The letter 𝑝 is used to denote price, which is the selling price per item.

• Often, price 𝑝 can be found as a function of 𝑞. That is, one can find a price function 𝑝(𝑞).

• The Revenue 𝑅 is the number of items sold times the selling price per item. That is 𝑅 = 𝑞𝑝.

• When the price 𝑝 is in the form of a price function 𝑝(𝑞), the equation 𝑅 = 𝑞𝑝 becomes the Revenue

Function 𝑅(𝑞) = 𝑞 ⋅ 𝑝(𝑞).

Now observe that it will always be true that 𝑅(0) = 0 ⋅ 𝑝(0) = 0

That tells us that in any situation involving Revenue, there will always be a known initial condition 𝑅(0) = 0.

This can be exploited to find solutions to derivative equations.

[Example 4] (Finding the Revenue Function from the Marginal Revenue)

Suppose that the Marginal Revenue is −3𝑞2 + 600𝑞 − 100,000. Find the Revenue function.

Solution

Recall the Marginal Revenue just means 𝑅′(𝑞). So we know that 𝑅′(𝑞) = −3𝑞2 + 600𝑞 + 100,000

And from the discussion above, we realize that there is a known initial condition 𝑅(0) = 0

So our job is to find the function 𝑅(𝑞) that satisfies both of these equations:

derivative equation: 𝑅′(𝑞) = −3𝑞2 + 600𝑞 + 100,000

initial condition: 𝑅(0) = 0

Initial Conditions involving Cost

In our class, when using math to model situations involving business, the Cost function 𝐶(𝑞) describes the cost

of making 𝑞 items. If a company makes no items, so 𝑞 = 0, the cost 𝐶(0) will in general not be zero, because

there will still be certain overhead expenses—payroll, rent, maintenance—that must be met. The number 𝐶(0)

is called the Fixed Cost.

[Example 5] (Finding the Cost Function from the Marginal Cost and Fixed Cost)

Suppose that the Marginal Cost is 5 + .08𝑞 and the Fixed Cost is 1000. Find the Cost function.

End of Outline for Day 42 Integrating with Initial Conditions (Section 28)

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Sample Lecture Outline for 2019 2020 Spring …Sample Lecture Outline for 2019 – 2020 Spring Semester MATH 1350 Day 1 X Discuss Course Organization Show them the Blackboard Page