Calculus (1) full course transcription

Abdelrahman Ramzy

Abstract
This is a full transcription of the Calculus(1) full course condensed by the great Dr. Linda Green. The full video is
Available at freeCodeCamp.org youtube channel. https://www.youtube.com/watch?v=HfACrKJ_Y2w.

1 Rational Expressions

zy
1.1 Simplifying
21
EX 1.1. Simplify 45
by reducing to the lowest terms

Solution

21
45
3 7
3 3 5
7
15
am
3𝑥6
EX 1.2. Simplify 𝑥2 4𝑥𝑥
by reducing to the lowest terms

Solution

3𝑥6 3
ˆ 𝑥
2
 3
𝑥2 4𝑥𝑥 ˆ 𝑥2ˆ𝑥
2 𝑥2
.R
1.2 Multiplying and Dividing
4 2
EX 1.3. Compute 3 5

Solution
A

4 2 42 8
3 5 35 15
4
EX 1.4. Compute 5
2
3

Solution

4
5 4 3 43 12 6
2 5 2 52 10 5
3
𝑥2  𝑥
𝑥 4
EX 1.5. Compute 𝑥 1
𝑥2 16
𝑥2  𝑥
𝑥 4 𝑥2 𝑥 𝑥2 16 𝑥ˆ𝑥
1 ˆ 𝑥4
ˆ 𝑥
4

𝑥 1 𝑥4 𝑥1 ˆ𝑥

4 ˆ𝑥

1
𝑥ˆ𝑥  4
𝑥2 16
 

1.3 Adding and Subtracting

7 4
EX 1.6. Subtract: 6

15

I dedicated more than 40 days in order to fully transcribe this course. So, if you think that this transcription is helpful, please consider supporting
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1
 Solution

• first we factorize the denominators:

6 2 3
15 3 5
• second we find LCD by multiplying the uncommon factors times only one factor of each common factors:
LCD 2 3 5 30
• finally: we should change the two fractions into terms of the LCD, by multiplying each fraction by the missing
factors of LCD:
7 5 4 2 35 8 27 9
 
6 5 15 2 30 30 30 10
3 5
EX 1.7. Subtract 2𝑥2 𝑥2 1

Solution

• 2𝑥  2 2ˆ𝑥  1

zy
𝑥2  1 ˆ𝑥  1ˆ𝑥  1

• LCD 2ˆ𝑥  1ˆ𝑥  1

• 3
2ˆ𝑥1

5
ˆ𝑥1ˆ𝑥1

3 ˆ𝑥1 5 2

2ˆ𝑥1 ˆ𝑥1 ˆ𝑥1ˆ𝑥1 2
3ˆ𝑥1
2ˆ𝑥1ˆ𝑥1

52
2ˆ𝑥1ˆ𝑥1
am
3𝑥310
2ˆ𝑥1ˆ𝑥1
3𝑥7
2ˆ𝑥1ˆ𝑥1

2 Difference Quotient
For a function 𝑦 𝑓 ˆ𝑥:
R
• Secant Line is a line between two points on the graph of the
function.
ˆ𝑏, 𝑓 ˆ𝑏
• The average rate of change for 𝑓 ˆ𝑥 on the interval (︀𝑎, 𝑏⌋︀ is
the slope of the secant line between the two points (𝑎, 𝑓 ˆ𝑎 and
A.

ˆ𝑏, 𝑓 ˆ𝑏, i.e. 𝑚

𝑟𝑖𝑠𝑒 𝛿𝑦 𝑓 𝑏 𝑓 𝑎
. ˆ  ˆ  𝑦 𝑓 ˆ𝑥
𝑟𝑢𝑛 𝛿𝑥 𝑏 𝑎 

⌋︂ ˆ𝑎, 𝑓 ˆ𝑎
EX 2.1. Find the average rate of change for 𝑓 ˆ𝑥 𝑥 on the interval
(︀1, 4⌋︀.

Solution
𝑎 𝑏
𝑓 ˆ𝑏𝑓 ˆ𝑎 𝑓 ˆ4𝑓 ˆ1 21 1
The average rate of change is the slope, so: 𝑚 𝑏𝑎 41 3 3
.
Definition 2.1. A difference quotient represents the average rate of change of 𝑓 ˆ𝑥 on the interval (︀𝑥, 𝑥  ℎ⌋︀

ˆ𝑥  ℎ, 𝑓 ˆ𝑥  ℎ

𝑦 𝑓 ˆ𝑥
ˆ𝑥, 𝑓 ˆ𝑥

𝑥 𝑥ℎ

2
 𝑓 ˆ𝑥ℎ𝑓 ˆ𝑥
i.e. the slope of the secant line between the two points (𝑥, 𝑓 ˆ𝑥 and ˆ𝑥  ℎ, 𝑓 ˆ𝑥  ℎ is 𝑚 ℎ
.

EX 2.2. Find and simplify the difference quotient for 𝑓 ˆ𝑥 2𝑥2  𝑥  3

Solution

𝑓 ˆ𝑥  ℎ 2ˆ𝑥  ℎ2  ˆ𝑥  ℎ  3 2𝑥2  4ℎ𝑥  2ℎ2  𝑥  ℎ  3

2 2 2
𝑓 ˆ𝑥  ℎ  𝑓 ˆ𝑥 
2𝑥  4ℎ𝑥  2ℎ  𝑥  ℎ 


3 
2𝑥  𝑥 


3
2
2ℎ  4ℎ𝑥  ℎ ℎˆ2ℎ  4𝑥  1
So, 𝑚 ℎ 2ℎ ℎ4𝑥 1
ˆ   
ℎˆ2ℎ  4𝑥  1

3 Graphs and Limits

zy
EX 3.1. Julia’s Suchi and salad buffet costs $10 per pound, but if you get exactly one pound your meal is free. Let
𝑓 ˆ𝑥 represent the price of your lunch in dollars as a function of its weight 𝑥 in pounds:

• Write and equation 𝑦 𝑓 ˆ𝑥 as a piecewise defined function

Solution

)︀
⌉︀
⌉︀0 if
am
𝑥 1
𝑓 ˆ𝑥 ⌋︀
⌉︀
⌉︀ 10𝑥 if 𝑥 x 1
]︀

• Draw a graph of 𝑦 𝑓 ˆ𝑥

Solution
.R

10

1 2 3

• Describe the behavior of𝑓 ˆ𝑥 when 𝑥 is near 1, but not equal to 1
A

Solution

lim 𝑓 ˆ𝑥 10 x 𝑓 ˆ1 0

𝑥 1

Definition 3.1. For real numbers 𝑎 and 𝐿, lim 𝑓 ˆ𝑥 𝐿 means 𝑓 ˆ𝑥 gets arbitrarily close to 𝐿 as 𝑥 gets arbitrarily
𝑥 𝑎
close to 𝑎

EX 3.2. Describe the behavior of 𝑦 𝑔 ˆ𝑥 when 𝑥 is near 2

3
 𝑦
3
2
1

 3  2  1 1 2 3 𝑥
1
 2
 3

Solution

lim 𝑔 ˆ𝑥 𝐷.𝑁.𝐸. Because lim 𝑔 ˆ𝑥 1 x lim 𝑔 ˆ𝑥 3

zy
𝑥 2 𝑥 2 𝑥 2

Definition 3.2.

• lim 𝑓 ˆ𝑥 𝐿 means 𝑓 ˆ𝑥 approaches 𝐿 as 𝑥 approaches 𝑎 from the left.

𝑥 𝑎

• lim 𝑓 ˆ𝑥 𝐿 means 𝑓 ˆ𝑥 approaches 𝐿 as 𝑥 approaches 𝑎 from the right.

𝑥 𝑎

EX 3.3. Describe the behavior of 𝑦

𝑦
am ℎˆ𝑥 when 𝑥 is near 2

3
2
1

 4 3
  2  1 1 2 3 𝑥
.R
 1
 2
 3

Solution
A

lim ℎˆ𝑥 ª , and lim ℎˆ𝑥 ª . So, lim ℎˆ𝑥 𝐷.𝑁.𝐸

𝑥 2 𝑥 2 𝑥 2

4 When Limits fail to exist

EX 4.1. For the function 𝑓 ˆ𝑥 graphed below:
𝑦
3
2
1

 3  2  1 1 2 3 𝑥
1
2
3

4
 Find the following:
lim  𝑓 ˆ𝑥 21 lim  𝑓 ˆ𝑥 1 lim 𝑓 ˆ𝑥 𝐷.𝑁.𝐸
𝑥  1 𝑥 1 𝑥  1
lim 𝑓 ˆ𝑥 2 lim 𝑓 ˆ𝑥 2 lim 𝑓 ˆ𝑥 2 𝑓 ˆ1 𝐷.𝑁.𝐸
𝑥 1 𝑥 1 𝑥 1
lim 𝑓 ˆ𝑥  ª lim 𝑓 ˆ𝑥  ª lim 𝑓 ˆ𝑥 𝐷.𝑁.𝐸
𝑥 2 𝑥 2 𝑥 2

Notes. Ways that limits fail to exist:

(1) Limit from the left x Limit from the right.
(2) The two Limits are equal but not bounded.
(3) Wild behavior, the following example shows the idea.

EX 4.2. What is lim sinˆ 𝜋𝑥  ? use graph or a table of values for evidence.
𝑥 0

Solution

zy
𝑦
1

 1
am
sinˆ 𝜋𝑥  goes up and down very wildly as 𝑥 0.

5 Limit Laws
Limit Laws:
(1) lim ˆ𝑓 ˆ𝑥  𝑔 ˆ𝑥 lim 𝑓 ˆ𝑥  lim 𝑔 ˆ𝑥.
.R
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
(2) lim ˆ𝑓 ˆ𝑥  𝑔 ˆ𝑥 lim 𝑓 ˆ𝑥  lim 𝑔 ˆ𝑥
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
(3) lim ˆ𝑐𝑓 ˆ𝑥 𝑐 lim 𝑓 ˆ𝑥
𝑥 𝑎 𝑥 𝑎
(4) lim ˆ𝑓 ˆ𝑥𝑔 ˆ𝑥 lim 𝑓 ˆ𝑥 lim 𝑔 ˆ𝑥
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
lim 𝑓 ˆ𝑥
(5) lim ˆ 𝑓 ˆ𝑥  𝑥 𝑎

𝑥 𝑎 𝑔 ˆ 𝑥 𝑥
lim 𝑔 ˆ𝑥
𝑎
A

2
𝑥 3𝑥6
EX 5.1. Find lim 𝑥9
𝑥 2

Solution

lim ˆ𝑥2 3𝑥6 lim ˆ𝑥2  lim ˆ3𝑥 lim ˆ6

𝑥2 3𝑥6 𝑥 2 𝑥 2 𝑥 2 𝑥 2
lim 𝑥9 lim ˆ𝑥9 lim ˆ𝑥 lim ˆ9
𝑥 2 𝑥 2 𝑥 2 𝑥 2
lim ˆ𝑥2 3 lim ˆ𝑥 lim ˆ6
𝑥 2 𝑥 2 𝑥 2 22 3 26 16
lim ˆ𝑥 lim ˆ9 29 11
𝑥 2 𝑥 2

6 The Squeeze Theorem

⌋︂
EX 6.1. For the function 𝑔 ˆ𝑥 below, suppose that for 𝑥-values near 1 4 4 B 𝑔 ˆ𝑥 B 3𝑥2  4𝑥  5. What can we say
about lim 𝑔 ˆ𝑥
𝑥 1

5
 10
𝑦 3𝑥2  4𝑥  5 𝑔 ˆ 𝑥
8

6
⌋︂
𝑦 4 𝑥
4

1 2 3 4

Solution

Although we don’t have the rule of the function 𝑔 ˆ𝑥 ,we can say that lim 𝑔 ˆ𝑥 4.
𝑥 2

zy
Theorem 6.1. (The Squeeze Theorem) Suppose 𝑓 ˆ𝑥 B 𝑔 ˆ𝑥 B ℎˆ𝑥 for 𝑥-values near 𝑎 (not necessarily for 𝑥 𝑎),
and let lim 𝑓 ˆ𝑥 lim ℎˆ𝑥 𝐿, then lim 𝑔 ˆ𝑥 𝐿
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
𝑦
ℎˆ𝑥 3

𝑔 ˆ 𝑥 2

𝑓 ˆ𝑥
1
am
 3  2  1 1 2 3 𝑥
 1
 2
 3

EX 6.2. Find lim 𝑥2 sinˆ 𝑥1 .

.R
𝑥 0

Solution

Attempting: lim 𝑥2 sinˆ 𝑥1  lim ˆ𝑥2  lim sinˆ 𝑥1  which is not

𝑥 0 𝑥 0 𝑥 0
helpful, because lim sinˆ 𝑥1  𝐷.𝑁.𝐸. 𝑦
A

𝑥 0
So, we use Squeeze Theorem:
 1 B sinˆ 𝑥1  B 1

𝑥2

𝑥2 B 𝑥2 sinˆ 𝑥1  B 𝑥2 𝑥
But since lim ˆ𝑥2  0 and lim ˆ𝑥2  0, then lim 𝑥2 sinˆ 𝑥1 .
𝑥 0 𝑥 0 𝑥 0

7 Calculating Limits using Algebraic tricks

EX 7.1. Find the following:
2
𝑥3 1 2
(1) lim 2 lim ˆ𝑥1ˆˆ𝑥𝑥1𝑥1 ˆ𝑥  1 lim 𝑥 𝑥𝑥11 23 .
𝑥 1 𝑥 1 𝑥 1 𝑥 1
2 2 2
(2) lim ˆ5𝑧𝑧 25 lim 2510𝑧𝑧𝑧 25 lim 𝑧 𝑧10𝑧 lim 𝑧ˆ𝑧𝑧 10 lim 𝑧  10 10.

𝑧 0 𝑧 0 𝑧 0 𝑧 0 𝑧 0
1 1 1 3 1 𝑟 3 3𝑟 3
𝑟 3 
𝑟 3 3  3 𝑟  3 3ˆ𝑟 3
(3) lim 𝑟
3
lim 𝑟
lim 𝑟
lim 𝑟 1
lim 1 1
.
𝑟 0 ⌋︂
𝑟 0 ⌋︂ ⌋︂
𝑟 0 𝑟 0 3ˆ𝑟 3 𝑟 𝑟 0 3ˆ𝑟 3 9
𝑥32 𝑥32 𝑥32 𝑥⌋︂
34 𝑥⌋︂
1 1 1
(4) lim 𝑥1
lim 𝑥1
⌋︂
𝑥32
lim lim lim ⌋︂
4
𝑥 1 𝑥 1 𝑥 1 ˆ𝑥1ˆ 𝑥32 𝑥 1 ˆ𝑥1ˆ 𝑥32 𝑥 1 𝑥32

6
 2𝑥10
(5) lim .
𝑥  5 ⋃︀𝑥5⋃︀

Solution

)︀
⌉︀
⌉︀𝑥  5 when 𝑥  5 A 0 𝑥 A 5 Ô
⋃︀𝑥  5⋃︀ ⌋︀
⌉︀
⌉︀
]︀
ˆ𝑥  5 when 𝑥5@0 𝑥 @ 5 Ô .So:
2𝑥10 2𝑥10 2ˆ𝑥5
lim  ⋃︀𝑥5⋃︀
lim  ˆ𝑥5
lim 
𝑥5 ˆ
lim  2  2 and
𝑥  5 𝑥  5 𝑥  5 𝑥  5
2ˆ𝑥5
lim 2𝑥10 lim 2𝑥10 lim lim  2 2.
𝑥 5 ⋃︀𝑥5⋃︀ 𝑥 5 ˆ𝑥5 𝑥 5 ˆ𝑥5 𝑥 5

8 When the Limit of the denominator is 0

4𝑥
EX 8.1. Find lim .
𝑥 3 𝑥3

zy
Solution

lim 4𝑥 ª , while lim 4𝑥

ª . So, lim4𝑥
ª 𝐷.𝑁.𝐸.
𝑥 3 𝑥3 𝑥 3 𝑥3 𝑥 3 𝑥3
EX 8.2. Find lim 5𝑥 .
𝑥 4 ⋃︀𝑥3⋃︀
am Solution

lim 5𝑥
𝑥4⋃︀ ⋃︀
lim  5𝑥
𝑥4 ˆ
Ô 


ª

Ô
𝑥  4 𝑥 4
lim 5𝑥 lim  ˆ𝑥5𝑥 
ª
𝑥 4 ⋃︀𝑥4⋃︀ 𝑥 4 4 

So, existence of this Limit depends on your definition of existing Limits, i.e. if you considered unbounded Limits
exist, then the above Limit does exist, and vice versa.
.R
Notes. Ways that limits fail to exist:
(1) If lim 𝑓 ˆ𝑥 x 0 and lim 𝑔 ˆ𝑥 0, then the two sided limits lim 𝑓 ˆ 𝑥
𝑔 ˆ𝑥
and lim 𝑓 ˆ𝑥
𝑔 ˆ𝑥
could be ª or ª.
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
(2) If the two sided limits are not equal, then lim 𝑓𝑔 𝑥𝑥 𝐷.𝑁.𝐸. And if they are equal, then the existence of lim 𝑓𝑔 𝑥𝑥
ˆ

ˆ 
 ˆ

ˆ



𝑥 𝑎 𝑥 𝑎
depends on your definition of existing Limits.
(3) If lim 𝑓 ˆ𝑥 0 and lim 𝑔 ˆ𝑥 0, then lim 𝑓𝑔 𝑥𝑥 could exist or not, and Limits of the form ˆ 00  are called
ˆ 
A

ˆ 
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
indeterminate.

9 Graphs and equations

EX 9.1. Find the equation of this line:
𝑦

3
ˆ1, 2
2

1 1 2 3 4 5 6𝑥
ˆ5, 1


 1

The equation of the line is 𝑦 𝑚𝑥  𝑏 such that 𝑚 is the slope, and 𝑏 is 𝑦-intercept:

7
 Solution

Δ𝑦 𝑦2 𝑦1
𝑠𝑙𝑜𝑝𝑒 𝑟𝑖𝑠𝑒𝑟𝑢𝑛 Δ𝑥 𝑥2 𝑥1
1 2
5 1


 4
3
𝑚,  




3 11
as for 𝑏, we can compute it by plugging in either ˆ5, 1 or ˆ1, 2 in the function 𝑦 

4
𝑥  𝑏. So, 𝑏 4
,
therefore, the equation of the line is 𝑦 43 𝑥  11
4
. 

EX 9.2. Find the equations of these lines:

𝑦
4

zy
 3  2  1 1 2 3𝑥
 1

Solution

Ô Ô 𝑦 Ô 𝑦
For the horizontal line: 𝑦 𝑚𝑥  𝑏
am 𝑦 ˆ0𝑥  𝑏
as for the vertical line: the equation is 𝑥 2
𝑏 3.5,

EX 9.3. Find the equation of the line through the points ˆ1, 2 and ˆ4, 3.

Solution

Solution 1: 𝑚 43 13  5  
Ô
𝑦 35 𝑥  𝑏, and after plugging any point in the equation 𝑦 35 𝑥  𝑏, we find
 

Ô
3 
11 5 11

that 𝑏 3 𝑦 3 𝑥 3 ,
.R
Solution 2: 𝑦  𝑦𝑜 𝑚ˆ𝑥  𝑥𝑜  such that ˆ𝑥𝑜 , 𝑦𝑜  is a point on the Line:
5
so, we calculate the slope the same previous way, and then we plug any point in the above equation 𝑦2 Ô
ˆ  1.


3
𝑥

10 Rational Functions
𝑃 ˆ 𝑥
A

Rational function is a function 𝑓 ˆ𝑥 𝑄ˆ𝑥

such that 𝑃 ˆ𝑥 and 𝑄ˆ𝑥 are polynomials.
3𝑥2 12
EX 10.1. The graph of the function ℎˆ𝑥 𝑥2 3𝑥10
is shown below:
𝑦
15

10

 15 10
 5
 5 10 15 𝑥
 5

 10

 15

• How is the graph of this function is different from the graphs of a polynomial?

Solution

8
 The difference is the end behavior of the function, since the polynomial functions march ª or ª (the end
behavior means the horizontal asymptotes).
• What is the end behavior of the graph?

Solution
There is a horizontal asymptote at 𝑦 3. Because the function ℎˆ𝑥 𝑥23𝑥3𝑥1210  3𝑥
2 2


𝑥2
3 as 𝑥 ª and as  

𝑥 ª. (In general: In order to find the horizontal asymptote, we take the term with the largest degree of
both the numerator and denominator, and then we compute their quotient as 𝑥 ª and as 𝑥 ª. Note that
we can find horizontal asymptote only when the two quotients (as 𝑥 ª and as 𝑥 ª) are equal and they
are real (namely not equal to ª or ª)).
• what is the behavior of the graph of ℎˆ𝑥 near 𝑥  5?

Solution
2
There is a vertical asymptote at 𝑥 5. Because ℎˆ𝑥 𝑥23𝑥3𝑥1210 3 𝑥𝑥 52 𝑥𝑥 22 , and at 𝑥 5 the denominator




ˆ

ˆ



ˆ

ˆ 





zy
is 0 while the numerator not 0. (In general: the vertical asymptotes are corresponding to the 𝑥-values which
make the numeratorx 0 and the denominator 0).
• What is going on at 𝑥 2?

Solution
There is a hole at 𝑥 2. Because at 𝑥 2 the denominator and the numerator are equal 0. But to find the 𝑦-value
at 𝑥 2, we can choose a value 𝑥 near 2, and then we will find that 𝑦 3 𝑥𝑥 52 𝑥𝑥 22 3 𝑥 2 3 2 2 12
ˆ  ˆ   ˆ   ˆ  

12
am 𝑥 5 2 5 ˆ7
. So,


the hole is ˆ2, 7 . (In general: Holes are corresponding to the 𝑥 -values which make both the numerator and
œ
ˆ 

œ
 ˆ   

denominator equal 0, and in order to find the corresponding 𝑦-value for each 𝑥 , we must compute the original
œ
function as 𝑥 𝑥 ).

EX 10.2. What are the horizontal asymptotes for these functions:

(1) 𝑓 ˆ𝑥 5𝑥4
3𝑥2 5𝑥7
Ô 3𝑥5𝑥 2
5
3𝑥
0 as 𝑥 ª and as 𝑥 ª . So the horizontal asymptote is 𝑦 0.
(2) 𝑔 ˆ𝑥 2𝑥3 4
3𝑥3 7𝑥
Ô 3𝑥
2𝑥 3
3
2
3
as 𝑥 ª and as 𝑥 ª . So the horizontal asymptote is 𝑦 2
3
.
(3) ℎˆ𝑥 𝑥2 4𝑥5
Ô 2𝑥
𝑥 2
𝑥
as 𝑥 𝑥
as 𝑥
.R
2𝑥1 2
ª ª , and 2
ª ª . So there is no horizontal
asymptote.
3𝑥2 3𝑥
EX 10.3. Find the vertical asymptotes, horizontal asymptotes and holes for 𝑞 ˆ𝑥 2𝑥2 5𝑥3𝑥

Solution

3𝑥2 3𝑥
• Holes: 𝑞 ˆ𝑥 = 3𝑥ˆ𝑥1 . 3ˆ01
A

2𝑥2 5𝑥3𝑥 𝑥ˆ2𝑥1ˆ𝑥3

So, there is a hole at 𝑥 0, and its 𝑦-value= ˆ 2 01ˆ03
 1

• Vertical asymptotes: the vertical asymptotes are 𝑥 1

2
and 𝑥  3.
2
• Horizontal asymptote: 3𝑥
2𝑥3
3
2𝑥
0 as 𝑥 ª and as 𝑥 ª . So the horizontal asymptote is 𝑦 0

11 Limits at Infinity and Graphs

EX 11.1. What happens to 𝑓 ˆ𝑥 as 𝑥 goes to ª? and as 𝑥 goes to ª?
𝑦
3

 15  10 5
 5 10 15 𝑥
1

2

9
 Solution

lim 𝑓 ˆ𝑥 1 and lim 𝑓 ˆ𝑥 2.

𝑥 ª 𝑥 ª

Notes: (1) Limit in which 𝑥 goes to ª or ª are called "Limits at Infinity", and they are corresponding to
Horizontal asymptotes.
(2) "Infinite Limits" are when 𝑓 ˆ𝑥 ª or 𝑓 ˆ𝑥 ª, and they are corresponding to Vertical asymptotes.
(3) We will not have Vertical asymptotes or Horizontal asymptotes when 𝑥 and 𝑓 ˆ𝑥 both go to ª at the same time.
EX 11.2. For the functions 𝑔 ˆ𝑥 and ℎˆ𝑥 drawn below, what are the limits at infinity?

𝑦
𝑦
25
20 2 𝑦 ℎˆ𝑥
15
1
10
5

zy
 15 10  5 5 10 15 𝑥
3
  2  1 1 2 3 𝑥  1

Solution

• lim 𝑔 ˆ𝑥
𝑥 ª 𝑥
am
0 and lim 𝑔 ˆ𝑥
ª
ª .

• lim ℎˆ𝑥 ª and lim ℎˆ𝑥 𝐷.𝑁.𝐸 because ℎˆ𝑥 goes up and down as 𝑥 ª .
𝑥 ª 𝑥 ª

EX 11.3. Find the following:

1 1 1
(1) lim 𝑥
0 (2) lim 𝑥3
0 (3) lim 𝑥
0
𝑥 ª 𝑥 ª 𝑥 ª
1
(4) lim ⌋︂
𝑥
0
𝑥 ª
.R
Note: lim 1
𝑥𝑟
0 whenever 𝑟 A 0. Also lim 1
𝑥𝑟
0 for 𝑟 A 0 (after avoiding values like 𝑟 1
2
). while for
𝑥 ª 𝑥 ª

𝑟 @ 0 for example 𝑓 ˆ𝑥 1

𝑥 2
we find that lim 1
lim 𝑥2
𝑥 2 𝑥 ª
ª .
𝑥 ª

12 Limits at Infinity and Algebraic tricks

A

5𝑥2 4𝑥
EX 12.1. Evaluate lim 2𝑥3 11𝑥2 12𝑥
𝑥 ª

Solution
5𝑥2 4𝑥 𝑥2 ˆ5 𝑥
4
 𝑥2 5 𝑥4
1 5 𝑥4
50
lim 2𝑥3 11𝑥2 12𝑥
lim 𝑥3 ˆ2 11 
12

lim 𝑥3 2 11 
12 lim 𝑥 2 11 
12 0 200
0.
𝑥 ª 𝑥 ª 𝑥 𝑥2 𝑥 ª 𝑥 𝑥2 𝑥 ª 𝑥 𝑥2

EX 12.2. Find the following:

3𝑥3 6𝑥2 10𝑥2 𝑥3 ˆ3 𝑥

6

10

2
 ˆ
6
3 𝑥 
10

2

• lim 2𝑥3 𝑥2 5
lim 1
𝑥3 ˆ2 𝑥
𝑥2 𝑥3

5

lim ˆ
𝑥2 𝑥3
1
2 𝑥 
5

3000
200
3
2
.
𝑥 ª 𝑥 ª 𝑥3 𝑥 ª 𝑥3

𝑥4 3𝑥2 6 𝑥4 ˆ1 𝑥32  𝑥64 

• lim 2
5𝑥 𝑥2
lim 1
𝑥2 ˆ5 𝑥 
2

lim 𝑥2 
1
5
ª .
𝑥 ª 𝑥 ª 𝑥2 𝑥 ª

Conclusion:
(1) When the degree of numerator @ degree of denominator Ô 𝑥 limor 𝑓 ˆ𝑥 is 0.
degree of denominator Ô
ª ª

(2) When the degree of numerator lim 𝑓 ˆ𝑥 is the quotient of highest power
𝑥 or ª ª

terms.
(3) When the degree of numerator A degree of denominator Ô 𝑥 lim
ª or ª
𝑓 ˆ𝑥 ª or ª.

10
13 continuity
Types of discontinuity:
(1) Jump Discontinuity at 𝑥 1 (2) Removable Discontinuity at 𝑥 4
𝑦 )︀ 𝑦
⌉︀2𝑥 when 𝑥 B 1
⌉︀
4 𝑓 ˆ𝑥 ⌋︀ 4
⌉︀
⌉︀
]︀
𝑥  2 when 𝑥 A 1
3 3
2
2
1
1
 4 3
 2  1 1 2 3 4 𝑥
 1  3  2  1 1 2 3 4 5 𝑥
 1 ˆ𝑥3 ˆ𝑥4
2
 2 𝑔 ˆ 𝑥 ˆ 𝑥  4
 3  2

zy
 4  3

(3) Infinite Discontinuity at 𝑥 2: (4) Wild Discontinuity at 𝑥 0:

𝑦 𝑦
4 1
3
am
2
1

 4 3 2 1 1 2 3 4 5 𝑥 𝑥
1
1
ℎˆ𝑥 𝑥 2
2

𝑦 cosˆ 𝑥1 
3

.R
4

 1

Therefore, to assure that a function is continuous, we must avoid:

1. Jump: by making sure that lim 𝑓 ˆ𝑥 exists.
𝑥 𝑎
2. Removable: by making sure that 𝑓 ˆ𝑎 exists.
3. 𝑓 ˆ𝑎 is defined at the wrong place: By making sure that lim 𝑓 ˆ𝑥 𝑓 ˆ𝑎. Check the example below:
𝑥 𝑎
A

𝑦
4
3
2
1

1 2 3 4 5 𝑥
Note that the third condition implies the first two conditions and excludes Infinite and Wild Discontinuity.
EX 13.1. What are the places where 𝑓 is not continuous, and why?
𝑦
4
3
2
1

 4  3  2 1 1 2 3 4 𝑥
 1

11
 Solution

• at 𝑥  3 because 𝑓 ˆ3 𝐷.𝑁.𝐸

• at 𝑥 1 because lim 𝑓 ˆ𝑥 𝐷.𝑁.𝐸
𝑥 1

• at 𝑥 2 because 𝑓 ˆ2 x lim 𝑓 ˆ𝑥

𝑥 2

• at 𝑥 3 because lim 𝑓 ˆ𝑥 𝐷.𝑁.𝐸

𝑥 3
Note that 𝑓 ˆ𝑥 is continuous at 𝑥 2.


EX 13.2. Consider the function 𝑓 ˆ𝑥 at 𝑥 2 and at 𝑥 1.

𝑦
4
3
2

zy
1

 4  3  2 1
 1 2 3 4 𝑥
 1

Solution

• at 𝑥 
am
2 we notice that the function is not continuous (because lim 𝑓 ˆ𝑥 𝐷.𝑁.𝐸). But lim  𝑓 ˆ𝑥
𝑥 2 𝑥 2

𝑓 ˆ2
and then we deduce that the function is continuous from the left at 𝑥  2.
• at 𝑥 1 we notice that the function is not continuous (because lim 𝑓 ˆ𝑥 𝐷.𝑁.𝐸). But lim 𝑓 ˆ𝑥 𝑓 ˆ1 and
𝑥 1 𝑥 1
then we deduce that the function is continuous from the right at 𝑥 1.
Definition 13.1. A function is continuous from the left if lim 𝑓 ˆ𝑥 𝑓 ˆ𝑎.
𝑥 𝑎
.R
Definition 13.2. A function is continuous from the right if lim 𝑓 ˆ𝑥 𝑓 ˆ𝑎.
𝑥 𝑎

14 Continuity in intervals and Continuous functions

Definition 14.1. We say that the function 𝑓 ˆ𝑥 is continuous on the interval ˆ𝑏, 𝑐 if it is continuous at 𝑥 𝑎 for
every 𝑎 > ˆ𝑏, 𝑐.
A

Note: If the interval is closed ˆ(︀𝑏, 𝑐⌋︀, then in addition to the above conditions, we must have 𝑓 ˆ𝑥 is continuous
from the right at 𝑥 𝑏, and from the left at 𝑥 𝑐. the same can be said as for half opened intervals.
EX 14.1. In what intervals 𝑔 ˆ𝑥 is continuous?
𝑦
3
2
1

 3  2  1 1 2 3 𝑥
 1
 2
 3

Solution
𝑔 ˆ𝑥 is continuous on ˆª, 1 8 (︀1, 1 8 ˆ1, ª.

12
EX 14.2. Give some kinds of functions which are continuous everywhere (namely continuous on ˆª, ª)?

Solution
Polynomials, 𝑦 sinˆ𝑥, 𝑦 cosˆ𝑥, 𝑦 ⋃︀𝑥⋃︀ ...............

EX 14.3. Give some kinds of functions which are continuous on their domains?

Solution

Polynomials, Rational functions, Trigonometric functions, Inverse Trigonometric functions, 𝑦 lnˆ𝑥, 𝑦 𝑒𝑥 .......
. And sums, differences products and quotient of continuous functions are continuous on their domains. Also
compositions of continuous functions are continuous on their domains.
EX 14.4. Find lim cosˆ𝑥
𝑥 0

zy
Solution
Since cosˆ𝑥 is continuous, then lim cosˆ𝑥 cosˆ0 1.
𝑥 0

2
𝑥 4
EX 14.5. Find lim cosˆ 2𝑥 4
𝜋 


𝑥 2
am 2
Solution
2
𝑥 4 ˆ 𝑥2ˆ𝑥2 ˆ 𝑥2 ˆ22 𝑥 4
As 𝑥 2 we find that 2𝑥4 2ˆ𝑥2
𝜋 2
𝜋 2
𝜋 2𝜋. Therefore, lim 𝑐𝑜𝑠ˆ 2𝑥 4
𝜋



𝑐𝑜𝑠ˆ2𝜋  1.
𝑥 2

Note that in this example we used the property "lim 𝑓 ˆ𝑔 ˆ𝑥 𝑓 ‹ lim 𝑔 ˆ𝑥 for continuous functions (namely, we
𝑥 𝑎 𝑥 𝑎
can pass the limit for continuous functions)".

15 Intermediate value Theorem

.R
Theorem 15.1. Intermediate value Theorem states that if 𝑓 ˆ𝑥 is continuous on (︀𝑎, 𝑏⌋︀, 𝑁 is a number between 𝑓 ˆ𝑎
and 𝑓 ˆ𝑏, then there has to be a number 𝑐 > ˆ𝑎, 𝑏 such that 𝑓 ˆ𝑐 𝑁
𝑦

𝑓 ˆ𝑏
A

𝑁
𝑓 ˆ𝑎 Possible places for 𝑐

ac c c b 𝑥
Note: Intermediate value Theorem is only applied for continuous functions. For example, 𝑁 in the following
graph has no corresponding 𝑐:
𝑦

𝑓 ˆ𝑏

𝑓 ˆ𝑎
a b 𝑥

13
EX 15.1. Prove that the polynomial 𝑃 ˆ𝑥 5𝑥4  3𝑥3  12𝑥  7 has a real root.

Solution

We apply Intermediate value Theorem: If we found two real numbers 𝑎 and 𝑏 such that 𝑃 ˆ𝑎 @ 0 and 𝑃 ˆ𝑏 A 0,
then there must exist a real number 𝑐 > ˆ𝑎, 𝑏 such that 𝑃 ˆ𝑐 𝑁 0 (since 𝑃 ˆ𝑎 @ 𝑁 0 @ 𝑃 ˆ𝑏. Now, since
𝑃 ˆ0 7 and 𝑃 ˆ1 1, then 𝑃 ˆ𝑥 must have a real root 𝑐 > ˆ0, 1.

16 Right angle Trigonometry

For a right triangle with sides 𝑎, 𝑏 and 𝑐 and angle 𝜃 as drawn, we define:

𝑐
𝑎

zy
𝜃
𝑏
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
sinˆ𝜃 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑎
𝑐
, cscˆ𝜃 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑐
𝑎
.
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑏 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑐
cosˆ𝜃 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑐
, sec ˆ𝜃  𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑏
.
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tanˆ𝜃 𝑎
, cotˆ𝜃 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑏
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑏
am 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎
.
EX 16.1. Find the exact value of all six trigonometric functions of angle 𝜃 in this right triangle

𝜃
5
𝑎
.R
2

Solution
⌋︂ ⌋︂
𝑎 52  2
2 21, therefore:
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
sinˆ𝜃 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
2
, cscˆ𝜃 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
5⌋︂ 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
5
2
.
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 21 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 5
A

cosˆ𝜃 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 5
, secˆ𝜃 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
⌋︂
21
.
⌋︂
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 2 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 21
tanˆ𝜃 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
⌋︂
21
, cotˆ𝜃 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 2
.
X
EX 16.2. A kite is flying at an angle of elevation 75 , with 100𝑚 of kite string let out. How high is the kite

Solution

𝑦 100𝑚

X
75

sinˆ75 X

𝑦
100
Ô 𝑦 100 sinˆ75 X
 96.6𝑚

14
17 sine and cosine of special angles
EX 17.1. Without using calculator, find sinˆ45 X
 and cosˆ45 X
 using a right triangle with hypotenuse 1.

Solution

X
45

𝑎 1

X
45
𝑎
Ô 2𝑎2 Ô 𝑎
⌋︂ ⌋︂
2 2
𝑎2  𝑎2 1 1 
2
, since 𝑎 represents a length, then 𝑎 must be positive, therefore 𝑎 2
,
⌋︂
2 ⌋︂
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 2
sinˆ45 X
 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
2
1 2
,
⌋︂
2 ⌋︂
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 2
cosˆ45 X
 2

zy
ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 1 2
.
EX 17.2. Without using a calculator, find sinˆ30 X
 and cosˆ30 X


Solution

X
60

1
am
2 1

𝑎 30 X

X
30
R
X
60
A.

By drawing the original triangle (black triangle) and equivalent triangle (red triangle), we find that their
combination is an equilateral⌉︂triangle, therefore:
Ô 𝑎 Ô 𝑎
⌋︂

𝑎2  ˆ 21 2 12 12  ˆ 12 2 2
3
, therefore:
1
1
sinˆ30 X
 2
1⌋︂ 2
and
3 ⌋︂
3
cosˆ30 X
 2
1 2
.

To summarize:
angle 𝜃 in degrees angle 𝜃 in radians cosˆ𝜃 ⌋︂
sinˆ𝜃
X 𝜋 3 1
30 6 2
⌋︂ 2
⌋︂
X 𝜋 2 2
45 4 2 2
⌋︂
X 𝜋 1 3
60 3 2 2

18 The Unit Circle

If we want to find sinˆ25  (for example), we can draw a right angled triangle which contains an angle of measure
X

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
X
25 and then we compute ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 . But we can never do this if we wish to compute sinˆ120  (for example). So, X

we can do this using Unit Circle.

Definition 18.1. Unit Circle is a circle with radius 1, and its center is the origin point

15
Definition 18.2. For angle 𝜃 (or any angle) that can’t be part of a right triangle, we find the point on the unit circle
𝑦-coordinates
at angle 𝜃 and define cosˆ𝜃 𝑥-coordinates, sinˆ𝜃 𝑦-coordinates and tanˆ𝜃 𝑥-coordinates .
EX 18.1. For the angle 𝜑 drawn below, Find cosˆ𝜑, sinˆ𝜑 and tanˆ𝜑
ˆ0.33, 0.93
𝜑

Solution
sinˆ𝜑 0.93
cosˆ𝜑 0.33
tanˆ𝜑  31
11

zy
19 Properties of trig functions
Periodic property:
sinˆ𝜃  2𝑛𝜋  sinˆ𝜃  sinˆ𝜃  360𝑛  sinˆ𝜃 such that 𝑛 > N,
X

cosˆ𝜃  2𝑛𝜋  cosˆ𝜃  cosˆ𝜃  360𝑛  cosˆ𝜃 such that 𝑛 > N.

X

EX 19.1. Find the following:

• cosˆ5𝜋  cosˆ𝜋  2 2𝜋 
am cosˆ𝜋  1


⌋︂

• cosˆ420 X
 cosˆ60 X
 360 X
 cosˆ60 X



2
3

Odd and Even property:

ˆcosˆ𝜃 , sinˆ𝜃 

𝜃
 𝜃
.R
ˆcosˆ𝜃 , sinˆ𝜃 

cosˆ𝜃 cosˆ𝜃. Therefore, cosˆ𝜃 is Even.

sinˆ𝜃 sinˆ𝜃. Therefore, sinˆ𝜃 is Odd.
EX 19.2. Determine if tanˆ𝜃 is Even or Odd function
A

Solution
sinˆ𝜃  sinˆ𝜃 
tanˆ𝜃 cosˆ𝜃  cosˆ𝜃 
 tanˆ𝜃. Therefore, tanˆ𝜃 is Odd function.

Pythagorean property:

𝑠𝑖𝑛ˆ𝜃  1
𝜃

𝑐𝑜𝑠ˆ𝜃 

From Pythagoras Theorem we can conclude that: cos2 ˆ𝜃  sin2 ˆ𝜃 1.

20 Graphing sine and cosine

EX 20.1. Graph 𝑦 cosˆ𝑡 and 𝑦 sinˆ𝑡 where 𝑡 is in radians.

16
 𝜋 𝜋 𝜋 𝜋 2𝜋 3𝜋 5𝜋 7𝜋 5𝜋 4𝜋 3𝜋 5𝜋 7𝜋 11𝜋
𝑡 0 6 4 3 2 3 4 6
𝜋 6 4 3 2 3 4 6
2𝜋
cosˆ𝑡 1 0.866 0.707 0.5 0 0.5 0.707 0.866 1 0.866 0.707 0.5 0 0.5 0.707 0.866 1
sinˆ𝑡 0 0.5 0.707 0.866 1 0.866 0.707 0.5 0 0.5 0.707 0.866 1 0.866 0.707 0.5 0

Solution
𝑦 cosˆ𝑡
1
0.5

 2𝜋 3𝜋 𝜋


𝜋 𝜋 𝜋 3𝜋 2𝜋

2 20.5 2 2

1
𝑦 sinˆ𝑡

From the graph, we notice that:

zy
𝜋
(1): The graph of 𝑦 cosˆ𝑥 is the same as the graph of 𝑦 sinˆ𝑥 shifted horizontally by 2
. Therefore,
cosˆ𝑥 sinˆ𝑥  𝜋2  and sinˆ𝑥 cosˆ𝑥  𝜋2 .
(2) The Domain is ˆª, ª.
(3) The Range is ˆ1, 1.
(4) 𝑦 cosˆ𝑥 is Even, 𝑦 sinˆ𝑥 is Odd.
(5) The Absolute Maximum value is 1, the Absolute Minimum value is 1.
am
Definition 20.1. Midline is the horizontal line half way between max and min points.
Definition 20.2. Amplitude is the vertical distance between a max point and the midline.
Definition 20.3. Period is the horizontal length of the smallest repeating unit

21 Graph of sinusoidal functions

EX 21.1. Graph the functions 𝑦 3 sinˆ2𝑥 and 𝑦 3 sinˆ2𝑥  1.
.R
Solution
First, let’s graph 𝑦 sinˆ𝑥 normally. After that we can think of the graph of 𝑦 sinˆ2𝑥, we know that the period
of 𝑦 sinˆ𝑥 is 2𝜋, therefore, the period of 𝑦 sinˆ2𝑥 should be 𝜋. which means that the graph of 𝑦 sinˆ2𝑥
is the same as the graph of 𝑦 sinˆ𝑥 but compressed horizontally by a factor of 12 . As for 𝑦 3 sinˆ2𝑥 its graph
should be the same as 𝑦 sinˆ2𝑥 graph but stretched vertically by a factor of 3. Finally, 𝑦 3 sinˆ2𝑥  1 graph
should be the same as 𝑦 3 sinˆ2𝑥 graph but shifted up by 1.
A

4
𝑦 3 sinˆ2𝑥  1
3
𝑦 3 sinˆ2𝑥
2
1

 2𝜋 
3𝜋 𝜋


𝜋 𝜋 𝜋 3𝜋 2𝜋
2 2 2 2
 1
𝑦 sinˆ𝑥
 2
𝑦 sinˆ2𝑥
 3
Midline Amplitude Period
𝑦 sinˆ𝑥 𝑦 0 1 2𝜋
𝑦 sinˆ2𝑥 𝑦 0 1 𝜋
𝑦 3 sinˆ2𝑥 𝑦 0 3 𝜋
𝑦 3 sinˆ2𝑥  1 𝑦 1 3 𝜋

17
 𝜋 𝜋
EX 21.2. graph the functions 𝑦 3 sinˆ2ˆ𝑥  4
 and 𝑦 3 sinˆ2ˆ𝑥  4
  1.

Solution
We can solve this example the same way as in the previous example, the only difference is noticing that the graph
of 𝑦 sinˆ𝑥  𝜋4  is the same as the graph of 𝑦 sinˆ𝑥 but shifted to the right by 𝜋4 .
3
𝑦 3 sinˆ2ˆ𝑥  𝜋4   1
2
𝑦 3 sinˆ2ˆ𝑥  𝜋4 
1

 2𝜋 
3𝜋 𝜋


𝜋 𝜋 𝜋 3𝜋 2𝜋
2 2 2 2
1
𝑦 sinˆ𝑥  𝜋4 
2
𝑦 sinˆ2ˆ𝑥  𝜋4 
3

zy
4
Midline Amplitude Period
𝑦sinˆ𝑥  𝜋4  𝑦 0 1 2𝜋
𝑦 sinˆ2ˆ𝑥  𝜋4  𝑦 0 1 𝜋
𝑦 3 sinˆ2ˆ𝑥  𝜋4  𝑦 0 3 𝜋
𝑦 3 sinˆ2ˆ𝑥  𝜋4   1 𝑦 1 3 𝜋
am
Note: if 𝑔 ˆ𝑥 3 sinˆ2ˆ𝑥  𝜋4 , then 𝑔 ˆ𝑥 𝑓 ˆ𝑥  𝜋4  such that 𝑓 ˆ𝑥 3 sinˆ2𝑥. Hence, we can:
(1) graph 𝑓 ˆ𝑥
(2) shift its graph by 𝜋4 to the right to get the graph of 𝑔 ˆ𝑥.

summary: For the graphs of 𝑦 𝐴 cosˆ𝐵𝑥  𝐶   𝐷 and 𝑦 𝐴 sinˆ𝐵𝑥  𝐶   𝐷 with 𝐵 positive:

* Midline is 𝑦 𝐷
* Amplitude is ⋃︀𝐴⋃︀
.R
* Period is 2𝜋
𝐵
* Horizontal shift: In order to find the horizontal shift, we need to change 𝑦 𝐴 sinˆ𝐵𝑥  𝐶   𝐷 to the form
𝐶 𝐶
𝑦 𝐴 sinˆ𝐵 ˆ𝑥  𝐵   𝐷, and then the horizontal shift is 𝐵 (the same can be said as for 𝑦 𝐴 cosˆ𝐵𝑥  𝐶   𝐷).

1
EX 21.3. Find the Midline, Amplitude, Period and Horizontal shift to the function 𝑦 3
cosˆ 12 𝑥  3  5
A

Solution
1
𝑦 cosˆ 12 𝑥  3  5 31 cosˆ 12 ˆ𝑥  ˆ6  5, Therefore:
3
* Midline is 𝑦 5
* Amplitude is 13
* Period is 4𝜋
* Horizontal shift is 6 units to the left.

22 Graph of tan, sec, cot, csc

EX 22.1. What is slope of the line at angle 𝜃 drawn below? sketch a rough graph of 𝑦 tanˆ𝑥 for 𝑥 between 𝑓 𝑟𝑎𝑐𝜋2
and 𝜋2 .

Solution

18
 sin 𝜃
𝑠𝑙𝑜𝑝𝑒 𝑟𝑖𝑠𝑒
𝑟𝑢𝑛 cos 𝜃
ˆ

ˆ

tanˆ𝜃. In order to draw the graph, we should notice:


* When the angle is 0, the slop is 0.

* As the angle increase toward 𝜋2 , the slope goes to ª.
* As the angle goes from 0 to  𝜋2 , the slope goes to ª.
* At exactly 𝜋2 and  𝜋2 , the slope is undefined.
4
3
2
1

𝜋 𝜋

2 2
 1
 2

zy
 3
 4

Note: tanˆ𝑥 is periodic with period 𝜋. Because if we rotated the line 180 we will get the same line with the
X

same slope.
EX 22.2. Consider this graph of 𝑦
am tanˆ𝑥 and find the following:
4
3
2
1


5𝜋  2𝜋 
3𝜋  𝜋 
𝜋 𝜋 𝜋 3𝜋 2𝜋 5𝜋
2 2 2 1

2 2 2
.R
2


3


4


* 𝑥-intercepts : ..... 2𝜋, 𝜋, 0, 𝜋, 2𝜋..... Ô 𝑘𝜋 such that 𝑘 > Z.

 

* vertical asymptotes : .... 5𝜋 , 3𝜋 , 𝜋2 , 𝜋2 , 3𝜋

 , 5𝜋 .... Ô 𝑘 𝜋2 such that 𝑘 is odd.
 
A

2 2 2 2
* domain : ˜𝑥 ⋃︀ 𝑥 x 𝑘 2 for 𝑘 odd integers.
𝜋

* range : ˆ ,  ª ª

* period : 𝜋
EX 22.3. sketch the graph of 𝑦 secˆ𝑥

Solution
1
In order to sketch the graph, we should remember that 𝑦 secˆ𝑥 cos 𝑥
. Therefore, when cosˆ𝑥 1,
ˆ 

secˆ𝑥 1, and as cosˆ𝑥 goes from 1 to 0, secˆ𝑥 will go to ª. Similarly, when cosˆ𝑥 1, secˆ𝑥 1, and as
cosˆ𝑥 goes from 1 to 0, secˆ𝑥 will go to ª. Hence, the graph of 𝑦 secˆ𝑥 should be as follows:
4
3
2 𝑦 cosˆ𝑥
1
5𝜋 2𝜋 3𝜋  𝜋 𝜋 𝜋 𝜋 3𝜋 2𝜋 5𝜋
2 1
   
2 2 2 2 2
2
3
4

19
 * 𝑥-intercepts : None
* vertical asymptotes : 𝑘 𝜋2 such that 𝑘 is odd.
* domain : ˜𝑥 ⋃︀ 𝑥 x 𝑘 𝜋2 for 𝑘 odd integers.
* range : ˆ , 1 ˆ1, 
ª  8 ª

* period : 2𝜋
EX 22.4. sketch the graph of 𝑦 cscˆ𝑥

Solution
The graph of 𝑦 cscˆ𝑥 can be sketched the same way as the graph of 𝑦 secˆ𝑥.
4
3
2 𝑦 sinˆ𝑥
1
5𝜋 2𝜋 3𝜋 𝜋 𝜋 𝜋 𝜋 3𝜋 2𝜋 5𝜋
2 1
   
2 2 2 2 2
2

zy
3
4

* 𝑥-intercepts : None
* vertical asymptotes : 𝑘 𝜋 such that 𝑘 > Z.
* domain : ˜𝑥 ⋃︀ 𝑥 x 𝑘 𝜋 for 𝑘 integers.
* range : ˆ , 1 ˆ1, 
ª  8 ª

* period : 2𝜋
EX 22.5. sketch the graph of 𝑦
am cotˆ𝑥

Solution
In order to graph 𝑦 cotˆ𝑥, we should remember that 𝑦 cotˆ𝑥 𝑡𝑎𝑛1 𝑥 . Therefore, cotˆ𝑥 has the same sign
ˆ 

as tanˆ𝑥, undefined when tanˆ𝑥 0, goes to ª as tanˆ𝑥 approaches 0 from above and goes to ª as tanˆ𝑥
approaches 0 from below. Hence, the graph of 𝑦 cotˆ𝑥 should be as following:
4
3
.R
2
1
5𝜋 2𝜋 3𝜋 𝜋 𝜋 𝜋 𝜋 3𝜋 2𝜋 5𝜋
2 1
   
2 2 2 2 2
2
3

4
A

* 𝑥-intercepts : None
* vertical asymptotes : 𝑘 𝜋 such that 𝑘 > Z.
* domain : ˜𝑥 ⋃︀ 𝑥 x 𝑘 𝜋 for 𝑘 integers.
* range : ˆ , ª ª

* period : 𝜋

23 Solving basic Trigonometric equations

EX 23.1. For the equation 2 cosˆ𝑥  1 0:
.
(1) Find the solutions in the interval (︀0, 2𝜋 ⌋︀ 2 cosˆ𝑥  1 0 Ô
2 cosˆ𝑥 1 Ô cosˆ𝑥 

2
1
. Therefore, the
solutions are 𝑥 3𝜋 2
or 𝑥 4𝜋
3
.
(2) Give a general formula for all solutions (not just those in the interval (︀0, 2𝜋 ⌋︀) . 3𝜋
2
 2𝑘𝜋 or 4𝜋
3
 2𝑘𝜋 for 𝑘
integers.
⌋︂
EX 23.2. For the equation 2 tanˆ𝑥 3  tanˆ𝑥:
⌋︂
Ô ⌋︂
Ô
⌋︂

(1) Find the solutions in the interval (︀0, 2𝜋 ⌋︀ 2 tanˆ𝑥 . 3  tanˆ𝑥 3 tanˆ𝑥 3 tanˆ𝑥 3
3
.
Therefore, the solutions are 𝑥 𝜋6 or 𝑥 7𝜋 6
.
.
(2) Give a general formula for all solutions (not just those in the interval (︀0, 2𝜋 ⌋︀) 𝜋6  𝑘𝜋 for 𝑘 integers.

20
24 Derivative and Rates of change
EX 24.1. Find the slope of the tangent line.
𝑦

𝑓 ˆ𝑥 𝑥2

ˆ1.5, 2.25

Solution
We can do this by finding the slope of a secant line which passes through ˆ1.5, 2.25and another close point which

zy
Δ𝑦 𝑓 𝑥2 𝑓 𝑥1 ˆ  ˆ 
appears in the following table. Note that 𝑠𝑙𝑜𝑝𝑒 Δ𝑥 𝑥2 𝑥1 

first point second point slope of the secant line

𝑓 3 𝑓 1.5 ˆ  ˆ 
1.5 3 3 1.5
4.5

𝑓 2 𝑓 1.5 ˆ  ˆ 
1.5 2 2 1.5
3.5

𝑓 1.6 𝑓 1.5 ˆ  ˆ 
1.5 1.6 1.6 1.5
3.1 
𝑓 1.51 𝑓 1.5
ˆ  ˆ 
1.5 1.51 3.01
1.5
1.5
1.49
1.45
am1.51 1.5
𝑓 1.49 𝑓 1.5
ˆ

1.49 1.5
𝑓 1.45 𝑓 1.5
ˆ

2.99

2.95



ˆ

ˆ


1.45 1.5 
𝑓 1.4 𝑓 1.5 ˆ  ˆ 
1.5 1.4 1.4 1.5
2.9 
𝑓 1 𝑓 1.5 ˆ  ˆ 
1.5 1 1 1.5
2.6


The calculations in the table similar to finding lim

𝑥 1.5
𝑓 ˆ𝑥𝑓 ˆ1.5
𝑥1.5
Ô slope of tangent line 𝑥 1.5
lim 𝑓 ˆ𝑥𝑓 ˆ1.5
𝑥1.5
.
lim 𝑓 ˆ𝑥𝑥1.5
𝑓 ˆ1.5
lim 𝑓 ˆ𝑥𝑥1.5
𝑓 ˆ1.5
.R
The limit can be written as derivative of 𝑓 ˆ𝑥 at 𝑥 1.5 or 𝑓 ˆ1.5
œ
and it equals
𝑥 1.5 𝑥 1.5
3 by the way.
𝑓 ˆ𝑥𝑓 ˆ𝑎
Definition 24.1. The derivative of a function 𝑦 𝑓 ˆ𝑥 at an 𝑥-value is given by: 𝑓 ˆ𝑎 œ
lim 𝑥𝑎
, and we
𝑥 𝑎
deduce that 𝑓 ˆ𝑥 is differentiable at 𝑥 𝑎 if the previous limit exists.

Note: Another version of the definition of derivative can can be concluded by substituting 𝑥 in the above formula
A

by 𝑎  ℎ which explained on the graph below:

𝑦

ˆ𝑥, 𝑓 ˆ𝑥

ˆ𝑎, 𝑓 ˆ𝑎

𝑎 ℎ 𝑥 𝑥
𝑓 ˆ𝑎ℎ𝑓 ˆ𝑎
Therefore, we get the formula: 𝑓 ˆ𝑎 œ
lim ℎ
.
ℎ 0
EX 24.2. The following expressions represent the derivative of some functions at some value 𝑎. For each example
find the function and the value 𝑎.

• lim
𝑥 1
ˆ 𝑥52 16
𝑥1
Ô Since 𝑥 1, then 𝑎  1, and we can write the limit as lim
𝑥 1

ˆ 𝑥52 ˆ152
𝑥ˆ1
. By
comparing with the formula 𝑓 ˆ𝑎 œ
lim 𝑓 ˆ𝑥𝑥𝑓𝑎 ˆ𝑎 , we find that 𝑓 ˆ𝑥 2
ˆ𝑥  5 .
𝑥 𝑎

21
 • lim
ℎ 0
32ℎ 9
ℎ
Ô We can write this limit as lim
ℎ 0
32ℎ 32
ℎ
, and by comparing with the formula 𝑓 ˆ𝑎 œ

lim 𝑓 ˆ𝑎ℎℎ𝑓 ˆ𝑎 , we find that 𝑓 ˆ𝑥 3𝑥 and 𝑎 2.

ℎ 0

25 Computing derivatives from definition

1
EX 25.1. Find the derivative of 𝑓 ˆ𝑥 ⌋︂
3𝑥
at 𝑥  1.

Solution
⌈︂ 1 ⌈︂ 1
 ⌋︂ 1 ⌋︂1 ⌋︂ 1 1
3ˆ1ℎ 3ˆ1 4ℎ 4ℎ
 
𝑓 ˆ1ℎ𝑓 ˆ1
𝑓 œ
ˆ1 lim ℎ
lim ℎ
lim ℎ
4
lim ℎ
2
which gives 00 . Therefore,
ℎ 0 ℎ 0 ℎ 0 ℎ 0
⌋︂ 0
2 1 ⌋︂4ℎ

Ô ℎlim0
⌈︂ 1 ⌋︂ 0
4ℎ

4ℎ 2 2  *
we simplify the numerator ℎ 44ℎℎ ℎ1 . Again we get 00 .
lim 22⌋︂ So, we multiply by the
ℎ 0
Ô ℎlim0 ℎ1
⌋︂ ⌋︂

2⌋︂ 4ℎ 2⌋︂4ℎ 4ˆ4ℎ
conjugate lim 1 ⌋︂ lim 1 ⌋︂ ℎ lim ⌋︂ 1
2 4ℎ 2 4ℎ ℎ 0ℎ 4 4ℎ2ˆ4ℎ ℎ 0 ℎ 4 4ℎ2ˆ4ℎ ℎ 0 4 4ℎ2ˆ4ℎ
1 1

zy
⌋︂
4 42ˆ4 16
.
EX 25.2. Find the equation of the tangent line to 𝑦 𝑥3  3𝑥 at 𝑥 2.

Solution
3 3
𝑓 ˆ2ℎ𝑓 ˆ2
𝑠𝑙𝑜𝑝𝑒 𝑓 ˆ2
œ

ℎ
lim lim ˆ2ℎ 3ˆ2ℎℎˆ2 3ˆ2 which gives 00 . So, we simplify the numerator
ℎ 0 ℎ 0
Ô 3 2 3 3
lim 2 12ℎ6ℎ ℎℎ 63ℎ2 6 lim 9ℎ6ℎℎ ℎ
2 3
lim 9  6ℎ  ℎ2 9. Therefore, the
ℎ 0
is 𝑦  𝑓 ˆ2 9ˆ𝑥  2
am
Ô 𝑦
ℎ 0
 2
ℎ 0
9𝑥  18 Ô 𝑦 9𝑥  16.
equation of the tangent line

26 Interpreting Derivatives
EX 26.1. The graph of 𝑦 𝑓 ˆ𝑥 represents my distance from campus on a bike ride heading due north
distance(miles)
.R
40

30

20
ˆ3, 𝑓 ˆ3
10 ˆ4, 𝑓 ˆ4
A

1 2 3 4 5 6 time (hours)
Δ𝑦
(1) Interpret the slope of secant line through the points ˆ3, 𝑓 ˆ3 and ˆ4, 𝑓 ˆ4 𝑠𝑙𝑜𝑝𝑒 Δ𝑥 change in distance
change in time
.
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 which is a speed in a direction. So, the slope of secant line gives average velocity over the interval
3 B 𝑥 B 4. ( note that 𝑠𝑝𝑒𝑒𝑑 ⋃︀𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ⋃︀).
.
(2) Interpret the slope of the tangent line at 𝑥 3 The velocity at an exact instant of time is called instantaneous
velocity. So, slope of the tangent line at 𝑥 3 is 𝑓 ˆ3 lim 𝑓 𝑥𝑥 𝑓3 3 œ
instantaneous velocity ˆ 


ˆ 

𝑥 3

In general:
* slope of secant line is average rate of change.
* slope of tangent line is instantaneous rate of change.
EX 26.2. Suppose 𝑓 ˆ𝑥 represents the temperature of your coffee in degrees Fahrenheit as a function of time in
minutes 𝑥 since you have set it on the counter. Interpret the following equations.
(1) 𝑓 ˆ0 140 Ô
At time 0, temperature is 140 X

(2) 𝑓 ˆ10  𝑓 ˆ0 20 Ô

Temperature goes down be 20 as time goes from 0 to 10 minutes. X

(3) 𝑓 1010𝑓 0
ˆ 
2
ˆ 
Ô
Temperature is decreasing by an average of 2 per minute as time changes from 0 to 10 X

minutes.
(4) 𝑓 ˆ15 0.5
œ
Ô
At exactly 15 minutes, the temperature is decreasing at a rate of 0.5 per minute. X

22
EX 26.3. Suppose 𝑔 ˆ𝑥 represent the fuel efficiency for a Toyota Prius in miles/gallon as a function of 𝑥, the speed
in miles/hour that it is traveling. Interpret the following equations.
(1) 𝑔 ˆ45 52 Ô
At 45 mph fuel efficiency is 52 mpg.
(2) 𝑔 ˆ40  𝑔 ˆ35 10 Ô
As speed increases from 35 to 40 mph fuel efficiency foes up by 10 mpg.
(3) 𝑔 40 𝑔 35
ˆ

5
 ˆ
2 
Ô
Average rate of change of fuel efficiency is 2 mpg per mph as speed increases form 35 to 40
mph.
(4) 𝑔 ˆ60 2
œ
Ô
At 60 mph, fuel efficiency is decreasing at a rate of 2 mpg per mph.

27 The Derivative as a function: Its graph and domain

1
EX 27.1. For the function 𝑓 ˆ𝑥 𝑥
, find the derivative 𝑓 ˆ𝑥 œ

Solution
1 1 𝑥ℎ𝑥
𝑥 ℎ ˆ𝑥ℎ𝑥
lim 𝑓 ˆ𝑥ℎℎ𝑓 ˆ𝑥

𝑓 ˆ𝑥
œ
lim ℎ
𝑥
lim ℎ
lim 1 ℎ
lim 1 1
𝑥2
.
ℎ 0 ℎ 0 ℎ 0 ℎ 0ℎ ˆ 𝑥ℎ𝑥 ℎ 0 ˆ𝑥ℎ𝑥

zy
EX 27.2. The height of an alien spaceship above the earth’s surface is graphed below. Graph the rate of change of
the height as a function of time.
height (km)
6
5 𝑦 𝑔 ˆ𝑥
4 𝑠𝑙𝑜𝑝𝑒 1
3
am
2
1

2 4 6 8 10 12 time (min)

Solution
.R
For any 𝑥 > ˆ0, 2, the slope of the tangent line (which is the derivative) equals 1. For 𝑥 > ˆ2, 3 the slope is 0.
For 𝑥 > ˆ3, 5 the slope is also 0. As 𝑥 increases from 5 to approximately 7 the slope starts at a value about 3 until it
goes to 0. As 𝑥 increases from 7 to approximately 10, the slope starts at 0 and goes to a minimum value about 1,
and then goes from 1 to 0. As 𝑥 increases from 10 to ª, the slope goes from 0 to ª.
As for 𝑥 0, the slope or derivative or 𝑔 ˆ0 lim 𝑔 0 ℎℎ 𝑔 4 𝐷.𝑁.𝐸, because the limit from the left 𝐷.𝑁.𝐸.
œ ˆ   ˆ 

ℎ 0
As for 𝑥 2, 𝑔 ˆ2 𝐷.𝑁.𝐸, because limit (slope) from the left equals 1, while limit from the right equals 0.
œ
A

Similarly, we deduce that 𝑔 ˆ5 𝐷.𝑁.𝐸. œ

𝑔 ˆ3ℎ𝑔 ˆ3 2 1 3
As for 𝑥 3, 𝑔 ˆ3 𝐷.𝑁.𝐸, because limit from the left equals lim
œ

ℎ
lim ℎ2 lim ℎ2 ª , while
ℎ 0 ℎ 0 ℎ 0
1 1
limit from the right equals lim 𝑔 ˆ3ℎ𝑔 ˆ3
lim 22
lim ℎ0 0. Note that 𝑔 ˆ3  ℎ 2 when ℎ @ 0, and
ℎ 0 ℎ ℎ 0 ℎ ℎ 0
𝑔 ˆ3  ℎ 12 when ℎ A 0.
Now, the graph of the rate of change or the derivative should be as follows:
slope
4
3
2
1

2 4 6 8 10 12
1


• Ways that a derivative can fail to exist at 𝑥 𝑎:

(1) 𝑓 ˆ𝑥 don’t exist at 𝑥 𝑎:

23
 𝑎
(2) 𝑓 ˆ𝑥 has a corner at 𝑥 𝑎:

𝑎
Note (for example) that 𝑓 ˆ𝑥 ⋃︀𝑥⋃︀ has no derivative at 𝑥 0.

zy
𝑎

Because 𝑓 ˆ𝑥 œ
1 when 𝑥 A 0, and 𝑓 ˆ𝑥 œ
 1 when 𝑥 @ 0. Therefore, 𝑓 ˆ0 𝐷.𝑁.𝐸.
œ

Also the derivative fail to exist when 𝑓 ˆ𝑥 has a cusp at 𝑥 𝑎:

am
𝑎
(3) 𝑓 ˆ𝑥 has a discontinuity at 𝑥 𝑎:
.R
𝑎
(4) When the slope of the secant lines (namely limits from the left or the right) fail to exist because it is infinite:
1
For example, the derivative fails to exist at 𝑥 0 when 𝑓 ˆ𝑥 𝑥 3 .
A

Because as secant lines approach 0 (from the left or the right) they became vertical, which means that the slope
goes to infinity. Therefore, 𝑓 ˆ0 𝐷.𝑁.𝐸
œ

Definition 27.1. A function is differentiable at 𝑥 𝑎 is 𝑓 ˆ𝑎 exists.

œ

Definition 27.2. A function is differentiable on an open interval ˆ𝑏, 𝑐 if 𝑓 ˆ𝑥 is differentiable at every point 𝑎 in
ˆ𝑏, 𝑐.

• Notes:
(1) If 𝑓 ˆ𝑥 is differentiable at 𝑥 𝑎, then 𝑓 ˆ𝑥 has to be continuous at 𝑥 𝑎.
(2) If 𝑓 ˆ𝑥 is continuous at 𝑥 𝑎, then 𝑓 ˆ𝑥 may or may not be differentiable at 𝑥 𝑎, such as 𝑓 ˆ𝑥 ⋃︀𝑥⋃︀, it is
continuous at 𝑥 0 but not differentiable.

28 Proof that differentiable functions are continuous

Theorem 28.1. A function is differentiable at 𝑥 𝑎, then it is continuous at 𝑥 𝑎.

24
 𝑓 ˆ𝑥𝑓 ˆ𝑎
Proof. Since 𝑓 ˆ𝑥 is differentiable at 𝑥 𝑎, then lim 𝑥𝑎
𝑓 ˆ𝑎 exists. By multiplying both sides by
œ

𝑥 𝑎
lim ˆ𝑥  𝑎 we get:
𝑥 𝑎
𝑓 ˆ𝑥𝑓 ˆ𝑎
lim ˆ𝑥  𝑎 lim 𝑥𝑎
lim ˆ𝑥  𝑎 𝑓 ˆ𝑎, We can use limit multiplication rule here because all the limits exist
œ

𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
(since lim ˆ𝑥  𝑎 exist and equal 0), so we get:
𝑥 𝑎
lim
𝑥 𝑎
𝑥𝑎ˆ𝑓 ˆ𝑥𝑓 ˆ𝑎
ˆ

𝑥𝑎
0 Ô 𝑥lim𝑎ˆ𝑓 ˆ𝑥  𝑓 ˆ𝑎 0, we can’t use difference rule of limits here because we don’t
know if lim 𝑓 ˆ𝑥 exists (which actually what we are trying to prove), so, we add lim 𝑓 ˆ𝑎 two both sides:
Ô 𝑥lim𝑎ˆ𝑓 ˆ𝑥 Ô 𝑥lim𝑎 𝑓 ˆ𝑥
𝑥 𝑎 𝑥 𝑎
lim ˆ𝑓 ˆ𝑥  𝑓 ˆ𝑎  lim 𝑓 ˆ𝑎 lim 𝑓 ˆ𝑎  𝑓 ˆ𝑎  𝑓 ˆ𝑎 lim 𝑓 ˆ𝑎 𝑓 ˆ𝑎.
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
Therefore, 𝑓 ˆ𝑥 is continuous at 𝑥 𝑎.

29 The Power rule and derivatives of Sums, Differences and Constant

Multiples
Rules:

zy
𝑑𝑓
(1) Derivative of a constant 𝑐: If 𝑓 ˆ𝑥 𝑐 then 𝑑𝑥
0.

(2) Derivative of 𝑓 ˆ𝑥 𝑥: 𝑓 ˆ𝑥

œ
1.
am
(3) Power rule: If 𝑦 𝑥𝑛 then 𝑦 𝑛𝑥𝑛 1 such that 𝑛 > R.
œ 

𝑑 𝑑
(4) Derivative of constant multiple: 𝑑𝑥 ˆ𝑐 𝑓 ˆ𝑥 𝑐 𝑑𝑥 ˆ𝑓 ˆ𝑥.
𝑑
(5) Derivative of sum or difference: If 𝑓 ˆ𝑥 and 𝑔 ˆ𝑥 are differentiable functions, then: 𝑑𝑥
ˆ𝑓 ˆ𝑥  𝑔 ˆ𝑥
.R
𝑑 𝑑
𝑑𝑥
𝑓 ˆ𝑥  𝑑𝑥 𝑔 ˆ𝑥.
EX 29.1. Find the derivative of these functions:

• 𝑦 𝑥15 Ô 𝑑𝑥𝑑𝑦
15𝑥14 .

𝑥 Ô 𝑓 ˆ𝑥 𝑥 Ô 𝑓 ˆ𝑥 31 𝑥 .
⌋︂ 1 2
• 𝑓 ˆ𝑥 3 3 œ 
3

𝑔 ˆ𝑥 𝑥 1 Ô 𝑔 ˆ𝑥 𝑥 3.7 Ô 𝑔 ˆ𝑥

A

• 3.7

3.7𝑥 4.7 . œ



• 𝑓 ˆ𝑥 5𝑥3 Ô 𝑑𝑥 𝑑 𝑑
ˆ5𝑥3  5 𝑑𝑥 ˆ𝑥3  5 3𝑥2 15𝑥2 .

• 𝑦 7𝑥3 5𝑥2 4𝑥 2 Ô 𝑑𝑥
  
𝑑𝑦 𝑑
𝑑𝑥
ˆ7𝑥3 5𝑥2 4𝑥    2 𝑑
𝑑𝑥
ˆ7𝑥3  
𝑑
𝑑𝑥
ˆ5𝑥2  
𝑑
𝑑𝑥
ˆ4𝑥 
𝑑
𝑑𝑥
ˆ2
2 2
7 3𝑥  5 2𝑥  4 1  0 21𝑥  10𝑥  4.

30 Introduction to Trigonometric identities

EX 30.1. Find solutions of the following equations:
(1) 𝑥2  6𝑥 7 Ô
𝑥2  6𝑥  7 0 Ô
ˆ𝑥  7ˆ𝑥  1 0, therefore, 𝑥 7 or 𝑥 1.
2
(2) 𝑥  6𝑥 7  ˆ𝑥  7ˆ𝑥  1 Ô
𝑥2  6𝑥 7  𝑥2  6𝑥  7 Ô
𝑥2  6𝑥 𝑥2  6𝑥, so, solutions are all real
numbers.
Definition 30.1. An equation is called identity if it holds for all values of the variable.
EX 30.2. Decide which of the following equations are identities:
(1) sinˆ2𝑥 2 sinˆ𝑥 Ô
Not identity, because sinˆ2 𝜋2  0 x 2 sinˆ 𝜋2  2.
(2) cosˆ𝜃  𝜋   cosˆ𝜃 Ô

25
 𝑦 cosˆ𝜃  𝜋 
1 𝑦  cosˆ𝜃

 2𝜋 
3𝜋 𝜋 
𝜋 𝜋 𝜋 3𝜋 2𝜋
2 2 2 2
 1
from the graph of cosˆ𝜃  𝜋  and  cosˆ𝜃 we conclude that cosˆ𝜃  𝜋   cosˆ𝜃 is identity.
(3) secˆ𝑥  sinˆ𝑥 tanˆ𝑥 cosˆ𝑥 1
cos 𝑥
 sinˆ𝑥
sin 𝑥
cos 𝑥
Ô 1
cos 𝑥

sin2 𝑥
cos 𝑥 ˆ
1 sin2 𝑥
cos 𝑥

ˆ

ˆ


 ˆ 
ˆ

ˆ





ˆ
ˆ


 cos2 ˆ𝑥
cosˆ𝑥
cosˆ𝑥.
Therefore, the above equation is identity.

31 The Pythagorean Identities

(1) cos2 ˆ𝑥  sin2 ˆ𝑥 1 proof

ˆ𝑐𝑜𝑠ˆ𝜃 , 𝑠𝑖𝑛ˆ𝜃 
𝑠𝑖𝑛ˆ𝜃  1

zy
𝜃

𝑐𝑜𝑠ˆ𝜃 

(2) tan2 ˆ𝑥  1 sec2 ˆ𝑥 proof

cos2 ˆ𝑥  sin2 ˆ𝑥 1 by dividing over cos2 ˆ𝑥 Ô cos cos𝑥
2
ˆ sin2 ˆ𝑥

2 ˆ 𝑥
1
cos2 ˆ𝑥
Ô cos
cos
2
ˆ
2 ˆ𝑥 
𝑥 sin2 ˆ𝑥
cos2 ˆ𝑥
1
cos2 ˆ𝑥
Ô
tan2 ˆ𝑥  1 sec2 ˆ𝑥.
(3) cot2 ˆ𝑥  1 csc2 ˆ𝑥 proof
am
cos2 ˆ𝑥  sin2 ˆ𝑥 1 by dividing over sin2 ˆ𝑥 Ô cos sin𝑥
2
ˆ 
2 ˆ𝑥
sin2 ˆ𝑥 1
sin2 ˆ𝑥
Ô cos
sin
2
ˆ
2 ˆ 𝑥 
𝑥 sin2 ˆ𝑥
sin2 ˆ𝑥
1
sin2 ˆ𝑥
Ô
cot2 ˆ𝑥  1 csc2 ˆ𝑥.

32 Sum and Difference formulas

Question: Is it true that sinˆ𝐴  𝐵  sinˆ𝐴  sinˆ𝐵  . Not true because sin 𝜋 ˆ 
𝜋
2
  1 x sinˆ𝜋   sinˆ 𝜋2 
0  1 1.
.R
Sum and Difference formulas:
(1) sinˆ𝐴  𝐵  sinˆ𝐴 cosˆ𝐵   cosˆ𝐴 sinˆ𝐵 .
(2) cosˆ𝐴  𝐵  cosˆ𝐴 cosˆ𝐵   sinˆ𝐴 sinˆ𝐵 .
(3) sinˆ𝐴  𝐵  sinˆ𝐴  ˆ𝐵  sinˆ𝐴 cosˆ𝐵   cosˆ𝐴 sinˆ𝐵  sinˆ𝐴 cosˆ𝐵   cosˆ𝐴 sinˆ𝐵 .
(4) cosˆ𝐴  𝐵  cosˆ𝐴  ˆ𝐵  cosˆ𝐴 cosˆ𝐵   sinˆ𝐴 sinˆ𝐵  cosˆ𝐴 cosˆ𝐵   sinˆ𝐴 sinˆ𝐵 .
A

EX 32.1. Find the exact value for sinˆ105 X



Solution
⌋︂ ⌋︂ ⌋︂ ⌋︂ ⌋︂
3 2 1 2 6 2
sinˆ105 X
 sinˆ60 X
 45 X
 sinˆ60 X
 cosˆ45   cosˆ60X  sinˆ45X 
X

2 2

2 2 4
.
EX 32.2. If cosˆ𝑣  0.7, cosˆ𝑤 0.9 find cosˆ𝑣  𝑤. Assume 𝑣 and 𝑤 are in the first quadrant.

Solution

10 7 10 9

𝑤 𝑣
⌋︂ ⌋︂
51 19

26
 9 7
cosˆ𝑣  𝑤 cosˆ𝑣  cosˆ𝑤  sinˆ𝑣  sinˆ𝑤 0.9 0.7  ⌋︂
19
⌋︂
51
0.3187.

33 Double angle formulas

Question: True or False sinˆ2𝜃 2 sinˆ𝜃 . False because:
2 𝑦 2 sinˆ𝜃 2 𝑦 sinˆ2𝜃
1 1
 2𝜋  1 2𝜋  2𝜋  1 2𝜋
 2  2

Double angle formulas:

(1) sinˆ2𝜃 sinˆ𝜃 cosˆ𝜃  cosˆ𝜃 sinˆ𝜃 2 sinˆ𝜃 cosˆ𝜃.
(2) cosˆ2𝜃 cosˆ𝜃 cosˆ𝜃  sinˆ𝜃 sinˆ𝜃 cos2 ˆ𝜃  sin2 ˆ𝜃.
(3) cosˆ2𝜃 cos2 ˆ𝜃  sin2 ˆ𝜃 1  sin2 ˆ𝜃  sin2 ˆ𝜃 1  2 sin2 ˆ𝜃.

zy
(4) cosˆ2𝜃 cos2 ˆ𝜃  sin2 ˆ𝜃 cos2 ˆ𝜃  ˆ1  cos2 ˆ𝜃 2 cos2 ˆ𝜃  1.
1
EX 33.1. Find cosˆ2𝜃 if cosˆ𝜃  ⌋︂
10
and 𝜃 terminates in third quadrant.

Solution
1 8
cosˆ2𝜃 2 cos2 ˆ𝜃  1 2ˆ ⌋︂
10
2  1 

10
.
am
Note: The sentence "𝜃 terminates in third quadrant" in the previous example is necessary if we wanted to compute
sinˆ𝜃, we will need that if we computed cosˆ2𝜃 using formula which involves sinˆ𝜃.
EX 33.2. Solve the equation 2 cosˆ𝑥  sinˆ2𝑥 0

Solution
2 cosˆ𝑥  𝑠𝑖𝑛ˆ2𝑥 0 Ô
2 cosˆ𝑥  2 sinˆ𝑥 cosˆ𝑥 0 cosˆ𝑥ˆ1  sinˆ𝑥 0. Therefore, either Ô
cosˆ𝑥 0 Ô
𝑥 𝜋2  𝜋𝑘 or:
1  sinˆ𝑥 0 Ô
sinˆ𝑥 1 𝑥 3𝜋 Ô
 2𝜋𝑘, this set of solutions are actually include in the first set (namely
.R
2
𝜋
the set 𝑥 2  𝜋𝑘).

34 Higher order Derivatives and notations

function first derivative notation second derivative notation 𝑛-th derivative notation
2 𝑛
𝑑𝑓
𝑓 ˆ𝑥 𝑓 ˆ𝑥, 𝑑𝑥
œ 𝑑
𝑓 ˆ𝑥, 𝐷𝑓 𝑓 ˆ𝑥, 𝑑𝑑𝑥𝑓2 , 𝑑𝑥
œœ 𝑑 𝑑𝑓
ˆ 𝑑𝑥  𝑓 𝑛 ˆ𝑥, 𝑑𝑑𝑥𝑛𝑓
ˆ 
A

, 𝑑𝑥
𝑑𝑦 𝑑2 𝑦 𝑛
𝑑 𝑦
𝑦 𝑦 , 𝑑𝑥œ
𝑦 , 𝑑𝑥2 œœ
𝑦 𝑛 , 𝑑𝑥
ˆ 
𝑛

𝑑𝑦
Some times we use Leibniz notation ‰ 𝑑𝑥 ⋂︀
𝑥 𝑎
Ž in order to evaluate the derivative at a certain value of 𝑥.

35 The derivative of 𝑒𝑥
Remember that 𝑒 2.71828182845905......, and the function 𝑦 𝑒𝑥 can be graphed as follows:
𝑎

Facts:
1 𝑛
(1) lim ‰1  𝑛
Ž 𝑒.
𝑛 ª

𝑒 ℎ
1
 𝑒0ℎ 𝑒0
(2) lim ℎ
1. This expression reminds us of the derivative, because it can be written as lim ℎ
1, which is
ℎ 0 ℎ 0
𝑥 𝑑 𝑥
the derivative of 𝑒 at 𝑥 0, which means that 𝑑𝑥
𝑒 ⋂︀𝑥 0 1.

27
 𝑑 𝑥 𝑒ℎ 1
(3) 𝑑𝑥
𝑒 𝑒𝑥 proof In order to prove this fact, we will assume that fact number 2 is true. Assume that lim ℎ
1,
ℎ 0
𝑥 ℎ 𝑥 𝑥 ℎ ℎ
from the definition derivative 𝑑 𝑥
𝑑𝑥
𝑒 lim 𝑒
ℎ
𝑒

lim 𝑒 ˆ𝑒ℎ 1 𝑒𝑥 lim 𝑒
ℎ
1

𝑒𝑥 .
ℎ 0 ℎ 0 ℎ 0
2
EX 35.1. Find the derivative of 𝑔 ˆ𝑥 𝑒𝑥2  2𝑒𝑥  𝑥𝑒2  𝑥𝑒

Solution
𝑑 𝑑 𝑒2 2

𝑑𝑥
𝑔 ˆ𝑥 𝑑𝑥
ˆ𝑒𝑥2  2𝑒 𝑥
 𝑥𝑒 2
 𝑥  2𝑒𝑥  2𝑒 𝑥
 𝑒2  𝑒2 𝑥𝑒 1
.

36 Proofs of the power rule, sum, difference and constant multiple rules
𝑑
Theorem 36.1. For a constant 𝑐, 𝑑𝑥
ˆ𝑐 0.
𝑑 𝑓 ˆ𝑥ℎ𝑓 ˆ𝑥
Proof. 𝑑𝑥
𝑓 ˆ𝑥 lim ℎ
lim 𝑐𝑐 lim 0 0.
ℎ 0 ℎ 0 ℎ ℎ 0ℎ

𝑑
Theorem 36.2. 𝑑𝑥
ˆ𝑥 1.

zy
𝑑 𝑥ℎ𝑥
Proof. 𝑑𝑥
ˆ𝑥 lim ˆ

ℎ
lim ℎ lim 1 1.
ℎ 0 ℎ 0ℎ ℎ 0

𝑑
Theorem 36.3. (the power rule) 𝑑𝑥
ˆ𝑥𝑛  𝑛𝑥𝑛 1 . 

𝑛 𝑛
Proof. (the proof is only for 𝑛 positive integers) 𝑑
𝑑𝑥
ˆ𝑥𝑛  lim ˆ𝑥ℎ𝑑𝑒𝑛
 𝑥

ℎ 0
𝑛ˆ𝑛1 𝑛2 2 𝑛1 𝑛ˆ𝑛1 𝑛2
𝑛𝑥𝑛1 ℎ ℎ ......𝑛𝑥ℎ𝑛1 ℎ𝑛 𝑥𝑛 𝑛2 𝑛 1

ℎ 0
𝑥
lim 
 𝑛

2 𝑥
ℎ
am 
 lim
ℎ 0

ℎˆ𝑛𝑥  2 𝑥 ℎ......𝑛𝑥ℎ ℎ 

ℎ𝑛2
𝑛𝑥𝑛 1 .

𝑥𝑛 𝑎𝑛 ˆ𝑥𝑎ˆ𝑥𝑛1 𝑥𝑛2 𝑎𝑥𝑛3 𝑎2 .....𝑥𝑎


  𝑎𝑛1 
Another proof. 𝑓 ˆ𝑎 œ
lim 𝑥𝑎
lim  ˆ 𝑥
𝑎

𝑥 𝑎 𝑥 𝑎 
𝑎𝑛 1
 𝑎𝑛 2 𝑎  𝑎𝑛 3 𝑎2  .....  𝑎𝑎
  𝑛1 𝑛𝑎 𝑛1
 𝑎 𝑛𝑎 .
𝑑
Theorem 36.4. (constant multiple rule) If 𝑐 is a constant, and 𝑓 is differentiable function, then 𝑑𝑥
ˆ𝑐𝑓 ˆ𝑥
𝑑
𝑐 𝑑𝑥 ˆ𝑓 ˆ𝑥.

𝑑 𝑐𝑓 ˆ𝑥ℎ𝑐𝑓 ˆ𝑥 𝑓 ˆ𝑥ℎ𝑓 ˆ𝑥 𝑑

Proof. ˆ𝑐𝑓 ˆ𝑥 lim 𝑐 lim 𝑐 𝑑𝑥 ˆ𝑓 ˆ𝑥.
R
𝑑𝑥 ℎ 0 ℎ ℎ 0 ℎ

𝑑 𝑑 𝑑
Theorem 36.5. (sum rule) If 𝑓 and 𝑔 are differentiable functions, then 𝑑𝑥
ˆ𝑓 ˆ𝑥  𝑔 ˆ𝑥 𝑑𝑥
𝑓 ˆ𝑔   𝑑𝑥 𝑔 ˆ𝑥.
𝑓 ˆ𝑥ℎ𝑔 ˆ𝑥ℎ𝑓 ˆ𝑥𝑔 ˆ𝑥
Proof. 𝑑
𝑑𝑥
ˆ𝑓 ˆ𝑥  𝑔 ˆ𝑥 lim ℎ
lim Š 𝑓 ˆ 𝑥ℎ𝑓 ˆ𝑥
ℎ

𝑔 ˆ𝑥ℎ𝑔 ˆ𝑥
ℎ

ℎ 0 ℎ 0
A.

𝑓 ˆ𝑥ℎ𝑓 ˆ𝑥 𝑔 ˆ𝑥ℎ𝑔 ˆ𝑥 𝑑 𝑑

lim ℎ
 lim
ℎ 𝑑𝑥
𝑓 ˆ𝑔   𝑑𝑥 𝑔 ˆ𝑥.
ℎ 0 ℎ 0

𝑑 𝑑 𝑑
Theorem 36.6. (difference rule) If 𝑓 and 𝑔 are differentiable functions, then 𝑑𝑥
ˆ𝑓 ˆ𝑥  𝑔 ˆ𝑥 𝑑𝑥
𝑓 ˆ𝑔   𝑑𝑥 𝑔 ˆ𝑥.
𝑑 𝑑 𝑑 𝑑 𝑑 𝑑
Proof. 𝑑𝑥
ˆ𝑓 ˆ𝑥  𝑔 ˆ𝑥 𝑑𝑥
ˆ𝑓 ˆ𝑥  ˆ𝑔 ˆ𝑥 𝑑𝑥
𝑓 ˆ𝑥  𝑑𝑥 ˆ𝑔 ˆ𝑥 𝑑𝑥
𝑓 ˆ𝑥  𝑑𝑥 𝑔 ˆ𝑥.

37 The product and quotient rule

𝑑 𝑑 𝑑
The product rule: If 𝑓 and 𝑔 are differentiable functions, then 𝑑𝑥
ˆ𝑓 ˆ𝑥𝑔 ˆ𝑥 𝑓 ˆ𝑥 𝑑𝑥 𝑔 ˆ𝑥  𝑔 ˆ𝑥 𝑑𝑥 𝑓 ˆ𝑥.
⌋︂ ⌋︂ ⌋︂ ⌋︂ ⌋︂ ⌋︂
EX 37.1. Find ˆ 𝑡 𝑒𝑡  œ
. ˆ 𝑡 𝑒𝑡  œ
𝑡 𝑑 𝑡
𝑑𝑡
𝑒  𝑒𝑡 𝑑
𝑑𝑡
𝑡
1
𝑡 𝑒𝑡  21 𝑡 2 𝑒𝑡 𝑡 𝑒𝑡  2 𝑡
𝑡
𝑒⌋︂
.
𝑑 𝑑
𝑑 𝑓 ˆ𝑥 𝑔 ˆ𝑥 𝑑𝑥 𝑓 ˆ𝑥𝑓 ˆ𝑥 𝑑𝑥 𝑔 ˆ 𝑥
The quotient rule: If 𝑓 and 𝑔 are differentiable functions, then Š
𝑑𝑥 𝑔 ˆ𝑥
 ˆ𝑔 ˆ𝑥
2 .

EX 37.2. Find 𝑑
Š
𝑧2
𝑑𝑧 𝑧 3 1
 . 𝑑
Š
𝑧2
𝑑𝑧 𝑧 3 1

ˆ 𝑧 3 1ˆ2𝑧 𝑧 2 ˆ3𝑧 2 
3
ˆ𝑧 1
2
2𝑧 4 2𝑧 3𝑧 4
3
ˆ𝑧 1
2 ˆ
2𝑧 𝑧 4
𝑧 3 12
.

28
38 Proof of the product and the quotient rules
𝑑 𝑓 ˆ𝑥ℎ𝑔 ˆ𝑥ℎ𝑓 ˆ𝑥𝑔 ˆ𝑥
Proof of the product rule: 𝑑𝑥
ˆ𝑓 ˆ𝑥𝑔 ˆ𝑥 lim ℎ
ℎ 0
𝑓 ˆ𝑥ℎ𝑔 ˆ𝑥ℎ𝑓 ˆ𝑥𝑔 ˆ𝑥ℎ𝑓 ˆ𝑥𝑔 ˆ𝑥ℎ𝑓 ˆ𝑥𝑔 ˆ𝑥 𝑔 ˆ𝑥ℎˆ𝑓 ˆ𝑥ℎ𝑓 ˆ𝑥𝑓 ˆ𝑥ˆ𝑔 ˆ𝑥ℎ𝑔 ˆ𝑥
lim ℎ
lim ℎ
ℎ 0 ℎ 0
𝑓 ˆ𝑥ℎ𝑓 ˆ𝑥 𝑔 ˆ𝑥ℎ𝑔 ˆ𝑥 𝑑 𝑑
lim 𝑔 ˆ𝑥  ℎ lim ℎ
 lim 𝑓 ˆ𝑥 lim ℎ
𝑔 ˆ𝑥 𝑑𝑥 𝑓 ˆ𝑥  𝑓 ˆ𝑥 𝑑𝑥 𝑔 ˆ𝑥. Note that the limit
ℎ 0 ℎ 0 ℎ 0 ℎ 0
lim 𝑔 ˆ𝑥  ℎ exists because 𝑔 ˆ𝑥 continuous (since it is differentiable).
ℎ 0
𝑑
𝑑 1 
𝑑𝑥 𝑓 ˆ𝑥
Theorem 38.1. (the reciprocal rule) ˆ
𝑑𝑥 𝑓 ˆ𝑥
 ˆ 𝑓 ˆ𝑥2
.
1 1
𝑓 ˆ𝑥ℎ 𝑓 ˆ𝑥


Proof. 𝑑
ˆ
1
𝑑𝑥 𝑓 ˆ𝑥
 lim ℎ
lim 𝑓 ˆ𝑥𝑓 ˆ𝑥ℎ lim ˆ𝑓 ˆ𝑥ℎ𝑓 ˆ𝑥 1
lim ˆ 𝑓 ˆ𝑥ℎ𝑓 ˆ𝑥
ℎ 0 ℎ 0 ℎ𝑓 ˆ𝑥𝑓 ˆ𝑥ℎ ℎ 0 ℎ 𝑓 ˆ𝑥𝑓 ˆ𝑥ℎ ℎ 0 ℎ
𝑑
1 
𝑑𝑥 𝑓 ˆ𝑥
lim 2 .
ℎ 0 𝑓 ˆ𝑥𝑓 ˆ𝑥ℎ ˆ𝑓 ˆ𝑥

𝑑
𝑑 𝑓 ˆ𝑥 𝑑 1 𝑑 1 1 𝑑 
𝑑𝑥 𝑔 ˆ𝑥
Proof of the quotient rule: Š
𝑑𝑥 𝑔 ˆ𝑥
 𝑑𝑥
Š𝑓 ˆ𝑥 𝑔 ˆ 𝑥
 𝑓 ˆ𝑥 𝑑𝑥 Š𝑔 𝑥 ˆ 
  𝑔 ˆ𝑥 𝑑𝑥 𝑓 ˆ𝑥 𝑓 ˆ𝑥 ˆ 𝑔 ˆ𝑥2


zy
𝑑 𝑑
1 𝑑 𝑔 ˆ 𝑥 𝑔 ˆ𝑥 𝑑𝑥 𝑓 ˆ𝑥𝑓 ˆ𝑥 𝑑𝑥 𝑔 ˆ𝑥
𝑔 ˆ𝑥 𝑑𝑥
𝑓 ˆ𝑥 𝑔 ˆ 𝑥 ˆ𝑔 ˆ𝑥
2 .

39 Some special trigonometric limits

#1: lim sin𝜃 𝜃
ˆ 
1. In another words, sinˆ𝜃  𝜃 as 𝜃 0.
𝜃 0

1
am 𝑦 sinˆ𝜃 
𝜃

2𝜋 2𝜋

 1
.R
#2: lim cos 𝜃𝜃
ˆ  1
1
𝜃 0
sinˆ𝜃 
𝑦 𝜃
1

2𝜋 2𝜋
A

 1

EX 39.1. Estimate sinˆ0.01769 without calculator . sin 0.01769

ˆ   0.01769.
EX 39.2. Find lim tanˆ7𝑥 = lim sinˆ7𝑥 1
lim 7𝑥 1 7
lim 1 7
.
𝑥 0 sinˆ4𝑥 𝑥 0 cosˆ7𝑥 sinˆ4𝑥 𝑥 0 cosˆ7𝑥 4𝑥 4 𝑥 0 𝑐𝑜𝑠ˆ7𝑥 4

tanˆ7𝑥 tanˆ7𝑥 sin 7𝑥 ˆ  1 7𝑥 4𝑥 sin 7𝑥*1 1 


ˆ * 1 4𝑥 

 * 17𝑥
 7
Another solution: Find lim = lim cos lim   4𝑥 4
𝑥 .
𝑥 0 sinˆ4𝑥 sinˆ4𝑥 7𝑥 sinˆ4𝑥 7𝑥 4𝑥 7𝑥 cos 7𝑥 sin 4
ˆ 
 ˆ  ˆ 
𝑥 0 𝑥 0  

40 Composition of function
Definition 40.1. The composition of two functions: 𝑔 X 𝑓 ˆ𝑥 is defined by 𝑔 X 𝑓 ˆ𝑥 𝑔 ˆ𝑓 ˆ𝑥.
EX 40.1. The table below define the functions 𝑓 and 𝑔. so find the following:

𝑥 1 2 3 4 5 𝑥 4 5 6 7 8 9
𝑓 ˆ𝑥 8 3 6 7 4 𝑔 ˆ𝑥 1 3 8 10 2 2

29
 (1) 𝑔 X 𝑓 ˆ4=𝑔 ˆ7 10.
(2) 𝑓 X 𝑔 ˆ4=𝑓 ˆ1 8.
(3) 𝑓 X 𝑓 ˆ2=𝑓 ˆ3 6.
(4) 𝑓 X 𝑔 ˆ6=𝑓 ˆ8 which 𝐷.𝑁.𝐸, therefore, 6 is not in the domain of 𝑓 X 𝑔.
Note: from the first and the second example we conclude that 𝑓 X 𝑔 x 𝑔 X 𝑓 .
EX 40.2. Let 𝑝ˆ𝑥 𝑥2  𝑥, 𝑞 ˆ𝑥  2𝑥. find:

(1) 𝑞 X 𝑝ˆ1=𝑞 ˆ𝑝ˆ1 𝑞 ˆ12  1 𝑞 ˆ2 2 2 4.

(2) 𝑞 X 𝑝ˆ𝑥=𝑞 ˆ𝑝ˆ𝑥 𝑞 ˆ𝑥2  𝑥 2 ˆ𝑥2  𝑥 2𝑥2  2𝑥.
(3) 𝑝 X 𝑞 ˆ𝑥=𝑝ˆ𝑞 ˆ𝑥 𝑝ˆ2𝑥 ˆ2𝑥2  ˆ2𝑥 4𝑥2  2𝑥.
(4) 𝑝 X 𝑝ˆ𝑥=𝑝ˆ𝑝ˆ𝑥 𝑝ˆ𝑥2  𝑥 ˆ𝑥2  𝑥2  𝑥2  𝑥 𝑥4  2𝑥3  2𝑥2  𝑥.
⌋︂
EX 40.3. ℎˆ𝑥 𝑥2  7, find two functions 𝑓 and 𝑔 such that ℎˆ𝑥 𝑓 X 𝑔 ˆ𝑥 .
⌋︂ ⌋︂
The two functions are 𝑓 ˆ𝑥 𝑥, 𝑔 ˆ𝑥 𝑥2  7. Another solution: 𝑓 ˆ𝑥 𝑥  7, 𝑔 ˆ𝑥 𝑥2 .

zy
41 Solving rational equations
Solving rational equations steps:
#1 Find LCD.
#2 Clear denominators.
#3 Simplify an solve.
#4 Plug in solutions.
EX 41.1. Solve 1
𝑥3
1
am
1
𝑥
.
#1 Find LCD: LCD ˆ𝑥  3 𝑥
#2 Clear denominators: ˆˆ𝑥  3 𝑥 𝑥1 3 ˆˆ𝑥  3 𝑥ˆ1  1
Ô 𝑥2
 ˆˆ𝑥  3 𝑥 1  ˆ𝑥  3
Ô Ô 𝑥 43 .
 𝑥
#3 Simplify and solve: 𝑥2 𝑥2  3𝑥  𝑥  2 0 4𝑥  3 

Ô
3
3  3  1  3
#4 Plug in solutions: 4
3 3 1 4 9 3
, which means that 𝑥 4
is a solution.
4

EX 41.2. Solve 4𝑐
𝑐5

1
𝑐1
2
3𝑐 3
𝑐2 4𝑐5
.
.R
#1 Find LCD: Factors of denominators are ˆ𝑐  5, ˆ𝑐  1, and 𝑐2  4𝑐  5 ˆ𝑐  5ˆ𝑐  1. So LCD ˆ𝑐  5ˆ𝑐  1.
2
#2 Clear denominators: ˆ𝑐  5ˆ𝑐  1ˆ 𝑐4𝑐5   ˆ𝑐  5ˆ𝑐  1ˆ 𝑐 11  ˆ𝑐  5ˆ𝑐  1ˆ 𝑐23𝑐 4𝑐 35 . 

Ô Ô Ô
   

#3 Simplify and solve: 4𝑐2  4𝑐  𝑐  5 3𝑐2  3 𝑐2  3𝑐  2 0 ˆ𝑐  1ˆ𝑐  2 0 𝑐 1, or 𝑐 2.

#4 Plug in solutions: After plugging in the solutions we find that 𝑐 1 is a solution, while 𝑐 2 is extraneous
solution.
A

42 Derivative of Trigonometric functions

If we measured the slope of the tangent line (which is the derivative) of 𝑦 sinˆ𝑥 in a some interval, we will find
that at 𝑥 0 the slope is 1. And as 𝑥 goes from 0 to 𝜋2 , the slope goes from 1 to 0. If we extended this process, we will
find that the behavior of the derivative of 𝑦 sinˆ𝑥 is similar to the behavior of the function 𝑦 cosˆ𝑥. Similarly,
we can show that the behavior of the derivative of 𝑦 cosˆ𝑥 is similar to the behavior of 𝑦  sinˆ𝑥. This proves
œ
(graphically) that sin ˆ𝑥 cosˆ𝑥 and cos ˆ𝑥  sinˆ𝑥. œ

𝑦 cosˆ𝜃 𝑦 sinˆ𝜃
𝑦  sinˆ𝜃 𝑦 cosˆ𝜃
1 1

2𝜋 2𝜋  2𝜋 2𝜋
 1  1

𝑑 𝑑 sin 𝑥 cosˆ𝑥 cosˆ𝑥sinˆ𝑥ˆ sinˆ𝑥 cos2 ˆ𝑥sin2 ˆ𝑥 1

EX 42.1. Find 𝑑𝑥
tanˆ𝑥= 𝑑𝑥 ˆ cos
ˆ

ˆ 𝑥
 cos2 ˆ𝑥 cos2 ˆ𝑥 cos2 ˆ𝑥
sec2 ˆ𝑥.
𝑑 𝑑 1 ˆ sinˆ𝑥 1 sinˆ𝑥
EX 42.2. Find 𝑑𝑥
secˆ𝑥= 𝑑𝑥 ˆ cos 𝑥 ˆ 
 cos2 ˆ𝑥 cosˆ𝑥 cosˆ𝑥
secˆ𝑥 tanˆ𝑥.

30
 𝑑 𝑑 cos 𝑥 sinˆ𝑥ˆ sinˆ𝑥cosˆ𝑥 cosˆ𝑥 sin2 ˆ𝑥cos2 ˆ𝑥 1
EX 42.3. Find 𝑑𝑥
cotˆ𝑥= 𝑑𝑥 ˆ sin
ˆ

ˆ 𝑥
 sin2 ˆ𝑥
ˆ

sin2 ˆ𝑥 sin2 ˆ𝑥

 csc2 ˆ𝑥.
𝑑 𝑑 1 ˆcosˆ𝑥 1 cosˆ𝑥
EX 42.4. Find 𝑑𝑥
cscˆ𝑥= 𝑑𝑥 ˆ sin 𝑥 ˆ 
 sin2 ˆ𝑥 sinˆ𝑥 sinˆ𝑥
 cscˆ𝑥 cotˆ𝑥.
Summary:
𝑑 𝑑 𝑑
𝑑𝑥
sinˆ𝑥 cosˆ𝑥 𝑑𝑥
tanˆ𝑥 sec2 ˆ𝑥 𝑑𝑥
secˆ𝑥 secˆ𝑥 tanˆ𝑥
𝑑 𝑑 2 𝑑
𝑑𝑥
cos ˆ𝑥  sinˆ𝑥
𝑑𝑥
cot ˆ𝑥  csc ˆ𝑥
𝑑𝑥
csc ˆ𝑥  cscˆ𝑥 cotˆ𝑥

EX 42.5. For 𝑔 ˆ𝑥 𝑥 cosˆ𝑥

𝑀 cotˆ𝑥
, find 𝑔 ˆ𝑥 œ
.
ˆ 𝑀 cotˆ𝑥ˆ𝑥ˆ sinˆ𝑥cosˆ𝑥ˆ𝑥 cosˆ𝑥ˆ csc2 ˆ𝑥
𝑔 ˆ𝑥
œ
cosˆ𝑥 2
ˆ𝑀  
sinˆ𝑥
cosˆ𝑥 cosˆ𝑥 1
𝑀 𝑥 sinˆ𝑥𝑀 cosˆ𝑥𝑥 sinˆ sin
𝑥 
ˆ𝑥cosˆ𝑥
 sinˆ𝑥
𝑥 cosˆ𝑥
sin2 ˆ𝑥 sin2 ˆ𝑥

cosˆ𝑥 2 sin2 ˆ𝑥
ˆ𝑀  
sinˆ𝑥
𝑀 𝑥 sin3 ˆ𝑥𝑀 cosˆ𝑥 sin2 ˆ𝑥𝑥 cosˆ𝑥 sin2 ˆ𝑥cos2 ˆ𝑥 sinˆ𝑥𝑥 cosˆ𝑥
ˆ𝑀 sinˆ𝑥cosˆ𝑥
2 .

43 Proofs of limit and derivative of sine and cosine

zy
Theorem 43.1. lim sin𝜃 𝜃 ˆ 
1.
𝜃 0
𝑝𝑟𝑜𝑜𝑓. From the opposite figure, It is obvious that:
the area of B the area of Á B the area of . But since:
area of 1
2
𝑏𝑎𝑠𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 cos 𝜃 2sin 𝜃 ˆ  ˆ 
𝜃

area of Á 𝜋𝑟2 2𝜋 𝜃
𝜋 12 2𝜋 𝜃 𝜃
2
area of 1
2
𝑏𝑎𝑠𝑒
cosˆ𝜃  sinˆ𝜃 
ℎ𝑒𝑖𝑔ℎ𝑡
B
1
2
1
𝜃
B tanˆ𝜃 
am
tanˆ𝜃 tan2 𝜃 . Then:
Ô cosˆ𝜃 sinˆ𝜃 B 𝜃 B
ˆ 

sinˆ𝜃 
Ô  ˆ𝜃
cos :

 1
 B sin 𝜃
𝜃
B 1 
 *Ô 1
2 2 2 cosˆ𝜃  ˆ  cos 𝜃 ˆ 

lim sin𝜃ˆ𝜃
𝜃 0
1 Ô lim sin𝜃ˆ𝜃
𝜃 0
1.
Note: When we divided over sinˆ𝜃 previously, the inequality didn’t flip. Because sinˆ𝜃 is positive when 𝜃 is
positive (namely 𝜃 0 ). And if 𝜃 0 , sinˆ𝜃 will be negative and the inequality will flip, but the theorem will
 

also be true.
EX 43.1. Proof that lim cos 𝜃𝜃
𝜃 0
ˆ  1
0 .
.R
2
sin2 ˆ𝜃 
lim cos 𝜃𝜃 ˆ  1 cosˆ𝜃 1
cosˆ𝜃 1
lim 𝜃cos
cosˆ
ˆ

ˆ
𝜃 1
𝜃 1
lim 𝜃 

ˆ cosˆ𝜃 1
lim  sinˆ𝜃 
𝜃
sinˆ𝜃 
cosˆ𝜃 1
 1 0
11
0.
𝜃 0 𝜃 0 𝜃 0 𝜃 0
EX 43.2. Proof that 𝑑
𝑑𝑥
sinˆ𝑥 cosˆ𝑥 .
sinˆ𝑥ℎsinˆ𝑥
𝑑
𝑑𝑥
sinˆ𝑥 ℎ
lim lim sinˆ𝑥 cosˆℎcosℎˆ𝑥 sinˆℎsinˆ𝑥 lim sinˆ𝑥ˆcosˆℎ1cosˆ𝑥 sinˆℎ
ℎ
ℎ 0 ℎ 0 ℎ 0
*0 1
A

cosˆℎ

1 sin
ˆℎ

*
lim sinˆ𝑥  lim cosˆ𝑥 cosˆ𝑥.
ℎ 0  ℎ

ℎ 0 ℎ
EX 43.3. Proof that 𝑑
𝑑𝑥
cosˆ𝑥  sinˆ𝑥 .
cosˆ𝑥ℎcosˆ𝑥
𝑑
𝑑𝑥
cosˆ𝑥 ℎ
lim lim 𝑐𝑜𝑠ˆ𝑥 cosˆℎsinℎˆ𝑥 sinˆℎcosˆ𝑥 lim cosˆ𝑥ˆcosˆℎ1sinˆ𝑥 sinˆℎ
ℎ
ℎ 0 ℎ 0 ℎ 0
*0
 1
lim cosˆ𝑥 cos
ˆℎ1 sin
ˆℎ

*
 lim  sinˆ𝑥  sinˆ𝑥.
ℎ 0  ℎ ℎ 0  ℎ

44 Rectilinear motion
EX 44.1. A particle moves up and down along a straight line. Its position in 𝑚𝑚 at time 𝑡 seconds is given by the
equation 𝑠ˆ𝑡 𝑡4  16
3
𝑡3  6𝑡2 . Find 𝑠 ˆ𝑡 and 𝑠 ˆ𝑡. what do the represent? œ œœ

Solution
𝑠 ˆ𝑡 4𝑡  16𝑡  12𝑡, and 𝑠 ˆ𝑡 represents velocity (𝑣 ˆ𝑡), because 𝑠 ˆ𝑡 𝑑𝑠
œ 3 2 œ

𝑑𝑡
is the change of position over œ

time.
𝑑
𝑠 ˆ𝑡 12𝑡2  32𝑡  12, and 𝑠 ˆ𝑡 represents acceleration (𝑎ˆ𝑡), because 𝑠 ˆ𝑡 𝑑𝑡
œœ œœ
ˆ𝑠 ˆ𝑡 is the change of œœ œ

velocity over time.

31
 • Notes:
#1 𝑣 ˆ𝑡 A 0 means the position increasing, so particle is moving up.
#2 𝑣 ˆ𝑡 @ 0 means the position decreasing, so particle is moving down.
#3 𝑣 ˆ𝑡 0 means particle is at rest.

• We know from physics that 𝐹 𝑚 𝑎, so if:

#1 𝑎ˆ𝑡 A 0 it looks like the particle is being pulled up (force in positive direction).
#2 𝑎ˆ𝑡 @ 0 it looks like the particle is being pulled down (force in negative direction).
#3 𝑎ˆ𝑡 0 means that there is no force on the particle in that instant, and Velocity continues as it is.

EX 44.2. use the table of values to describe the particle’s motion at time 1.5 seconds and at time 2.5 seconds.
𝑡 𝑠ˆ𝑡 𝑠 ˆ𝑡 𝑠 ˆ𝑡
œ œœ

1.5 0.5625 4.5 9

2.5 6.77 7.5 7

zy
Solution
𝑠ˆ𝑡
10

 5
1 2
am 3 4 𝑡

 10

• At 𝑡 1.5 seconds:
# 𝑠ˆ𝑡 A 0 Ô particle is above position 0.
# 𝑠 ˆ𝑡 @ 0 Ô its position is decreasing or the particle is moving down.
œ

# 𝑠 ˆ𝑡 @ 0 Ô velocity is decreasing (namely getting more and more negative).

.R
œœ
Therefore, the particle is
moving down faster and faster. Because 𝑠𝑝𝑒𝑒𝑑 ⋃︀𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ⋃︀ is increasing.
• At 𝑡 2.5 seconds:
# 𝑠ˆ𝑡 @ 0 Ô particle is below position 0.
# 𝑠 ˆ𝑡 @ 0 Ô its position is decreasing or the particle is moving down.
œ

# 𝑠 ˆ𝑡 A 0 Ô velocity is increasing (namely less negative). Therefore, the particle is slowing down. Because
A

œœ

𝑠𝑝𝑒𝑒𝑑 ⋃︀𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ⋃︀ is decreasing.

Notes:
(1) When velocity and acceleration have the same sign Ô particle is speeding up.
(2) When velocity and acceleration have different signs Ô particle is slowing down.
EX 44.3. A particle moves up and down along a straight line. Its position in 𝑚𝑚 at time 𝑡 seconds is given by the
equation 𝑠ˆ𝑡 𝑡4  16
3
𝑡3  6𝑡2 .
𝑠ˆ𝑡
10
𝑎ˆ𝑡 𝑣 ˆ𝑡 𝑠ˆ𝑡
5

1 2 3 4 𝑡
 5

 10

32
 (1) When the particle is at rest ? The particle is at rest when 𝑠 ˆ𝑡 0 œ
Ô 4𝑡3  16𝑡2  12𝑡 0 Ô 4𝑡ˆ𝑡2  4𝑡  3
0 Ô 4𝑡ˆ𝑡  3ˆ𝑡  1 0. Therefore, the particle is at rest at 𝑡 0, 3, 1.
(2) When the particle is moving up ? moving down ?
* The particle is moving up when 𝑠 ˆ𝑡 𝑣ˆ𝑡 A 0.
œ

* The particle is moving down when 𝑠 ˆ𝑡 𝑣ˆ𝑡 @ 0.œ

𝑣 ˆ𝑡                          
0 0.45 1 2.22 3
Therefore, particle is moving down when 𝑡 in ˆª, 0 8 ˆ1, 3. Particle is moving up when 𝑡 in ˆ0, 1 8 ˆ3, ª.
(3) when is the particle speeding up? slowing down ?
* The particle is speeding up when 𝑎ˆ𝑡 and 𝑣ˆ𝑡 both positive or negative.
* The particle is slowing down when 𝑎ˆ𝑡 and 𝑣ˆ𝑡 have opposite signs.
Ô Ô
⌋︂

at 𝑎ˆ𝑡 0 12𝑡2  32𝑡  12 0 𝑡 34  37  0.45, 2.22.

𝑣 ˆ𝑡                          
𝑎ˆ𝑡
                          
0 0.45 1 2.22 3
⌋︂ ⌋︂

zy
Therefore, the particle is slowing down when 𝑡 in ˆª, 0 8 ˆ 34 
3
7
, 1 8 ˆ 43 
3
7
, 3. The particle is speeding
⌋︂ ⌋︂
4 7 4 7
up when 𝑡 in ˆ0,  3 3
 8 ˆ1, 3

3
 8 ˆ3, ª.
Note: In the previous example we determined the curves of 𝑠ˆ𝑡, 𝑣 ˆ𝑡 𝑠 ˆ𝑡 and 𝑎ˆ𝑡
œ
𝑠 ˆ𝑡 according to the
œœ

facts:
(1) 𝑣 ˆ𝑡 A 0 when 𝑠ˆ𝑡 increasing.
(2) 𝑎ˆ𝑡 A 0 when 𝑣 ˆ𝑡 increasing.
EX 44.4. A particle moves up and down along a straight line. Its position in 𝑚𝑚 at time 𝑡 seconds is given by the
equation 𝑠ˆ𝑡 𝑡4  16
3
𝑡3  6𝑡2
am
(1) what is the net change in position for the particle between 1 and 4 seconds ? Net change is 𝑠ˆ4  𝑠ˆ1
32 5
3

3
9𝑚𝑚.
(2) what is the total distance traveled by the particle between 1 and 4 seconds ? The total distance is not 𝑠ˆ4  𝑠ˆ1.
Because the particle changes direction at 𝑡 1 and 𝑡 3.
𝑠ˆ𝑡
10
.R
5

1 2 3 4 𝑡
5

A

 10
32 5
Therefore, the total distance ⋃︀𝑠ˆ4  𝑠ˆ3⋃︀  ⋃︀𝑠ˆ3  𝑠ˆ1⋃︀ ⋂︀ 3  ˆ27⋂︀  ⋂︀27  3 ⋂︀ 66.3𝑚𝑚.

45 Marginal cost
EX 45.1. Suppose that the total cost of producing 𝑋 tie-dyed T-shirts is 𝑐ˆ𝑥:

(1) Is 𝑐ˆ𝑥 increasing or decreasing ? 𝑐ˆ𝑥 is increasing. Because the cost of T-shirts will increase as the number of
T-shirts increase.
(2) Is 𝑐 ˆ𝑥 increasing or decreasing ? The reasonable representation of 𝑐 ˆ𝑥 is:
œ œ

𝑎
Because (for example) the cost of one T-shirt when we are making 1000 T-shirts is less than the cost of one T-shirt
when we are making few T-shirts. Therefore 𝑐 ˆ𝑥 is decreasing. œ

33
(3) Interpret the following :
1. 𝑐ˆ204  𝑐ˆ200 : Additional cost for making 204 T-shirts instead of 200 T-shirts.
2. 𝑐 204 4 𝑐 200 : This equals the average rate of change of 𝑐ˆ𝑥 (its unit probably $/T-shirt).
ˆ  ˆ 

3. 𝑐 ˆ200 : This is the instantaneous rate of change of 𝑐ˆ𝑥 in $/T-shirt.

œ

Note: 𝑐ˆ𝑥 is called the cost function. 𝑐 ˆ𝑥 is called the marginal cost.
œ

⌋︂
EX 45.2. Suppose that the cost function for producing ipads is 𝑐ˆ𝑥 500  300 𝑥. Find and interpret:
(1) 𝑐ˆ401  𝑐ˆ400 $7.5 which means that it costs $7.5 to go from 400 to 401 ipads.
(2) 𝑐 ˆ400 : 𝑐 ˆ𝑥 300 21 𝑥 2 , hence 𝑐 ˆ400 $7.5  𝑐 401 1 𝑐 400 which is the rate of change. 𝑐 ˆ𝑥 is called
œ œ  œ ˆ  ˆ  œ

the marginal cost and represents the rate in which the cost function is increasing with each additional item.

46 Logarithms
Definition 46.1. log𝑎 ˆ𝑏 means 𝑎𝑐 𝑏, 𝑎 is called the base of logarithm. The base is required to be positive
number.
EX 46.1. Evaluate the following expressions by hand by writing them using exponents instead of logs:

zy
(1) log2 ˆ16=4 because 24 16.
(2) log2 ˆ2=1 because 21 2.
(3) log2 ˆ 21 =  1 because 2 1 12 . 

(4) log2 ˆ 18 =  3 because 2 3 18 . 

(5) log2 ˆ1=0 because 20 1.

(6) log10 ˆ1, 000, 000=6 because 106 1, 000, 000.
(7) log10 ˆ0.001=  3 because 10 3 0.001. 

(8) log10 ˆ0 𝐷.𝑁.𝐸.

(9) log10 ˆ100 𝐷.𝑁.𝐸.
am
Note: It is possible to take the 𝑙𝑜𝑔 of numbers that are larger than 0, but not numbers that are smaller than or
equal 0. In other words, the domain of the function 𝑓 ˆ𝑥 log𝑎 ˆ𝑥 is ˆ0, ª.
Note: lnˆ𝑥 means log𝑒 ˆ𝑥, and called the natural 𝑙𝑜𝑔. While logˆ𝑥 means log10 ˆ𝑥, and called the common
𝑙𝑜𝑔.
EX 46.2. Rewrite using exponents:
(1) log3 ˆ 19  2 Ô
3 2 19 . 

Ô
.R
(2) logˆ13 1.11394 101.11394 13.
1
(3) lnˆ 𝑒  1 Ô
𝑒  1 1𝑒 .ˆ

EX 46.3. Rewrite the following using 𝑙𝑜𝑔𝑠. Do not solve for any variables.
(1) 3𝑢 9.78 Ô
log3 ˆ9.78 𝑢.
(2) 𝑒3𝑥 7 4  𝑦

Ô
lnˆ4  𝑦  3𝑥  7.
A

47 log functions and their graphs

EX 47.1. Graph 𝑦 log2 ˆ𝑥 by plotting points.

Solution

𝑥 𝑓 ˆ𝑥
1
4
log2 ˆ 14  2
1
2
log2 ˆ 12  1
1 log2 ˆ1 0
2 log2 ˆ2 1
4 log2 ˆ4 2
8 log2 ˆ8 𝐷.𝑁.𝐸

It obvious from the graph that:

34
 (1) Domain is ˆ0, ª.
(2) Range is ˆª, ª.
(3) The vertical asymptote is 𝑦-axis which is the axis 𝑥 0.
EX 47.2. Graph 𝑦 lnˆ𝑥  5. Find the domain, range and asymptotes.

Solution

𝑦 lnˆ𝑥  5
𝑎
ˆ1, 5

𝑦 lnˆ𝑥

ˆ1, 0

zy
𝑦 lnˆ𝑥 𝑦 lnˆ𝑥  5
Domain: ˆ0, ª ˆ0, ª
Range: ˆª, ª ˆª, ª
Vertical asymptotes: 𝑥 0 𝑥 0
EX 47.3. Graph 𝑦 logˆ𝑥  2. Find the domain, range and asymptotes.
am Solution

𝑦 logˆ𝑥  2
ˆ1, 0

2 ˆ1, 0

𝑦 logˆ𝑥
.R
𝑦 logˆ𝑥 𝑦 logˆ𝑥  2
Domain: ˆ0, ª ˆ2, ª
Range: ˆª, ª ˆª, ª
Vertical asymptotes: 𝑥 0 𝑥  2
EX 47.4. Find the domain of 𝑓 ˆ𝑥 lnˆ2  3𝑥 . We need 2  3𝑥 A 0 Ô 2 A 3𝑥 Ô 𝑥 @ 23 . Therefore, domain is
A

2
ˆª, 3 .

48 Rules for combining logs and exponents

EX 48.1. Evaluate:

(1) log10 ˆ103 =3.

(2) log𝑒 ˆ𝑒4.2 =4.2.
(3) 10log10 1000 =1000.
ˆ 

(4) 𝑒log𝑒 9.6 =9.6.

ˆ 

From the previous example we can conclude the following:

(1) For any base 𝑎, log𝑎 ˆ𝑎𝑥  𝑥.
(2) For any base 𝑎, 𝑎log𝑎 𝑥 𝑥. ˆ 

(3) Log and Exponent with the same base undo each other.
EX 48.2. Find:

(1) 3log3 ˆ 1.4

=1.4.

35
(2) lnˆ𝑒𝑥 = log𝑒 ˆ𝑒𝑥  𝑥.
(3) 10log 3𝑧 =10log10 3𝑧 3𝑧.
ˆ  ˆ 

EX 48.3. True or False: lnˆ10𝑥  𝑥 . False, because lnˆ10𝑥  log𝑒 ˆ10𝑥  x 𝑥. For example, at 𝑥 1,
log𝑒 ˆ101  2.3 x 1.

49 Log rules
The following table contains some 𝑙𝑜𝑔 rules and their corresponding exponent rules.
exponent rule 𝑙𝑜𝑔 rule name of the rule
𝑎0 1 log𝑎 ˆ1 0 

𝑎𝑚 𝑎𝑛 𝑎𝑚 𝑛 log𝑎 ˆ𝑥𝑦  log𝑎 ˆ𝑥  log𝑎 ˆ𝑥


product rule
𝑎𝑚
𝑎𝑛
𝑎𝑚 𝑛 
log𝑎 ˆ 𝑥𝑦  log𝑎 ˆ𝑥  log𝑎 ˆ𝑦  quotient rule
𝑚 𝑛 𝑚𝑛
ˆ𝑎  𝑎 log𝑎 ˆ𝑥𝑛  𝑛 log𝑎 ˆ𝑥 power rule
EX 49.1. Rewrite the following as a sum or difference of logs:
𝑥
(1) logˆ 𝑦𝑧 = logˆ𝑥  logˆ𝑦𝑧  logˆ𝑥  ˆlogˆ𝑦   logˆ𝑧 .

zy
(2) logˆ5 2𝑡 = logˆ5  logˆ2𝑡  logˆ5  𝑡 logˆ2.

EX 49.2. Rewrite the following as a single log:

(1) log5 ˆ𝑎  log5 ˆ𝑏  log5 ˆ𝑐= log5 ˆ𝑎  log5 ˆ𝑐  log5 ˆ𝑏 log5 ˆ𝑎𝑐  log5 ˆ𝑏 log5 ˆ 𝑎𝑐
𝑏
.
am
(2) lnˆ𝑥  1  lnˆ𝑥  1  2 lnˆ𝑥2  1= lnˆˆ𝑥  1ˆ𝑥  1  lnˆˆ𝑥2  12  lnˆ ˆ
2
𝑥 1
𝑥2 12
 lnˆ 𝑥21 1 .


50 The chain rule

EX 50.1. Write the following functions as a composition of functions:
⌈︂
(1) ℎˆ𝑥 sinˆ𝑥
R
inner function: 𝑔 ˆ𝑥 sin
⌋︂
ˆ𝑥.
outer function: 𝑓 ˆ𝑢 𝑢.

5ˆtanˆ𝑥  secˆ𝑥3
A.

(2) 𝐾 ˆ𝑥

inner function: 𝑔 ˆ𝑥 tanˆ𝑥  secˆ𝑥.

outer function: 𝑓 ˆ𝑢 5𝑢3 .

𝑥2 
(3) ℎˆ𝑥 𝑟ˆ𝑥 𝑒sin ˆ

inner function: 𝑔 ˆ𝑥 𝑥2 .

outer function: 𝑓 ˆ𝑢 𝑒𝑠𝑖𝑛 𝑢
ˆ
.
or
inner function: 𝑔 ˆ𝑥 sinˆ𝑥2 .
outer function: 𝑓 ˆ𝑢 𝑒𝑢 .
or
inner function: 𝑔 ˆ𝑥 𝑥2 .
middle function: ℎˆ𝑣  sinˆ𝑣 .
outer function: 𝑓 ˆ𝑢 𝑒𝑢 .
Theorem 50.1. (The chain rule) If 𝑔 is differentiable at 𝑥, 𝑓 is differentiable at 𝑔 ˆ𝑥. Then ˆ𝑓 X 𝑔  ˆ𝑥 𝑓 ˆ𝑔 ˆ𝑥 𝑔 ˆ𝑥.
œ œ œ

or
𝑑𝑦
If 𝑢 𝑔 ˆ𝑥, 𝑦 𝑓 ˆ𝑢 𝑓 ˆ𝑔 ˆ𝑥, then 𝑑𝑢 𝑑𝑥
𝑔 ˆ𝑥 and 𝑑𝑢 œ
𝑓 ˆ𝑢 𝑓 ˆ𝑔 ˆ𝑥. Therefore, the above formula can be
œ œ

36
written as follows:
𝑑𝑦 𝑑𝑦 𝑑𝑢
𝑑𝑥 𝑑𝑢 𝑑𝑥
.
⌈︂
EX 50.2. Find the derivative of ℎˆ𝑥 sinˆ𝑥.

Solution

1
ℎˆ𝑥 ˆsinˆ𝑥 2 , hence:
inner function is 𝑔 ˆ𝑥 sinˆ𝑥, 𝑔 ˆ𝑥 cosˆ𝑥. œ

1 1
outer function is 𝑓 ˆ𝑢 𝑢 2 , 𝑓 ˆ𝑢 12 𝑢 2 . œ 

1
Therefore: ℎ ˆ𝑥 𝑓 ˆ𝑔 ˆ𝑥 𝑔 ˆ𝑥 12 ˆsinˆ𝑥
œ œ œ 
2 cosˆ𝑥.
EX 50.3. Find the derivative of 𝑘 ˆ𝑥 5ˆtanˆ𝑥  secˆ𝑥3 .

Solution

zy
inner function is 𝑔 ˆ𝑥 tanˆ𝑥  secˆ𝑥, 𝑔 ˆ𝑥 sec2 ˆ𝑥  secˆ𝑥 tanˆ𝑥. œ

outer function is 𝑓 ˆ𝑢 5𝑢3 , 𝑓 ˆ𝑢 15𝑢2 . œ

Therefore: 𝑘 ˆ𝑥 𝑓 ˆ𝑔 ˆ𝑥 𝑔 ˆ𝑥 15ˆtanˆ𝑥  secˆ𝑥2 ˆsec2 ˆ𝑥  secˆ𝑥 tanˆ𝑥.
œ œ œ

𝑥2 
EX 50.4. Find the derivative of 𝑟ˆ𝑥 𝑒sin ˆ
am Solution

𝑥2  𝑥2  𝑑 𝑥2  𝑑 𝑥2 
𝑟 ˆ𝑥
œ
𝑒sin ˆ
𝑒sin ˆ

𝑑𝑥
ˆsinˆ𝑥2  𝑒sin ˆ
cosˆ𝑥2  𝑑𝑥 ˆ𝑥2  𝑒sin ˆ
cosˆ𝑥2  2𝑥.

51 More on the chain rule

EX 51.1. Show that ˆ5𝑥  œ
lnˆ5 5𝑥 . Hint, rewrite 5𝑥 as 𝑒ln ˆ 5 𝑥
.
.R
Solution

ˆ5𝑥 œ ˆ𝑒lnˆ5 𝑥 œ ˆ𝑒ˆlnˆ5 𝑥 œ 𝑒 ˆ lnˆ5 𝑥

lnˆ5 ˆ𝑒ˆlnˆ5 𝑥 5𝑥 lnˆ5.
𝑑
Theorem 51.1. 𝑑𝑥
ˆ𝑎𝑥  lnˆ𝑎 𝑎𝑥 .
⌋︂
A

EX 51.2. Find the derivative of 𝑦 sinˆ5𝑥 2cos ˆ 5𝑥  1

Solution

𝑑𝑦 𝑑 𝑑 1 1

𝑑𝑥
sinˆ5𝑥 𝑑𝑥 ˆˆ2cos 5𝑥  1 2   𝑑𝑥 ˆsinˆ5𝑥 ˆ2cos 5𝑥  1 2
ˆ  ˆ 

1 1
sinˆ5𝑥 12 ˆ2cos 5𝑥  1 2 𝑑𝑥
ˆ 𝑑
ˆ2cos 5𝑥  1  cosˆ5𝑥 5 ˆ2cos 5𝑥  1 2
 ˆ  ˆ 

1 1
sinˆ5𝑥 12 ˆ2cos 5𝑥  1 2 lnˆ2 2cos 5𝑥 𝑑𝑥
ˆ   𝑑
ˆcosˆ5𝑥  5 cosˆ5𝑥ˆ2cos 5𝑥  1 2 ˆ  ˆ 

1 1
sinˆ5𝑥 21 ˆ2cos 5𝑥  1 2 lnˆ2 2cos 5𝑥 ˆ sinˆ5𝑥 5  5 cosˆ5𝑥ˆ2cos 5𝑥  1 2 .
ˆ   ˆ  ˆ 

𝑑
EX 51.3. Using the following table of values, find 𝑑𝑥
ˆ𝑓 X 𝑔 ⋂︀𝑥 1 .
𝑥 𝑓 ˆ𝑥 𝑔 ˆ𝑥 𝑓 ˆ𝑥œ
𝑔 ˆ𝑥 œ

1 3 2 9 5

2 4 5 10 3

3 1 3 2 1

4 0 2 6 0

Solution

37
 𝑑
𝑑𝑥
ˆ𝑓 X 𝑔  𝑓 ˆ𝑔 ˆ𝑥 𝑔 ˆ𝑥, therefore:
œ œ

𝑑
𝑑𝑥
ˆ𝑓 X 𝑔 ⋂︀ 𝑓 ˆ𝑔 ˆ1 𝑔 ˆ1 𝑓 ˆ2
œ œ œ
ˆ5 10 ˆ5 50.

𝑥 1

proof of chain rule: (This proof is pseudo, because 𝑔 ˆ𝑥  𝑔 ˆ𝑎 might be 0)
𝑓 𝑔 𝑥 𝑓 𝑔 𝑎
𝑑
𝑑𝑥
ˆ𝑓 X 𝑔  lim
X

𝑥 𝑎
ˆ



lim 𝑓 𝑔 𝑥 𝑥 𝑓𝑎 𝑔 𝑎
X ˆ 


𝑔 𝑥 𝑔 𝑎
𝑔 𝑥 𝑔 𝑎
ˆ
lim 𝑓 𝑔𝑔𝑥𝑥
ˆ

ˆ
 ˆ ˆ  ˆ 


ˆ

ˆ



ˆ ˆ

ˆ
𝑓 ˆ𝑔 ˆ𝑎 𝑔 ˆ𝑥𝑔 ˆ𝑎


𝑔 ˆ𝑎
 𝑥𝑎
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎
𝑓 ˆ𝑔 ˆ𝑥𝑓 ˆ𝑔 ˆ𝑎 𝑔 ˆ𝑥𝑔 ˆ𝑎 𝑓 ˆ𝑔 ˆ𝑥𝑓 ˆ𝑔 ˆ𝑎 𝑓 ˆ𝑔 ˆ𝑥𝑓 ˆ𝑔 ˆ𝑎
lim 𝑔ˆ𝑥𝑔ˆ𝑎 lim 𝑥𝑎 lim 𝑔ˆ𝑥𝑔ˆ𝑎 𝑔 ˆ𝑥
œ
lim 𝑔 ˆ𝑥𝑔 ˆ𝑎
𝑔 œ ˆ𝑥
𝑥 𝑎 𝑥 𝑎 𝑥 𝑎 𝑔 ˆ𝑥 𝑔 ˆ𝑎
lim 𝑓 ˆ𝑢𝑢𝑔𝑓ˆˆ𝑎𝑔ˆ𝑎 𝑔 œ ˆ𝑥 𝑓 œ ˆ𝑔 ˆ𝑎 𝑔 œ ˆ𝑎.
𝑢 𝑔 ˆ𝑎

52 Implicit differentiation
Implicitly defined curves:

zy
am
2𝑦 2  𝑥2 2𝑥𝑦 3  2𝑥𝑦 𝑥6  𝑦 6 4𝑥2 𝑦 2  3𝑦 2 𝑥4  3𝑥2 𝑦 4

Explicitly defined curves:

.R
A

𝑥5
 5𝑥4  15𝑥 sinˆ5𝑥
3
𝑦 2 2
𝑦 5𝑥

EX 52.1. Find the equation of the tangent line for 9𝑥2  4𝑦 2 25 at the point ˆ1, 2.

Solution

Method 1. we solve for 𝑦:

Ô Ô
⌋︂
2 2
9𝑥2  4𝑦 2 25 𝑦 2 25 49𝑥 𝑦  252 9𝑥 . But since the tangent line at the point ˆ1, 2 (namely, 𝑦 is
 

positive), then we take the positive solution of the equation:

1
𝑦 12 ˆ25  9𝑥2  2 Ô 𝑑𝑦 1 1 1
ˆ25  9𝑥2  2 ˆ18𝑥
𝑑𝑦
⋂︀ 9
𝑠𝑙𝑜𝑝𝑒. Ô 

Ô
𝑑𝑥 2 2 𝑑𝑥 𝑥 1 8
9
Therefore, the equation of the tangent line is 𝑦  2 8 ˆ𝑥  1 𝑦 89 𝑥  25
8
 

Method 2. Implicit differentiation:

𝑑
𝑑𝑥
𝑑
ˆ9𝑥2  4𝑦 2  𝑑𝑥 ˆ25 Ô
𝑑
9 𝑑𝑥 𝑑
ˆ𝑥2   4 𝑑𝑥 ˆ𝑦  0. Next, we will derive 𝑦 and consider it a function of 𝑥 (because
it is a function 𝑥 for a small pieces of the curve):

38
 𝑑𝑦
18𝑥  4 2𝑦 𝑑𝑥 Ô 𝑑𝑥
𝑑𝑦 

4
9 𝑥
𝑦
, and by plugging the point ˆ1, 2 we get 𝑑𝑦
𝑑𝑥


8
9
.

𝑦 9 𝑥  25
8 8

9𝑥2  4𝑦 2 25

EX 52.2. Find 𝑦 if 𝑥3 𝑦 2  sinˆ𝑥𝑦 

œ
𝑥3  𝑦 3

zy
Solution

𝑑 𝑑
ˆ𝑥3 𝑦 2  sinˆ𝑥𝑦  𝑑𝑥 ˆ𝑥3  𝑦 3  Ô 𝑑𝑥𝑑 ˆ𝑥3𝑦2 
𝑑
ˆsinˆ𝑥𝑦 
𝑑 𝑑
ˆ𝑥3   𝑑𝑥 ˆ𝑦 3  Ô
Ô
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑑𝑦 2 𝑑𝑦
𝑥3 2𝑦 𝑑𝑥  3𝑥 𝑦
2 2
 cosˆ𝑥𝑦  ˆ𝑥  𝑦 2
3𝑥  3𝑦 𝑑𝑥
Ô
𝑑𝑥
3 𝑑𝑦 𝑑𝑦 2 𝑑𝑦
𝑥 2𝑦 𝑑𝑥  cosˆ𝑥𝑦  𝑥 𝑑𝑥  3𝑦 𝑑𝑥 3𝑥2  3𝑥 𝑦 2 2
 cosˆ𝑥𝑦  𝑦
𝑑𝑦 3𝑥2 3𝑥2 𝑦 2 cosˆ𝑥𝑦  𝑦
𝑑𝑥 𝑥3 2𝑦 cosˆ𝑥𝑦  𝑥3𝑦 2
.
am
𝑥3 𝑦 2  sinˆ𝑥𝑦  𝑥3  𝑦 3
.R

53 Derivative of exponential function

A

EX 53.1. Find 𝑑
𝑑𝑥
ˆ5𝑥  .𝑑
𝑑𝑥
ˆ5𝑥 
𝑑
𝑑𝑥
ˆˆ𝑒lnˆ5 𝑥 
𝑑
𝑑𝑥
ˆ𝑒lnˆ5 𝑥  𝑒lnˆ5 𝑥 lnˆ5.
𝑑 𝑑
In general: 𝑑𝑥
ˆ𝑎𝑥  𝑎𝑥 lnˆ𝑥, therefor 𝑑𝑥 ˆ𝑒𝑥  𝑒 𝑥
lnˆ𝑥 𝑥
𝑒 .

54 Derivative of log function

𝑑
EX 54.1. Find 𝑑𝑥 .
ˆlog𝑎 ˆ𝑥 𝑦 log𝑎 ˆ𝑥 Ô 𝑎𝑦 𝑥 Ô 𝑑𝑥𝑑 ˆ𝑎𝑦  𝑑
𝑑𝑥
ˆ𝑥 Ô lnˆ𝑎 𝑎𝑦 𝑑𝑦
𝑑𝑥
1 Ô
𝑑𝑦 1 𝑑 1
𝑑𝑥 ln 𝑎 𝑎𝑦
ˆ 
. Hence, 𝑑𝑥 ˆlog𝑎 ˆ𝑥 ln 𝑎 𝑥 . ˆ 

𝑑 1 1
Conclusion: 𝑥
ˆlnˆ𝑥 lnˆ𝑒 𝑥 𝑥
.
𝑑
EX 54.2. Find 𝑑𝑥
ˆlnˆ⋃︀𝑥⋃︀.

Solution

)︀
⌉︀
⌉︀𝑥 when 𝑥 A 0
The function 𝑦 lnˆ⋃︀𝑥⋃︀ is related to the function 𝑦 lnˆ𝑥. Because ⋃︀𝑥⋃︀ ⌋︀ . Therefore,
⌉︀
⌉︀𝑥 when 𝑥 @ 0
]︀

39
 )︀
⌉︀
⌉︀lnˆ𝑥 when 𝑥A0
lnˆ⋃︀𝑥⋃︀ ⌋︀ . Hence,
⌉︀
⌉︀ lnˆ𝑥 when 𝑥@0
]︀
)︀ 𝑑 )︀ 1
𝑑
⌉︀
⌉︀ ˆlnˆ𝑥 when 𝑥 A 0 ⌉︀
⌉︀ 𝑥 when 𝑥A0 𝑑 1
ˆlnˆ⋃︀𝑥⋃︀ ⌋︀ 𝑑𝑥 ⌋︀ 1 . Therefore, ˆlnˆ⋃︀𝑥⋃︀ .
𝑑𝑥 ⌉︀
⌉︀
𝑑
ˆlnˆ𝑥 when 𝑥 @ 0 ⌉︀
⌉︀ ˆ1 when 𝑥 @ 0 𝑑𝑥 𝑥
]︀ 𝑑𝑥 ]︀ 𝑥

55 Logarithmic differentiation
𝑑
EX 55.1. Find 𝑑𝑥
ˆ𝑥𝑥 

Solution

𝑦 𝑥𝑥 Ô lnˆ𝑦 𝑥 lnˆ𝑥 Ô 𝑑𝑥𝑑 ˆlnˆ𝑦 𝑑𝑥𝑑 ˆ𝑥 lnˆ𝑥 Ô

1
𝑦
𝑑𝑦
𝑑𝑥
𝑥 1
𝑥
 1 lnˆ𝑥 Ô 𝑑𝑥
𝑑𝑦
𝑦 ˆ1 lnˆ𝑥 Ô 𝑑𝑦

𝑥
𝑥𝑥 ˆ1  lnˆ𝑥.

zy
𝑑 1
EX 55.2. Find 𝑑𝑥
ˆtanˆ𝑥 𝑥

Solution

𝑦 ˆtanˆ𝑥 𝑥
1
Ô lnˆ𝑦 1
lnˆtanˆ𝑥
𝑥
Ô 𝑑𝑥𝑑 ˆlnˆ𝑦 𝑑𝑥𝑑 ˆ 𝑥1 lnˆtanˆ𝑥 Ô
1
𝑦
𝑑𝑦
𝑑𝑥
1
𝑥
1
tanˆ𝑥
sec ˆ𝑥  𝑥 2 lnˆtanˆ𝑥
2 
Ô 𝑑𝑥 𝑑𝑦
𝑦 ˆ 𝑥1 tan1 𝑥 sec2 ˆ𝑥 ln tan
𝑥
𝑥

ˆ
2
ˆ 


tanˆ𝑥
1
 𝑥
1 1
ˆ 𝑥 tanˆ𝑥
am
2
sec ˆ𝑥  lnˆtanˆ𝑥
𝑥2
.
ˆ 

𝑥 cosˆ𝑥
EX 55.3. Find the derivative of 𝑦 2
ˆ𝑥 5
5

Solution

lnˆ𝑦  lnˆ 𝑥𝑥cos 𝑥

2 𝑥 5
ˆ 
Ô lnˆ𝑦 lnˆ𝑥 cosˆ𝑥  lnˆˆ𝑥2  𝑥5  lnˆ𝑥  lnˆcosˆ𝑥  5 lnˆ𝑥2  𝑥 Ô
Ô
ˆ  
.R
𝑑 𝑑 𝑑 𝑑
𝑑𝑥
ˆlnˆ𝑦  𝑑𝑥 ˆlnˆ𝑥  𝑑𝑥 ˆlnˆcosˆ𝑥  5 𝑑𝑥 ˆlnˆ𝑥2  𝑥
1
𝑦 𝑑𝑥
𝑑𝑦
𝑥
1

1
cosˆ𝑥
1
ˆ sinˆ𝑥  5ˆ 𝑥2 𝑥 ˆ2𝑥  1 Ô 𝑑𝑦
𝑑𝑥
2
ˆ𝑥 5
5
ˆ
𝑥 cosˆ𝑥 𝑥
1

sinˆ𝑥
cosˆ𝑥

5ˆ2𝑥1
𝑥2 𝑥
.

56 Inverse function
Suppose 𝑓 ˆ𝑥 is a function defined by the chart:
A

𝑥 2 3 4 5
𝑓 ˆ𝑥 3 5 6 1
Inverse function for 𝑓 ˆ𝑥, written as 𝑓 1 ˆ𝑥, and it undoes what 𝑓 ˆ𝑥 does. In other words 𝑓
 1

ˆ3 2, 𝑓 1

ˆ5 3,
𝑓 1 ˆ6 4 and 𝑓 1 ˆ1 5.
 

Key fact 1. Inverse function reverse the roles of 𝑦 and 𝑥.

1
By graphing 𝑓 ˆ𝑥 and 𝑓 
ˆ𝑥 on the same axes we notice the following
6

2 4 6

1
Key fact 2. The graph of 𝑦 𝑓 
ˆ𝑥 is obtained from the graph 𝑦 𝑓 ˆ𝑥 by reflecting over the line 𝑦 𝑥.

40
𝑓 1 X 𝑓 ˆ2 𝑓 1 ˆ𝑓 ˆ2 𝑓 1 ˆ3
  
2 𝑓 X 𝑓 1 ˆ3 𝑓 ˆ𝑓 1 ˆ3 𝑓 2 3
  ˆ

𝑓 1 X 𝑓 ˆ3 𝑓 1 ˆ𝑓 ˆ3 𝑓 1 ˆ5

  
3 𝑓 X 𝑓 1 ˆ5 𝑓 ˆ𝑓 1 ˆ5 𝑓 3 5
  ˆ

𝑓 1 X 𝑓 ˆ4 𝑓 1 ˆ𝑓 ˆ4 𝑓 1 ˆ6

  
4 𝑓 X 𝑓 1 ˆ6 𝑓 ˆ𝑓 1 ˆ6 𝑓 4 6
  ˆ

𝑓 1 X 𝑓 ˆ5 𝑓 1 ˆ𝑓 ˆ5 𝑓 1 ˆ1

  
5 𝑓 X 𝑓 1 ˆ1 𝑓 ˆ𝑓 1 ˆ1 𝑓 5 1
  ˆ

Key fact 3. 𝑓 1 X 𝑓 ˆ𝑥 𝑥 and 𝑓 X 𝑓

  1
ˆ𝑥 𝑥. This is the mathematical way of saying that 𝑓 and 𝑓  1
undo each other.

EX 56.1. 𝑓 ˆ𝑥 𝑥3 , guess what the inverse of 𝑓 ˆ𝑥 should be.

Solution
⌋︂ ⌋︂ ⌋︂
The inverse⌋︂of 𝑓 ˆ𝑥 𝑥3 should be 𝑓 1
ˆ𝑥 3
𝑥. Because 𝑓 ˆ𝑓  1
ˆ𝑥 𝑓 ˆ 3 𝑥 ˆ 3
𝑥3 𝑥 and 𝑓 1
ˆ𝑓 ˆ𝑥
3
𝑓 1 ˆ𝑥3 

𝑥3 𝑥.

EX 56.2. Find the inverse of the function 𝑓 ˆ𝑥 53𝑥𝑥 

zy
Solution

5𝑥
In order to find the inverse, we reverse the roles of 𝑥 and 𝑦 of the function 𝑦 :
Ô Ô Ô Ô
3𝑥
𝑥 53𝑦𝑦

3𝑥𝑦 5  𝑦 3𝑥𝑦  𝑦 5 𝑦 ˆ3𝑥  1 5 𝑦 3𝑥5 1 
𝑓 1
ˆ𝑥.

Note. Not all functions have an inverse. For example, 𝑓 ˆ𝑥 𝑥2 has no inverse function. Because the function
𝑓 ˆ𝑥 𝑥2 sends both 2 and 2 to 4. Hence the inverse of 𝑓 ˆ𝑥 𝑥2 should send 4 to both 2 and 2, and in this case
it will not be a function
am
Key fact 4. A function has an inverse function if and only if the graph of 𝑓 ˆ𝑥 satisfies the horizontal line test
(i.e. every horizontal line intersects the graph of 𝑦 𝑓 ˆ𝑥 in at most one point).

✗ ✗ 𝑎 ✓ ✓
.R

Definition 56.1. A function is one-to-one if it passes the horizontal line test. Equivalently, a function is one-to-one
if for any two different 𝑥-values 𝑥1 and 𝑥2 , 𝑓 ˆ𝑥1  and 𝑓 ˆ𝑥2  are different numbers. Sometimes, this is said 𝑓 ˆ𝑥 is
one-to-one if, whenever 𝑓 ˆ𝑥1  𝑓 ˆ𝑥2  then 𝑥1 𝑥2 .
A

⌋︂
1 1
EX 56.3. Find 𝑝 
ˆ𝑥, where 𝑝ˆ𝑥 𝑥  2 drawn. Graph 𝑝 
ˆ𝑥 on the same axes as 𝑝ˆ𝑥.

Solution

In order to graph 𝑝 1 ˆ𝑥, we just reflect 𝑝ˆ𝑥 over the line 𝑦 𝑥

 2 4 6
⌋︂
1
We can find 𝑝 ˆ𝑥 by reversing the roles of 𝑥 and 𝑦 in the function 𝑦

𝑥  2:
𝑥
⌋︂
𝑦2 Ô 𝑥2 𝑦  2 Ô
𝑦 𝑥2  2 𝑝 1 ˆ𝑥 (note that for


⌋︂ the function ˆ𝑥, we must have 𝑥 B 0. Because

𝑝 1⌋︂ 

the domain
⌋︂ of 𝑝 1
ˆ
𝑥  is just the 𝑦-values of the function 𝑦 𝑥  2, and 𝑦 𝑥  2 C 0). For the function .
𝑝ˆ𝑥 𝑥  2 what is:
(1) the domain of 𝑝ˆ𝑥 ? 𝑥 C 2 Ô
(︀2, ª

41
(2) the range of 𝑝ˆ𝑥 ? 𝑦 C 0 (︀0, ªÔ
(3) the domain of 𝑝ˆ𝑥 1 ? 𝑥 C 0
Ô
(︀0, ª
(4) the range of 𝑝ˆ𝑥 1 ? 𝑦 C 2

Ô
(︀2, ª

1
Key fact 5. For any invertible function 𝑓 ˆ𝑥, the domain of 𝑓 
ˆ𝑥 is the range of 𝑓 ˆ𝑥, and the range of
𝑓 1 ˆ𝑥 is the domain of 𝑓 ˆ𝑥.


57 Inverse trigonometric functions

Inverse sin function:

2𝜋

zy
𝑦 sinˆ𝑥

 2𝜋 𝜋 . 𝜋 2𝜋

 𝜋

2𝜋

am
Since 𝑦 sinˆ𝑥 satisfies Horizontal line test in (︀ 2𝜋 , 𝜋2 ⌋︀, then:


Restricted sinˆ𝑥 has: Domain (︀ 2𝜋 , 𝜋2 ⌋︀ 

Range (︀1, 1⌋︀.
arcsinˆ𝑥 has: Domain (︀1, 1⌋︀ Range (︀ 2𝜋 , 𝜋2 ⌋︀.


Hence, arcsinˆ𝑥 is the angle between 2𝜋 and 𝜋2 whose sine is 𝑥. i.e. 𝑦


arcsinˆ𝑥 means: 𝑥 sinˆ𝑦  such that
B 𝑦 B 𝜋2 .
.R
𝜋
2

Inverse cos function:

2𝜋
A

𝜋
𝑦 cosˆ𝑥
.
 2𝜋 𝜋 . 𝜋 2𝜋

 𝜋

2𝜋


Since 𝑦 cosˆ𝑥 satisfies Horizontal line test in (︀0, 𝜋 ⌋︀, then:

Restricted cosˆ𝑥 has: Domain (︀0, 𝜋 ⌋︀ Range (︀1, 1⌋︀.
arccosˆ𝑥 has: Domain (︀1, 1⌋︀ Range (︀0, 𝜋 ⌋︀.
Hence, arccosˆ𝑥 is the angle between 0 and 𝜋 whose cosine is 𝑥. i.e. 𝑦 arccosˆ𝑥 means: 𝑥 cosˆ𝑦  such that
0 B 𝑦 B 𝜋.

42
Inverse tan function:

2𝜋

𝜋
𝑦 tanˆ𝑥

2𝜋 𝜋
 𝜋 2𝜋

 𝜋

 2𝜋

zy
Since 𝑦 tanˆ𝑥 satisfies Horizontal line test in (︀ 2𝜋 , 𝜋2 ⌋︀, then: 

Restricted tanˆ𝑥 has: Domain (︀ 2𝜋 , 𝜋2 ⌋︀ Range (︀ª, ª⌋︀.



arcsinˆ𝑥 has: Domain (︀ª, ª⌋︀ Range (︀ 2𝜋 , 𝜋2 ⌋︀. 

𝜋 𝜋
Hence, arctanˆ𝑥 is the angle between 2 and 2 whose tan is 𝑥. i.e. 𝑦 
arctanˆ𝑥 means: 𝑥 tanˆ𝑦  such that
𝜋
2
B 𝑦 B 𝜋2 .

am
Warning: sin 1 ˆ𝑥 is the inverse function of sinˆ𝑥, namely sin 1
ˆ𝑥 x 1
sinˆ𝑥
. The same can be said as for
tan 1 ˆ𝑥 and cos 1 ˆ𝑥
 

58 Derivative of inverse functions

𝑑 1
EX 58.1. Find 𝑑𝑥
sin ˆ𝑥
.R
Solution

𝑦 
sin 1
ˆ𝑥 means 𝑥 sinˆ𝑦  Ô 𝑑𝑥𝑑 ˆ𝑥 𝑑
𝑑𝑥
ˆsinˆ𝑥 Ô 1 cosˆ𝑦  𝑑𝑦
𝑑𝑥
Ô 𝑑𝑥
𝑑𝑦 1
cosˆ𝑦 
Ô
𝑑𝑦 1 𝑑𝑦
cosˆsin1 ˆ𝑥
𝑑𝑥
, and yet we didn’t find explicitly. So, we can find the explicit derivative as follows:
𝑑𝑥
Since sinˆ𝑦  𝑥, then we can draw the following triangle:
A

1 𝑥

𝑦
⌋︂
1  𝑥2
⌋︂
Ô 𝑑𝑥
⌋︂
1𝑥2 𝑑𝑦 1
Therefore, cosˆ𝑦  1
1  𝑥2 ⌋︂ .
1𝑥2

𝑑 1
EX 58.2. Find 𝑑𝑥
cos 
ˆ𝑥

Solution

1
𝑦 cos 
ˆ𝑥 means cosˆ𝑦  𝑥, hence, we can draw the following triangle:

43
 ⌋︂
1 1  𝑥2

𝑦
𝑥
Therefore, by taking the derivative of cosˆ𝑦  𝑥, we find that:
𝑑
𝑑𝑥
𝑑
ˆ𝑥 𝑑𝑥 ˆcosˆ𝑦  Ô
1 ˆ sinˆ𝑦  𝑑𝑥 𝑑𝑦 𝑑𝑦
𝑑𝑥
1
sin 𝑦
Ô 

ˆ 
Ô 𝑑𝑥
𝑑𝑦
⌋︂
1

1𝑥2
.

𝑑 1
EX 58.3. Find 𝑑𝑥
tan 
ˆ𝑥

zy
Solution

1
𝑦 tan 
ˆ𝑥 means tanˆ𝑦  𝑥, hence, we can draw the following triangle:

⌋︂
1  𝑥2 𝑥
am
𝑦
1
Therefore, by taking the derivative of tanˆ𝑦  𝑥, we find that:
Ô Ô 𝑑𝑥 Ô 𝑑𝑥
.R
𝑑 𝑑 𝑑𝑦 𝑑𝑦 1 𝑑𝑦 1
𝑑𝑥
ˆ𝑥 𝑑𝑥 ˆtanˆ𝑦  1 ˆsec2 ˆ𝑦  𝑑𝑥 sec2 ˆ𝑦  1𝑥2
.

1 1 1 1 1 1
Similarly: sec 
ˆ𝑥 ⌋︂ . csc 
ˆ𝑥 ⌋︂ . cot 
ˆ𝑥 1𝑥2
.
𝑥 𝑥2 1 𝑥 𝑥2 1

Summary:
𝑑
sin 1 ˆ𝑥  1 𝑑
tan  1
ˆ𝑥
1 𝑑
sec 1

ˆ𝑥
1
A

⌋︂ ⌋︂
𝑑𝑥 1𝑥2 𝑑𝑥 1𝑥2 𝑑𝑥 𝑥 𝑥2 1
𝑑 1 𝑑 1 𝑑 1

𝑑𝑥
cos 1 ˆ𝑥 
⌋︂
𝑑𝑥
cot 1
ˆ𝑥 1𝑥2 𝑑𝑥
csc 1

ˆ𝑥 ⌋︂
1𝑥2 𝑥 𝑥2 1

1 𝑎𝑥
EX 58.4. Find the derivative of 𝑦 tan 
ˆ 𝑎𝑥 .

Solution

𝑑𝑦 1 𝑑 𝑎𝑥 1 ˆ 𝑎𝑥 1ˆ𝑎𝑥 ˆ1 ˆ𝑎𝑥ˆ𝑎𝑥 𝑎𝑥𝑎𝑥 2𝑎 𝑎

𝑑𝑥 1ˆ 𝑎𝑥 2 ˆ
𝑑𝑥 𝑎𝑥
 1ˆ 𝑎 𝑥 2 ˆ𝑎𝑥
2 𝑎𝑥2 ˆ𝑎𝑥2 𝑎2 2𝑎𝑥𝑥2 𝑎2 2𝑎𝑥𝑥2 2𝑎2 2𝑥2 𝑎2 𝑥2
.
𝑎𝑥  𝑎 𝑥  ˆ

59 Related rates - distance

EX 59.1. A tornado is 20 miles west of us, heading due east towards Philips Hall at rate of 40 mph. You hop on your
bike and ride due south at a speed of 12 mph. How fast is the distance between you and the tornado changing after
15 minutes?

Solution

44
Step 1. Draw a picture:

𝑏
𝑐

Step 2. Write down equations that relate the quantities of interest:

zy
𝑎2  𝑏2 𝑐.

Step 3. Derive both sides of the equation with respect to time (𝑡):
𝑑
𝑑𝑡
ˆ𝑎2  𝑏2  𝑑
𝑑𝑡
ˆ𝑐2  Ô 2𝑎 𝑑𝑎𝑑𝑡  2𝑏 𝑑𝑏
𝑑𝑡
2𝑐 𝑑𝑐
𝑑𝑡
.
am
Step 4. Plug in numbers and solve the quantity in interest:
⌋︂
At 𝑡 15𝑚𝑖𝑛 0.25ℎ𝑜𝑢𝑟𝑠, 𝑎 10, 𝑏 3 𝑐 102  32 Ô 𝑐 109. Also 𝑑𝑎 Ô
𝑑𝑡
⌋︂
40 because the

𝑑𝑏
distance is decreasing, and 𝑑𝑡 12 because the distance is increasing. Now, by plugging in numbers we get:
2 10 ˆ40  2 3 ˆ12 2 109 𝑑𝑐
⌋︂
𝑑𝑡
800  72 Ô
⌋︂
2 109 𝑑𝑐 𝑑𝑡
𝑑𝑐
𝑑𝑡
800 72
2 109
Ô
 35𝑚𝑝ℎ. 
⌋︂


É
! Don’t plug in numbers until the end.
É
! Use negative numbers for quantities the are decreasing.
.R

60 Related rates - water flows into a cone

EX 60.1. Water flows into a tank at rate of 3 cubic meters per minute. The tank is shaped like a cone with a height
of 4 meters and a radius of 5 meters at the top. Find the rate at which the water level is rising in the tank when the
water height is 2 meters.
A

Solution

𝑑ℎ
Note that rate at which the water is rising is 𝑑𝑡
. Now, we can solve the example as follows:
Step 1. Draw a picture:

5𝑚

𝑟 4𝑚

45
Step 2. Write down equations that relate the quantities of interest:

Volume of the cone 31 ˆerea of base height Ô 𝑣 1

𝜋𝑟2 ℎ. But from triangles similarity, we can conclude that
Ô Ô
3
𝑟 ℎ
5 4
𝑟 54 ℎ. Therefore, by plugging in 𝑟 5
4
ℎ in the volume equation we get: 𝑣 31 𝜋 ˆ 54 ℎ2 ℎ 𝑣 4825
𝜋ℎ3 .

Step 3. Derive both sides of the equation with respect to time (𝑡):

𝑣 25
48
𝜋ℎ3 Ô 𝑑𝑣𝑑𝑡 25
48
𝜋 3ℎ2 𝑑ℎ
𝑑𝑡
Ô 𝑑𝑣𝑑𝑡 75𝜋ℎ2
48
𝑑ℎ
𝑑𝑡
Ô 𝑑ℎ𝑑𝑡 𝑑𝑣
𝑑𝑡
48
75𝜋ℎ2
.

Step 4. Plug in numbers and solve the quantity in interest:

Since 𝑑𝑣
𝑑𝑡
3 and ℎ 2, then 𝑑ℎ
𝑑𝑡
3 48
75𝜋 22
Ô 𝑑ℎ𝑑𝑡 12
25𝜋
 0.15𝑚⇑𝑠.

61 Related rates - changing angle

zy
EX 61.1. A lighthouse that is half a mile west of shore has a rotating light that makes 2 revolutions per minute in
the counterclockwise direction. The shoreline runs north-south, and there is a cave on shore directly east of the
lighthouse. How fast is the beam of light moving along the shore at a point 1 mile north of the cave.

Solution

Step 1. Draw a picture:

am
𝜓

ℎ 𝑥
.R
𝜃 Cave
Lighthouse 1
2
mile
A

Step 2. Write down equations that relate the quantities of interest:

Our equation can’t be ℎ2 ˆ 12 2  𝑥2 . Because we don’t know much about ℎ (i.e. we don’t know much about
the derivative of ℎ with respect to time). Therefore, we will use the equation:
tanˆ𝜃 𝑥1
2
Ô
tanˆ𝜃 2𝑥.

Step 3. Derive both sides of the equation with respect to time (𝑡):
𝑑
𝑑𝑡
ˆtanˆ𝜃 
𝑑
𝑑𝑡
ˆ2𝑥 Ô sec2 ˆ𝜃 𝑑𝜃
𝑑𝑡
2 𝑑𝑥
𝑑𝑡
Ô 𝑑𝑥𝑑𝑡 1 1 𝑑𝜃
2 cos2 ˆ𝜃  𝑑𝑡
Ô 𝑑𝑥𝑑𝑡 1 ℎ𝑦𝑝
2

2 𝑎𝑑𝑗 2
𝑑𝜃
𝑑𝑡
Ô 𝑑𝑥𝑑𝑡 1
2 ˆ
ℎ2
1 2
2
𝑑𝜃
𝑑𝑡

Step 4. Plug in numbers and solve the quantity in interest:

At a point 1 mile north of the cave:

𝑥 1, therefore, ℎ2 12  ˆ 21 2 . Also 𝑑𝜃
𝑑𝑡
4𝜋 rad/min. Because the lighthouse rotates 2 times/minute. Hence,
𝑑𝑥 1 12 ˆ 21 2
𝑑𝑡 2 ˆ
1 2 4𝜋 10𝜋 miles/min.
2

46
62 Solving right triangles
EX 62.1. Solve the right triangle:

𝑐
X
𝐴
49

23 𝑏

Solution

𝐴 180  ˆ49 X X
 90  41 .
X X

𝑏
tanˆ49  23
X
Ô 𝑏 23 tanˆ49  26.46.
X

⌈︂
23
cosˆ49  𝑐X
Ô 𝑐 23 cosˆ49  35.06
X
or 𝑐 232  ˆ26.462 35.06.

zy
EX 62.2. Solve the right triangle:

𝜑
15
𝑥
am
𝜃
10

Solution

cosˆ𝜃 1015
Ô
𝜃 cos 1 ˆ 10
15
 
𝜃 Ô X
48.19 .
𝜑 ⌋︂180  ˆ90  48.19
X

⌋︂
X
 41.81 .
X X
R
𝑥 152  102 5 5.

63 Minimum and Maximum values

A.

Definition 63.1. A function 𝑓 ˆ𝑥 has an absolute maximum at 𝑥 𝑐 if 𝑓 ˆ𝑐 C 𝑓 ˆ𝑥 for all 𝑥 in the domain of 𝑓 ˆ𝑥.
The point ˆ𝑐, 𝑓 ˆ𝑐 is called an abs.max point, 𝑓 ˆ𝑐 is called the abs.max value.
𝑦

ˆ𝑐, 𝑓 ˆ𝑐

Definition 63.2. A function 𝑓 ˆ𝑥 has an absolute minimum at 𝑥 𝑐 if 𝑓 ˆ𝑐 B 𝑓 ˆ𝑥 for all 𝑥 in the domain of 𝑓 ˆ𝑥.
The point ˆ𝑐, 𝑓 ˆ𝑐 is called an abs.min point, 𝑓 ˆ𝑐 is called the abs.min value.

47
𝑦

ˆ𝑐, 𝑓 ˆ𝑐
𝑥

Definition 63.3. Absolute maximum and minimum values can be called global max and min values.
EX 63.1. Find abs.max and abs.min values.
10

zy
1 2 3 4
5


 10
am Solution

abs.min value 8, abs.min point: ˆ3, 8.

abs.max value 10, abs.min point: ˆ4, 10.
Definition 63.4. A function 𝑓 ˆ𝑥 has a local maximum at 𝑥 𝑐 if 𝑓 ˆ𝑐 C 𝑓 ˆ𝑥 for all 𝑥 near 𝑐. The point ˆ𝑐, 𝑓 ˆ𝑐
is called a local maximum point, 𝑓 ˆ𝑐 is called a local maximum value.
𝑦
.R

ˆ𝑐, 𝑓 ˆ𝑐
A

Definition 63.5. A function 𝑓 ˆ𝑥 has a local minimum at 𝑥 𝑐 if 𝑓 ˆ𝑐 B 𝑓 ˆ𝑥 for all 𝑥 near 𝑐. The point ˆ𝑐, 𝑓 ˆ𝑐
is called a local minimum point, 𝑓 ˆ𝑐 is called a local minimum value.
𝑦

ˆ𝑐, 𝑓 ˆ𝑐

Definition 63.6. local maximum and local minimum values can also be called relative max and min values.

48
EX 63.2. Mark all local max and min points. Mark all global max and min points. What is the abs.max value? What
is the abs.min value?
6
5
4
3
2
1

2 4 6 8 10 12

Solution

Note. Dr Linda Green labeled the points in different colors.

zy
Local max are labeled by green dots. Local min are labeled by red dots. There is no global max value because the
function goes up to ª. The global min point is labeled by black dot, and the global min value  0.5. Note that the
point ˆ0, 4 doesn’t count as a local max because the function is not defined on the left side of zero, it is also not
global max point. Also we should notice that the points in the interval (︀2, 3⌋︀ are all local min points.
6
5
4
3
am
2
1

2 4 6 8 10 12

EX 63.3. What do you notice about the derivative of a function at its local max and min values?
.R
6

2
A

2 4 6 8 10 12
 2

Solution

We notice that either 𝑓 ˆ𝑐 œ

0 or 𝑓 ˆ𝑐 𝐷.𝑁.𝐸.
œ

6
𝑓 œ ˆ𝑐  0
4

2
𝑓 œ ˆ𝑐 𝐷.𝑁.𝐸

2 4 6 8 10 12
 2 𝑓 œ ˆ𝑐 0

Definition 63.7. A number 𝑐 is critical number for a function 𝑓 if: 𝑓 ˆ𝑐 𝐷.𝑁.𝐸 or 𝑓 ˆ𝑐 œ œ
0.
Notes:
(1) If 𝑓 has a local max or min at 𝑥 𝑐, then 𝑐 must be a critical number for 𝑓 . ˆ𝑐, 𝑓 ˆ𝑐 is a critical point for 𝑓 .

49
(2) If 𝑐 is a critical number, then 𝑓 may or may not have a local max or min value at 𝑐. For example:
𝑓 ˆ𝑥 𝑥3 has a derivative 𝑓 ˆ𝑥 2𝑥2 , and 𝑓 ˆ0 0. So, 0 is critical for 𝑓 , while 𝑓 has no local max or min at
œ œ

𝑥 0.

64 First derivative and second derivative test

Definition 64.1. 𝑓 ˆ𝑥 has a local max at 𝑥 𝑐 if 𝑓 ˆ𝑐 C 𝑓 ˆ𝑥 for all 𝑥 in an open interval around 𝑐.
Definition 64.2. 𝑓 ˆ𝑥 has a local min at 𝑥 𝑐 if 𝑓 ˆ𝑐 B 𝑓 ˆ𝑥 for all 𝑥 in an open interval around 𝑐.

# First derivative test: If 𝑓 is a continuous function near 𝑥 𝑐 and if 𝑐 is a critical number, then:

zy
𝑓 ˆ𝑥 for 𝑥 @ 𝑐
œ
𝑓 ˆ𝑥 for 𝑥 A 𝑐
œ
extreme point at 𝑥 𝑐?
  local max
  local min
  no local extreme points
  no local extreme points
am
# Second derivative test: If 𝑓 is continuous near 𝑥 𝑐, then if 𝑓 ˆ𝑐
min at 𝑥 𝑐. If 𝑓 ˆ𝑐 0 and 𝑓 ˆ𝑐 @ 0, then 𝑓 has a local max at 𝑥 𝑐.
œ œœ
œ
0 and 𝑓 œœ
ˆ𝑐 A 0, then 𝑓 has a local

Note. If 𝑓 ˆ𝑐 0 or 𝐷.𝑁.𝐸 then second derivative is inconclusive.

œœ

65 Extreme value examples

𝑥1
EX 65.1. Find the absolute maximum and minimum values 𝑔 ˆ𝑥 𝑥2 𝑥2
on the interval (︀0.4⌋︀.
.R
Solution

abs.max and abs.min can occur at critical numbers or at end points. So, we need to compute 𝑔 ˆ𝑥 for critical numbers
and end points and compare them to find the abs.max and abs.min values. First we find critical numbers:
𝑔 ˆ𝑥
œ 𝑥2 𝑥 2 1 𝑥 1 2𝑥 1
ˆ  

𝑥2 𝑥 2 2
ˆ


 

2


2
𝑔 ˆ𝑥 𝑥 𝑥 𝑥2 2 2𝑥𝑥 2𝑥2 2𝑥 1
ˆ ˆ  
Ô œ
𝑔 ˆ𝑥 𝑥2 2𝑥 3


𝑥2 𝑥 2 2


ˆ
. 

 



 
Ô œ 

ˆ


 


𝑔 ˆ𝑥 𝐷.𝑁.𝐸 when ˆ𝑥  𝑥  2 0 which can’t happen because 𝑥 > (︀0.4⌋︀.

œ 2 2
A

𝑔 ˆ𝑥 0 when 𝑥2  2𝑥  3 0, which happens at 𝑥 3 and 𝑥 1. But the value 𝑥 1 as canceled, because
œ

1 ¶ (︀0, 4⌋︀. Hence, we need to compute 𝑔 ˆ𝑥 for the critical value 𝑥 3 and the end points 𝑥 0 and 𝑥 4 and
compare them to find abs.max and abs.min values.

𝑥 𝑔 ˆ𝑥
1
3 7
1
0 
2
3
4 22

1 1
Therefore, abs.max 7
and abs.min 
2
. This is a graph of 𝑔 ˆ𝑥, and our concerned interval is colored in
black.

50
 0.5

 4  2 2 4 6
 0.5

EX 65.2. Find absolute extreme values for 𝑓 ˆ𝑥 ⋃︀𝑥⋃︀  𝑥2 on the interval (︀2, 2⌋︀

Solution

We need to to compute 𝑓 ˆ𝑥 at critical numbers and end values and compare them. But first we need to redefine
𝑓 ˆ𝑥 as piecewise function:

zy
)︀ )︀
⌉︀ if 𝑥 C 0 ⌉︀ if 𝑥 A 0
Ô 𝑓 ˆ𝑥
2
⌉︀𝑥  𝑥 ⌉︀1  2𝑥
𝑓 ˆ𝑥 ⌋︀ œ
⌋︀ .
⌉︀
⌉︀ 𝑥  𝑥
2
if 𝑥 @ 0 ⌉︀
⌉︀ 1  2𝑥 if 𝑥 @ 0
]︀ ]︀
1 1
Hence, 𝑓 ˆ𝑥 œ
0 when 1  2𝑥 0 or when 1  2𝑥 0, which happens at 𝑥 2
or at 𝑥 
2
. Note that 𝑓 ˆ𝑥
œ

𝐷.𝑁.𝐸 at 𝑥 0.

𝑥
2
𝑓 ˆ𝑥
2
am
1 1
2 4
0 0
1 1

2 4
 2  2
1
Therefore, abs.max 4
and abs.min  2. This is a graph of 𝑓 ˆ𝑥, and our concerned interval is colored in
black.
R
 2  1 1 2
A.

 2

 4

66 The mean value theorem

Theorem 66.1. (the mean value theorem) Let 𝑓 ˆ𝑥 be a function defined on (︀𝑎, 𝑏⌋︀ such that:
(1) 𝑓 ˆ𝑥 is defined on (︀𝑎, 𝑏⌋︀.
(2) 𝑓 ˆ𝑥 is differentiable on ˆ𝑎, 𝑏.
Then there is a number 𝑐 in (︀𝑎, 𝑏⌋︀ such that the average rate of change of 𝑓 on (︀𝑎, 𝑏⌋︀ is equal to the derivative of 𝑓 at
𝑐. i.e. 𝑓 𝑏𝑏 𝑎𝑓 𝑎
ˆ 


ˆ
𝑓 ˆ𝑐.  œ

51
𝑓 ˆ𝑎

𝑓 ˆ𝑎
ˆ𝑐, 𝑓 ˆ𝑐

𝑎 𝑏

Note. The number 𝑐 is not unique.

zy
𝑎 𝑏
EX 66.1. Verify the mean value theorem for 𝑓 ˆ𝑥 2𝑥3  8𝑥  1 on the interval (︀1, 3⌋︀.
am
Solution

# 𝑓 is continuous on (︀1, 3⌋︀ and 𝑓 is differentiable on ˆ1, 3. Because 𝑓 is polynomial. So, we can verify the theorem.
# In order to verify the theorem we need to prove that 𝑓 ˆ𝑐 𝑓 33 𝑓1 1 for some 𝑐 in (︀1, 3⌋︀:œ ˆ  ˆ 

Ô


𝑓 ˆ𝑥 6𝑥2  8, and 𝑓 33 1𝑓 1

œ 31ˆ 5
 ˆ  ˆ 
18. Hence, we need to find 𝑐 in (︀1, 3⌋︀ such that 𝑓 ˆ𝑐 18 6𝑐2  8 œ

2 ⌉︂

⌉︂
18 Ô 𝑐2 26
6
Ô 𝑐 
26
6
. But 𝑐 
26
6
is in (︀1, 3⌋︀, which verifies the theorem.
.R
EX 66.2. If 𝑓 is a differentiable function and 𝑓 ˆ1 7 and 3 B 𝑓 ˆ𝑥 B 2 on the interval (︀1, 6⌋︀, then what is the
œ

biggest and smallest values that are possible for 𝑓 ˆ6.

Solution

The mean value theorem tells us that 𝑓 66 1𝑓 ˆ  ˆ1

𝑓 ˆ𝑐 for some 𝑐 > (︀1, 6⌋︀. But since 3 B 𝑓 ˆ𝑥 B 2 on the interval
œ œ
A

(︀1, 6⌋︀, then:

3 B
𝑓 6 𝑓 1
ˆ 

6 1

𝑓 ˆ𝑐 B 2
ˆ 
3 B
œ 𝑓 6
5
Ô ˆ  7
B 2 Ô 15 B 𝑓 ˆ6  7 B 10 Ô 8 B 𝑓 ˆ6 B 3.
Theorem 66.2. (Rolle’s Theorem) If 𝑓 ˆ𝑥 is a function defined on (︀𝑎, 𝑏⌋︀ and:
(1) 𝑓 ˆ𝑥 is defined on (︀𝑎, 𝑏⌋︀.
(2) 𝑓 ˆ𝑥 is differentiable on ˆ𝑎, 𝑏.
(3) 𝑓 ˆ𝑎 𝑓 ˆ𝑏.
Then there is a number 𝑐 > (︀𝑎, 𝑏⌋︀ such that 𝑓 ˆ𝑐 0. œ

𝑎 𝑏

52
Note. Roll’s Theorem actually follows from "The mean value Theorem". Because "mean value theorem" tells us that
𝑓 𝑏 𝑓 𝑎
ˆ 

𝑏 𝑎

ˆ
𝑓 ˆ𝑐. But when 𝑓 ˆ𝑎 𝑓 ˆ𝑏 we will get 𝑏 0𝑎 𝑓 ˆ𝑐
 œ
𝑓 ˆ𝑐 0. 
œ
Ô œ

67 Proof of the mean value Theorem for integrals

Theorem 67.1. (mean value Theorem for integrals) For a continuous function 𝑓 ˆ𝑥 on an interval (︀𝑎, 𝑏⌋︀, there is a
number 𝑐 with 𝑎 B 𝑐 B 𝑏 such that 𝑓 ˆ𝑐 R
𝑎
𝑏
𝑓 ˆ𝑥𝑑𝑥
.
𝑏𝑎

Proof 1. (in this proof we use Intermediate value theorem (IVT)).

* If 𝑓 ˆ𝑥 is constant on (︀𝑎, 𝑏⌋︀, then R𝑎 𝑓𝑏 𝑥𝑎 𝑑𝑥 that constant 𝑓 ˆ𝑐 for all 𝑐 > (︀𝑎, 𝑏⌋︀. Hence, the theorem is true for
𝑏
ˆ 

𝑓 ˆ𝑥 constan. Note that R𝑎 𝑏 𝑎

𝑏
𝑓 𝑥 𝑑𝑥 ˆ 
𝑓𝑎𝑣𝑒 . 

* If 𝑓 is not constant on (︀𝑎, 𝑏⌋︀, then 𝑓 have a minimum value 𝑚 and maximum value 𝑀 . Hence, for any 𝑥 > (︀𝑎, 𝑏⌋︀:
𝑚 B 𝑓 ˆ𝑥 B 𝑀 Ô R𝑎𝑏 𝑚𝑑𝑥 B R𝑎𝑏 𝑓 ˆ𝑥𝑑𝑥 B R𝑎𝑏 𝑀 𝑑𝑥 Ô 𝑚ˆ𝑏  𝑎 B R𝑎𝑏 𝑓 ˆ𝑥𝑑𝑥 B 𝑀 ˆ𝑏  𝑎 Ô
𝑚B R B 𝑀 Ô 𝑚 B 𝑓𝑎𝑣𝑒 B 𝑀 .
𝑏
𝑎 𝑓 𝑥 𝑑𝑥
ˆ 

Ô 𝑓 ˆ𝑥1  B 𝑓 ˆ𝑐 B 𝑓 ˆ𝑥2, then by IVT

𝑏 𝑎


Now, since 𝑓 is continuous and 𝑚 𝑓 ˆ𝑥1  B 𝑓𝑎𝑣𝑒 𝑓 ˆ𝑐 B 𝑀 𝑓 ˆ𝑥2 

zy
𝑥1 B 𝑐 B 𝑥2 for some 𝑥1 and 𝑥2 in (︀𝑎, 𝑏⌋︀. Hence, 𝑐 > (︀𝑎, 𝑏⌋︀.

Proof 2. (in this proof we use mean value theorem (MVT)).

𝑥
R
Let 𝑔 ˆ𝑥 𝑎 𝑓 ˆ𝑡𝑑𝑡 such that 𝑓 the function given to us in the statement of the theorem, then:
𝑎
(1) 𝑔 ˆ𝑎 𝑎 𝑓 ˆ𝑡𝑑𝑡 0. R
𝑏
(2) 𝑔 ˆ𝑏 𝑎 𝑓 ˆ𝑡𝑑𝑡. R
(3) 𝑔 ˆ𝑥 𝑓 ˆ𝑥.
œ
am
Now, from the fundamental theorem of calculus, 𝑔 ˆ𝑥 is differentiable and continuous on (︀𝑎, 𝑏⌋︀, and:

But from MVT we know that there must be 𝑎 B 𝑐 B 𝑏 such that:

(4) 𝑔 ˆ𝑐 𝑔 𝑏𝑏 𝑔𝑎 𝑎 .
œ ˆ 


ˆ 

Now, by substituting from (1), (2) and (3) in (4) we get: 𝑓 ˆ𝑐 R
𝑎
𝑏
𝑓 ˆ𝑡𝑑𝑡0
which proves the theorem.
𝑏𝑎

68 Polynomial and rational inequalities

.R
EX 68.1. Solve 𝑥2 @ 4.

Solution

Since 𝑥2 @ 4, then 𝑥2  4 @ 0. Now, in order to solve the inequality, we need to solve the equation 𝑥2  4 0 Ô
ˆ𝑥  2ˆ𝑥  2 0, therefore, 𝑥 2 or 𝑥 2. Hence, by plotting the inequality on the number line we find that:
A

 0  0 
 2 2

Therefore, the solution of the inequality is 2 @ 𝑥 @ 2 or 𝑥 > ˆ2, 2.

EX 68.2. Solve 𝑥3 C 5𝑥2  6𝑥.

Solution

Since 𝑥3 C 5𝑥2  6𝑥, then 𝑥3  5𝑥2  6𝑥 C 0. Now, in order to solve the inequality, we need to solve the equation
𝑥3  5𝑥2  6𝑥 0 Ô
𝑥ˆ𝑥  6ˆ𝑥  1 0, therefore, 𝑥 0 or 𝑥 6 or 𝑥 1. Hence, by plotting the inequality on
the number line we find that:

 0  0  0 
1 0 6

Therefore, the solution of the inequality is 𝑥 > (︀1, 0⌋︀ 8 (︀6, ª⌋︀.
𝑥2 6𝑥9
EX 68.3. Solve 𝑥1
B 0.

53
 Solution

𝑥2 6𝑥9
B 0. Hence, in order to solve the inequality, we need to solve the equation 𝑥2  6𝑥  9 0 Ô ˆ𝑥  32 0,
Ô 𝑥
𝑥1
therefore, 𝑥 3. We also need to solve 𝑥  1 0 1, hence, at 𝑥 1 the expression 𝐷.𝑁.𝐸. Now, by
plotting the inequality on the number line we find that:

 0  𝐷.𝑁.𝐸 
 3 1

Therefore, the solution of the inequality is 𝑥 > ˆª, 1.

69 Derivative and the shape of the graph

Definition 69.1. The function 𝑓 ˆ𝑥 is increasing if 𝑓 ˆ𝑥1  @ 𝑓 ˆ𝑥2  whenever 𝑥1 @ 𝑥2 .

zy
Definition 69.2. The function 𝑓 ˆ𝑥 is decreasing if 𝑓 ˆ𝑥1  A 𝑓 ˆ𝑥2  whenever 𝑥1 @ 𝑥2 .
# Increasing and decreasing function test:
* If 𝑓 ˆ𝑥 A 0 for all 𝑥 on an interval, then:
œ

𝑓 is increasing on this interval.

* If 𝑓 ˆ𝑥 @ 0 for all 𝑥 on an interval, then:
œ

𝑓 is decreasing on this interval.

am
Definition 69.3. 𝑓 ˆ𝑥 is concave up on an interval ˆ𝑎, 𝑏 if: the tangent lines lie below the curve of the function.
Definition 69.4. 𝑓 ˆ𝑥 is concave down on an interval ˆ𝑎, 𝑏 if: the tangent lines lie above the curve of the function.

# Concavity test:
* If 𝑓 ˆ𝑥 A 0 for all 𝑥 on an interval, then:
.R
œœ

𝑓 is concave up on this interval.

* If 𝑓 ˆ𝑥 @ 0 for all 𝑥 on an interval, then:
œœ

𝑓 is concave down on this interval.

A

Definition 69.5. 𝑓 ˆ𝑥 has an inflection point at 𝑥 𝑐 if:

𝑓 ˆ𝑥 is continuous at 𝑐 and it changes concavity at 𝑐. i.e. 𝑓 changes from concave up to concave down at 𝑐 or 𝑓
changes from concave down to concave up at 𝑐.
Inflection point test:
If 𝑓 ˆ𝑥 changes sign at 𝑥
œœ
𝑐, then 𝑓 has an inflection point at 𝑥
𝑐.

Note. If 𝑓 ˆ𝑥 changed sign then it gone through 0 or 𝑓 ˆ𝑥 𝐷.𝑁.𝐸.

œœ œœ

É
! 𝑓 ˆ𝑥 0 or 𝐷.𝑁.𝐸 doesn’t guarantee an inflection point.
œœ

EX: If 𝑓 ˆ𝑥 𝑥4 Ô
𝑓 ˆ𝑥 4𝑥3
œ
Ô
𝑓 ˆ𝑥 12𝑥2 , hence, 𝑓 ˆ0
œœ œœ
0 but
there is no inflection point at 𝑥 0.

𝑦 𝑥4

54
70 Linear approximation
EX 70.1. Suppose 𝑓 ˆ𝑡 is the temperature in degrees at time 𝑡 (in hours), where 𝑡 0 represents midnight. Suppose
that 𝑓 ˆ6 60 and 𝑓 ˆ6 3 ⇑ℎ𝑟. What is your best estimate for temperature at 7  00 am? and 8  00 am?
X œ X

Solution

we can say that: 𝑓 ˆ7  𝑓 ˆ6  𝑓 ˆ6 œ

ˆ7  6 63 , 𝑓 ˆ8  𝑓 ˆ6  𝑓 ˆ6
X œ
ˆ8  6 X
66 .

temp

𝑦 𝑓 ˆ𝑡 
tangent line
slope is 3X ⇑ℎ𝑟
66
3X

zy
63
3X
60

6 7 8 time
# Linear approximation:

* Approximation principle:
am
𝑓 ˆ𝑥  𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥.
œ

* Linear approximation formula when ∆𝑥 𝑥  𝑎: 𝑓 ˆ𝑎  Δ𝑥

tangent line at x=a
𝑓 ˆ𝑥  𝑓 ˆ𝑎  𝑓 ˆ𝑎ˆ𝑥  𝑎.
œ

𝑓 œ ˆ𝑎


𝑓 ˆ𝑎  𝑓 œ ˆ𝑎Δ𝑥
slope

Note that the expressions 𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥 and œ

𝑓 ˆ𝑎∆𝑥
œ

𝑓 ˆ𝑎  𝑓 ˆ𝑎ˆ𝑥  𝑎 sometimes referred to as 𝐿ˆ𝑥,

œ

𝑓 ˆ𝑎
and it is called the "Linearization of 𝑓 at 𝑎".
∆𝑥
.R
Hence the "Linearization of 𝑓 at 𝑎" 𝐿ˆ𝑎
𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥 𝑓 ˆ𝑎  𝑓 ˆ𝑎ˆ𝑥  𝑎.
œ œ

𝑎 𝑥 𝑎  Δ𝑥
Note. Equation of tangent line at 𝑥 𝑎 is 𝑦  𝑦𝑜 𝑚ˆ𝑥  𝑥𝑜  Ô 𝑦  𝑓 ˆ𝑎 𝑓 ˆ𝑥ˆ𝑥  𝑎
œ
Ô
𝑦 𝑓 ˆ𝑎  𝑓 ˆ𝑥ˆ𝑥  𝑎 𝐿ˆ𝑥 at 𝑎.
œ

⌋︂
EX 70.2. Use the Approximation to estimate 59 without a calculator.
A

Solution

Let 𝑓 ˆ𝑥
⌋︂
𝑥 Ô
𝑓 ˆ𝑥 œ 1 1
𝑥 2 , 𝑎 64, ∆𝑥 59  64 5. Now, since 𝑓 ˆ𝑥  𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥, then: œ

Ô
2 ⌋︂ ⌋︂
𝑓 ˆ59  𝑓 ˆ64  𝑓 ˆ𝑎∆𝑥 59 
1
œ
64  12 64 2 ˆ5 7.6875.


EX 70.3. Use Linearization of 𝑦 sinˆ𝑥 to estimate sinˆ33 X

 without calculator.

Solution

Let 𝑓 ˆ𝑥 sinˆ𝑥, 𝑎 30 X 𝜋

, ∆𝑥 33  30 X 11𝜋

𝜋
X
. Now, since 𝑓 ˆ𝑥  𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥, then: œ

Ô
6 60 6
𝑓 ˆ 11𝜋   𝑓 ˆ 6   𝑓 ˆ 6 ∆𝑥 sinˆ 60   sinˆ 6   cosˆ 𝜋6  ˆ 11𝜋   0.5453.
𝜋 𝜋 œ 11𝜋 𝜋 𝜋

60 60 6

𝑑
Note. We used radians because 𝑑𝑥
sinˆ𝑥 cosˆ𝑥 only works when 𝑥 in radians.

71 The differential
We knew previously that that 𝑓 ˆ𝑎  ∆𝑥  𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥, which means that 𝑓 ˆ𝑎  ∆𝑥 is unknown but can be
œ

approximated by the expression 𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥. Now, if we subtracted 𝑓 ˆ𝑎 from both sides of 𝑓 ˆ𝑎  ∆𝑥 
œ

55
𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥 we will get 𝑓 ˆ𝑎  ∆𝑥  𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥. We will use the last expression to show that:
œ œ

(the change of the function  the differential). But first, let’s define the differential.

# Definitions:
• The differential 𝑑𝑥 ∆𝑥, we can think of the differential as a small change in the value of 𝑥.
• The differential 𝑑𝑓 𝑓 ˆ𝑥𝑑𝑥
œ
𝑓 ˆ𝑥∆𝑥, the differential 𝑑𝑓 sometimes is written as 𝑑𝑦 (namely, 𝑑𝑓
œ
𝑑𝑦).
• The change in 𝑓 ∆𝑓 𝑓 ˆ𝑥  ∆𝑥  𝑓 ˆ𝑥, the change in 𝑓 at some value 𝑎 can be written as ∆𝑓
𝑓 ˆ𝑎  ∆𝑥  𝑓 ˆ𝑎. Note that the change in 𝑓 can be written as ∆𝑦 (namely, ∆𝑓 ∆𝑦 )
We showed at the beginning of this section that 𝑓 ˆ𝑎  ∆𝑥  𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥. Hence, and according to the above
œ

definitions 𝑓 ˆ𝑎  ∆𝑥  𝑓 ˆ𝑎  𝑓 ˆ𝑎∆𝑥 means the that ∆𝑓  𝑑𝑓 , which means that (the change of the function 
œ

the differential). Note that ∆𝑓  𝑑𝑓 can be written as ∆𝑦  𝑑𝑦.

EX 71.1. For 𝑓 ˆ𝑥 𝑥 lnˆ𝑥, Find:

(1) 𝑑𝑓 : 𝑑𝑓 𝑓 ˆ𝑥𝑑𝑥 ˆ𝑥 𝑥1  1 lnˆ𝑥𝑑𝑥 ˆ1  lnˆ𝑥𝑑𝑥.

œ

zy
(2) 𝑑𝑓 when 𝑥 2 and ∆𝑥 0.3 : 𝑑𝑓 ˆ1  lnˆ2 ˆ0.3  0.5079.
(3) ∆𝑓 when 𝑥 2 and ∆𝑥 0.3 : ∆𝑓 𝑓 ˆ𝑥  ∆𝑥  𝑓 ˆ𝑥 ˆ𝑥  ∆𝑥 lnˆ𝑥  ∆𝑥  𝑥 lnˆ𝑥 Ô
∆𝑓 ˆ2  ˆ0.3 lnˆ2  ˆ0.3  2 lnˆ2  0.4842.
EX 71.2. Suppose that the radius of a sphere is measured as 8𝑐𝑚 with possible error of 0.5𝑐𝑚.

(1) Use the differential to estimate the resulting error in computing the volume .
am
8𝑐𝑚
.R

We know that the volume of the sphere is 𝑉 4

3
𝜋𝑟3 , and the error in volume is ∆𝑉 . But since ∆𝑉  𝑑𝑉 , then the
estimated resulting error is 𝑑𝑉 43 𝜋 3𝑟2 𝑑𝑟 4𝜋𝑟 𝑑𝑟 𝑑𝑉 2
Ð𝑟ÐÐÐÐÐ
8,𝑑𝑟 0.5
4𝜋82 ˆ0.5 128𝜋  402.1𝑐𝑚3 .
(2) Estimate the relative error for volume . 4𝜋82 ˆ0.5
Relative error in general 𝑓Δ𝑓𝑥 . Hence, the relative error for volume is Δ𝑉
𝑉
 𝑑𝑉
𝑉 4 3 0.1875.
3 𝜋8
ˆ 
A

72 L’Hospital’s Rule
𝑓 ˆ𝑥 0
Definition 72.1. A limit of the form lim is called indeterminate form if:
𝑥 𝑎 𝑔 ˆ𝑥 0
lim 𝑓 ˆ𝑥 0 and lim 𝑔 ˆ𝑥 0.
𝑥 𝑎 𝑥 𝑎
EX: lim 𝑥22 .
𝑥 2 𝑥 2

𝑓 ˆ𝑥 ª
Definition 72.2. A limit of the form lim is called indeterminate form if:
𝑥 𝑎 𝑔 ˆ𝑥 ª

lim 𝑓 ˆ𝑥 ª and lim 𝑔 ˆ𝑥 ª .

𝑥 𝑎 𝑥 𝑎
3𝑥2 2
EX: lim 2𝑥
2 .
𝑥 ª

Theorem 72.3. (L’Hospital’s Rule) Suppose 𝑓 and 𝑔 are differentiable and 𝑔 ˆ𝑥 x 0 in an open interval around 𝑎 œ

(except possibly at 𝑎). If lim 𝑓𝑔 𝑥𝑥 is 00 or indeterminate form, then:

ˆ

ˆ



ª

ª
𝑥 𝑎
œ œ
lim 𝑓 ˆ𝑥 lim 𝑓𝑔œ ˆˆ𝑥𝑥 , provided that lim 𝑓𝑔œ ˆˆ𝑥𝑥 exists.
𝑥 𝑎 𝑔 ˆ𝑥 𝑥 𝑎 𝑥 𝑎

EX 72.1. Find lim

𝑥 ª
𝑥
3𝑥
.

56
lim
𝑥 ª
𝑥
3𝑥
Ð ª

ª
. So, we use L’Hospital’s Rule:
𝑥 1
lim 3𝑥
lim lnˆ3 3𝑥
0.
𝑥 ª 𝑥 ª

EX 72.2. Find lim sinˆ𝑥𝑥

𝑥 0 ˆsinˆ𝑥
3 .
lim sinˆ𝑥𝑥3
𝑥 0 ˆsinˆ𝑥
Ð 0
0
. So, we use L’Hospital’s Rule:

lim sin 𝑥 𝑥3 lim 3 sincos𝑥 𝑥2 cos

𝑥 0 sin 𝑥
ˆ
ˆ

ˆ
𝑥 0



1
𝑥
lim cos 𝑥
𝑥 0 3 sin 𝑥 ˆ ˆ
ˆ




ˆ  ˆ
ˆ

ˆ
1



2 lim 1 

cos
𝑥 0 ˆ 𝑥
* 1 lim cosˆ𝑥1

𝑥 0 3ˆsinˆ𝑥
2 Ð 0
0
. So, we use L’Hospital’s Rule
again:
lim 3cos 𝑥 1
sin 𝑥 2
ˆ
ˆ
lim 6 sin sin
ˆ



𝑥
𝑥 cos 𝑥
lim 6 cos1 𝑥 

ˆ 
ˆ 

ˆ 


ˆ 


6
1
.
𝑥 0 𝑥 0 𝑥 0

73 L’Hospital’s Rule - Additional Indeterminate Forms

The new Indeterminate Forms here are 0 ª , ª0 , 00 and 1 . ª

EX 73.1. Find lim sinˆ𝑥 lnˆ𝑥

zy
𝑥 0

Solution

lnˆ𝑥 sinˆ𝑥 0
lim sinˆ𝑥 lnˆ𝑥 0 ˆª (Indeterminate Form), lim ª
and lim (Indeterminate Form).
𝑥 0
1 ª
1 0
𝑥 0 sinˆ𝑥 𝑥 0 lnˆ𝑥
So, we use L’Hospital’s Rule:
lim sinˆ𝑥 lnˆ𝑥
𝑥 0
lim
𝑥 0
lnˆ𝑥
1
sinˆ𝑥
am
lim lnˆ𝑥
𝑥 0 cscˆ𝑥
1
lim  cscˆ𝑥 cotˆ𝑥
𝑥
𝑥 0
1
lim 1
𝑥
cosˆ𝑥
𝑥 0 sinˆ𝑥 sinˆ𝑥
2
 sin ˆ 𝑥
lim
𝑥 0 𝑥
1
cosˆ 𝑥
:

.R
lim
𝑥 0 𝑥
2
 sin ˆ𝑥
Ð 0
0
. At this point, we can find the limit using two methods:
(1) using L’Hospital’s Rule:
2
lim sin𝑥 𝑥
 ˆ 

𝑥 0
 2 sinˆ𝑥 cosˆ𝑥  201
lim 0
𝑥 0 1 1
sinˆ𝜃 
(2) using the fact that lim 𝜃 1 (check Theorem 43.1.):
A

𝜃 0
2
 sin ˆ𝑥
lim
𝑥 0 𝑥
sinˆ𝑥
1
sinˆ𝑥
:
lim
 lim  1 0 0
𝑥 0
 𝑥 𝑥 0
1 𝑥
EX 73.2. Find lim ‰1  𝑥
Ž
𝑥 ª

Solution

lim ‰1 
𝑥 ª
𝑥
Ð
1 𝑥
Ž 1 ª
which is indeterminate form. So, we can find the limit as follows:

Ž ÐÐ Ô ÐЪ Ð
lim
1 𝑥 ln ˆ 𝑥 𝑥
𝑦 ‰1  lnˆ𝑦  lnˆ‰1  𝑥1 Ž 
𝑥
lnˆ𝑦  𝑥 lnˆ1  𝑥1  lim lnˆ𝑦  lim 𝑥 lnˆ1  𝑥1  ª 0
𝑥 𝑥 ª ª

which is indeterminate form. So, we will need to rewrite the expression in different way, and then we will use
L’Hospital’s Rule:
1
Ô 𝑥lim lnˆ𝑦
1
1 1
1 𝑥2
lnˆ1 𝑥 
lim lnˆ𝑦  lim 𝑥 lnˆ1  𝑥1  lim 1 lim 𝑥
1 lim 1
1
1 𝑥
1. Hence, lim lnˆ𝑦  1,
𝑥 ª 𝑥 ª ª 𝑥 ª 𝑥 𝑥 ª
𝑥2
𝑥 ª 𝑥 ª

but we need to find lim 𝑦. In order to do that, we need to think of lim 𝑦 as lim 𝑒ln ˆ 𝑦
and we proved that
𝑥 ª 𝑥 ª 𝑥 ª

lnˆ𝑦  1
lim lnˆ𝑦  1. Therefore, lim 𝑦 lim 𝑒 𝑒 𝑒.
𝑥 ª 𝑥 ª 𝑥 ª

57
74 Newton’s Method
EX: Find a solution to the equation 𝑒𝑥 4𝑥 .
Attempting: If we take lnˆ of both sides we will get 𝑥 lnˆ4𝑥. But this will not help. Instead of that we
can approximate the solutions by looking at the graph of 𝑦 𝑒𝑥 and 𝑦 4𝑥:
20 𝑦 𝑒𝑥
15
𝑦 4𝑥
10

 3  2  1 1 2 3 4
 5

zy
 10
and then we will find that the solutions are approximately 0.3 and 2.1. But we can get better approximations using
Newton’s Method. But before proceeding to Newton’s Method we need to change the question "Find a solution to the
equation 𝑒𝑥 4𝑥" into the equivalent question "Find a zero to the function 𝑦 𝑒𝑥  4𝑥".

Solution using Newton’s Method:

EX: Find a zero to the function 𝑦 𝑒𝑥  4𝑥 .
Newton’s Method consists of the following three steps:
am
(1) Make a guess (let it be 𝑥1 ).
(2) Find the equation of the tangent line through your guess (namely through ˆ𝑥1 , 𝑓 ˆ𝑥1 ): So, the tangent line
equation should be 𝑦  𝑦𝑜 𝑚ˆ𝑥  𝑥𝑜  𝑦  𝑓 ˆ𝑥1  𝑓 ˆ𝑥1 ˆ𝑥  𝑥1 .Ô œ

(3) Find 𝑥-intercept of the tangent line (let it be 𝑥2 ):

In order to find 𝑥-intercept of the tangent line we need plug in 𝑦 0 in the the tangent line equation Ô
 𝑓 ˆ𝑥1  𝑓 ˆ𝑥1 ˆ𝑥  𝑥1 
œ
𝑥 𝑥1  𝑓𝑓œ 𝑥𝑥11
𝑥 𝑥2
Ô
𝑥2 𝑥1  𝑓𝑓œ 𝑥𝑥11 . ˆ

ˆ



ÐÐÐ ˆ

ˆ



.R
By noticing the graph below, we will find that the new value 𝑥2 is a better guess than the 𝑥1 . So, by repeating
these steps and taking 𝑥2 as our new guess we will get a much much better guess (𝑥3 , such that 𝑥3 is the 𝑥-intercept
of the tangent line through ˆ𝑥2 , 𝑓 ˆ𝑥2 ) and so on.
𝑦 𝑒𝑥  4𝑥
A

ˆ𝑥1 , 𝑓 ˆ𝑥1 

ˆ𝑥2 , 𝑓 ˆ𝑥2 
𝑥2 𝑥1

In generally, as we repeat this process over and over again, our 𝑛  1-th guess should be:
𝑥𝑛 1 𝑥𝑛  𝑓𝑓 œ 𝑥𝑛𝑛 . And this is the equation of Newton’s Method.

ˆ

ˆ


Now, let’s
𝑥𝑛
solve our problem using Newton’s Method equation, the equation in this example should be 𝑥𝑛 1


𝑥𝑛  𝑒 𝑒𝑥𝑛 4𝑥4𝑛 . So:



1. Let 𝑥1 3. 𝑥1 3
2. Then 𝑥2 𝑥1  𝑒 𝑒𝑥1 4𝑥4 1 3  𝑒𝑒3 443 2.49. 





2.49
3. Then 𝑥3 2.49  𝑒 𝑒2.49
4 2.49
4
 ˆ



2.23.
4. Then 𝑥4 2.15860801.
5. Then 𝑥5 2.15331858.
6. Then 𝑥6 2.15329236.
7. Then 𝑥7 2.15329236.

58
Hence, our 7-th guess is is 𝑥7 2.15329236 which is the approximation of the first zero of the function. In order to
find the second zero of the function, we can take 𝑥1 0 and repeat this process.

75 Antiderivatives
EX 75.1. If 𝑔 ˆ𝑥 œ
3𝑥2 , what could 𝑔 ˆ𝑥 be?

Solution

𝑔 ˆ𝑥 𝑥3 or 𝑔 ˆ𝑥 𝑥3  7 for example, or 𝑔 ˆ𝑥 𝑥3  𝑐 such that 𝑐 is any constant.

Definition 75.1. A function 𝐹 ˆ𝑥 is called an antiderivative of 𝑓 ˆ𝑥 on an interval ˆ𝑎, 𝑏 if 𝐹 ˆ𝑥 œ
𝑓 ˆ𝑥 on ˆ𝑎, 𝑏.
So, in the above example, 𝑥3 is an antiderivative of 3𝑥2 . Also 𝑥3  𝑐 is an antiderivative of 3𝑥2 .
EX 75.2. What are all antiderivatives of 𝑓 ˆ𝑥 3𝑥2

zy
Solution

All the antiderivatives of 𝑓 ˆ𝑥 3𝑥2 are the family 𝐹 ˆ𝑥 𝑥3  𝑐. But could there be any other antiderivatives ?
In fact, the answer is NO, and one way to think about this intuitively, is if you have two functions with the same
derivative, it is like having two runners in a race that always speed up and slow down at exactly the same times. If
one of those runners starts ahead of the other then the distance between them will always stay exactly the same. That
am
distance is the vertical distance drawn below on the graph, and that is the constant 𝑐.
𝑦 𝑥3  𝑐

𝑦 𝑥3
.R

In general: If 𝐹 ˆ𝑥 is antiderivative for 𝑓 ˆ𝑥, then all other antiderivatives for 𝑓 ˆ𝑥 can be written in the form
𝐹 ˆ𝑥  𝑐.

Table of antiderivatives:
Function 𝑓 ˆ𝑥 antiderivative 𝐹 ˆ𝑥 Function 𝑓 ˆ𝑥 antiderivative 𝐹 ˆ𝑥
A

1 𝑥𝑐 sec2 ˆ𝑥 tanˆ𝑥  𝑐

𝑥2
𝑥 2
 𝑐 secˆ𝑥 tanˆ𝑥 secˆ𝑥  𝑐
𝑥𝑛1
𝑥𝑛 ˆ𝑛 x 1 𝑛 1
 𝑐

1
1𝑥2
arctanˆ𝑥  𝑐
𝑥 1 𝑥1

ln ⋃︀𝑥⋃︀  𝑐 ⌋︂
1
arcsinˆ𝑥  𝑐
1𝑥2
𝑛1
𝑒𝑥 𝑒𝑥  𝑐 𝑎 𝑥 𝑛
𝑎 𝑥𝑛 1  𝑐 

sinˆ𝑥  cosˆ𝑥  𝑐 𝑎 𝑓 ˆ𝑥 𝑎 𝐹 ˆ𝑥  𝑐

cosˆ𝑥 sinˆ𝑥  𝑐 𝑓 ˆ𝑥  𝑔 ˆ𝑥 𝐹 ˆ𝑥  𝐺ˆ𝑥  𝑐
5 1
EX 75.3. Find the general antiderivative for 𝑓 ˆ𝑥 1𝑥2
 ⌋︂
2 𝑥
.

Solution

Ô 𝐹 ˆ𝑥 ⌋︂
1
1 1 1
𝑓 ˆ𝑥 5 1𝑥2

2
𝑥 
2 . Hence, 𝐹 ˆ𝑥 5 arctanˆ𝑥 
21
𝑥2
1  𝑐 5 arctanˆ𝑥  𝑥  𝑐.
2

76 Finding Antiderivatives using Initial conditions

EX 76.1. If 𝑔 ˆ𝑥 œ
𝑒𝑥  3 sinˆ𝑥, and 𝑔 ˆ2𝜋  5, find 𝑔 ˆ𝑥.

59
 Solution

𝑔 ˆ𝑥 𝑒𝑥  3 cosˆ𝑥  𝑐. But we know that 𝑔 ˆ2𝜋  5 Ô 𝑒2𝜋  3 cosˆ2𝜋   𝑐 5 Ô 𝑒2𝜋  3𝑐 5 Ô 𝑐 2  𝑒2𝜋 .
Hence, 𝑔 ˆ𝑥 𝑒𝑥  3 cosˆ𝑥  2  𝑒2𝜋 .
⌋︂ 1
EX 76.2. 𝑓 œœ
ˆ𝑥 𝑥ˆ𝑥  𝑥
. Find an equation for 𝑓 ˆ𝑥, if 𝑓 ˆ1 0 and 𝑓 ˆ0 2.

Solution

⌋︂
Ô 𝑓 Ô 𝑓 ˆ𝑥 Ô 𝑓 ˆ𝑥 Ô
⌋︂ 5 1
𝑥 3 1 𝑥2 𝑥2 2 5 1
𝑓 œœ
ˆ𝑥 𝑥 𝑥 𝑥
œœ
ˆ𝑥 𝑥2  𝑥 
2 œ
5  1  𝑐 œ

5
𝑥2  2𝑥 2  𝑐
2 2

Ô 𝑓 ˆ𝑥 354 𝑥 43 𝑥 𝑐𝑥 𝑑. Now, since 𝑓 ˆ0

7 3
2 𝑥2 𝑥2 7 3
𝑓 ˆ𝑥 5 7  2 3  𝑐𝑥  𝑑 2  2   2, then 𝑑 2. And since
2 Ô 𝑐 105
2 2
4 4 82 4 4 82 7 3
𝑓 ˆ1 0, then 0 35

3
 𝑐 . Therefore, 𝑓 ˆ𝑥 35

𝑥 3
𝑥 105
𝑥 2  2   2.
EX 76.3. You stand at the edge of a cliff at height 30 meters. You throw a tomato straight up in the air at a speed of
20 meters per second. How long does it take the tomato to reach the ground? What is its velocity at impact?

zy
Solution

We know that the acceleration to the gravity is 𝑎ˆ𝑡 9.8𝑚⇑𝑠2 . The negative sign here because the gravity is
pulling objects down towards the ground in the negative direction. We also have an initial condition 𝑣 ˆ0 20𝑚⇑𝑠,
and velocity is positive because we are throwing the tomatoes up. And we were told that 𝑠ˆ0 30𝑚. Therefore, we
can solve this example as follows:
𝑎ˆ𝑡 𝑠 ˆ𝑡 9.8
20 Ô
œœ
Ô
am
𝑠 ˆ𝑡 9.8𝑡  𝑐1 . But we know that 𝑣 ˆ0 𝑠 ˆ0 20𝑚⇑𝑠, hence, 𝑠 ˆ0 9.8ˆ0  𝑐1
œ

𝑐1 20. So, we can rewrite velocity equation as follows:

œ œ

𝑠 ˆ𝑡 9.8𝑡  20
œ
2
Ô
𝑠ˆ𝑡 9.8 𝑡2  20𝑡  𝑐2 . But we know that 𝑠ˆ0 30
2
𝑠ˆ0 9.8 02  20ˆ0  𝑐2 Ô
30 Ô
𝑐2 30. So, we can rewrite position equation as follows:
2
𝑠ˆ𝑡 9.8 𝑡2  20𝑡  30.
Now, we need to know "How long does it take the tomato to reach the ground?". In order to do this, we need to know
2
when 𝑠ˆ𝑡 9.8 𝑡2  20𝑡  30 0, which happens at 𝑡 1.17 and at 𝑡 5.25. The negative time doesn’t make
sense, so we left with the solution 𝑡 5.25𝑠𝑒𝑐𝑜𝑛𝑑𝑠.
.R
We also need to know "What is its velocity at impact?". In order to do this we need to compute 𝑣 ˆ5.25, because the
time of impact is 𝑡 5.25𝑠𝑒𝑐𝑜𝑛𝑑𝑠. Hence, 𝑣 ˆ5.25 𝑠 ˆ5.25 9.8ˆ5.25  20 31.45𝑚⇑𝑠. œ

77 Proof that two antiderivatives differ by a constant

Theorem 77.1. If 𝐹 ˆ𝑥 is one antiderivative of 𝑓 ˆ𝑥, then any other antiderivative of 𝑓 ˆ𝑥 can be written in the
A

form 𝐹 ˆ𝑥  𝐶 for some constant 𝐶.

The above theorem can be proved using the following two lemmas:
Lemma 77.2. If 𝑔 ˆ𝑥 œ
0 on the interval ˆ𝑎, 𝑏, then 𝑔 ˆ𝑥 𝐶 for some constant 𝐶.

Proof. MVT tells us that for any 𝑥1 , 𝑥2 in ˆ𝑎, 𝑏: 𝑔 𝑥𝑥22 ˆ  𝑔 ˆ 𝑥1 

𝑔 ˆ𝑥3  for some 𝑥3 between 𝑥1 ans 𝑥2 . But since
œ

Ô Ô
𝑥1


𝑔 ˆ𝑥3  0, then 𝑔 ˆ𝑥2   𝑔 ˆ𝑥1  0

œ
𝑔 ˆ𝑥2  𝑔 ˆ𝑥1  𝑔 ˆ𝑥 is constant.
Lemma 77.3. If 𝑔1 ˆ𝑥 and 𝑔2 ˆ𝑥 are two functions defined on ˆ𝑎, 𝑏 and 𝑔1 ˆ𝑥 œ
𝑔2 ˆ𝑥 on ˆ𝑎, 𝑏, then 𝑔1 ˆ𝑥
œ

𝑔2 ˆ𝑥  𝐶 for some constant 𝐶.

Proof. 𝑔1 ˆ𝑥 𝑔2 ˆ𝑥

œ œ
Ô
𝑔1 ˆ𝑥  𝑔2 ˆ𝑥 0. But since the derivative of the difference is the difference of the
œ œ

derivative, then ˆ𝑔1 ˆ𝑥  𝑔2 ˆ𝑥 0. But from the above lemma we know that if 𝑔 ˆ𝑥 0 then 𝑔 ˆ𝑥 𝐶. Hence,
œ œ

ˆ𝑔1 ˆ𝑥  𝑔2 ˆ𝑥 0 𝑔1 ˆ𝑥  𝑔2 ˆ𝑥 𝐶

œ
Ô 𝑔1 ˆ𝑥 𝑔2 ˆ𝑥  𝐶. Ô
So, we have proved that any two functions with the same derivative have to differ by a constant, which is exactly the
statement of Theorem 77.1.

60
78 Summation notation
5
EX 78.1. Q 2𝑖 .
𝑖 1

Solution

The above sigma notation is used to write a sum. The letter 𝑖 is called the index, the number 1 is called lower limit
5
(starting index) and the number 5 is called upper limit (ending index). We can evaluate the expression Q 2𝑖 by
𝑖 1
summing up 2𝑖 for all values of 𝑖 starting from 1 to 5. Hence,
5
Q 2𝑖 21  22  23  24  25 62.
𝑖 1
7
EX 78.2. Q 1𝑗 .
𝑗 3

zy
Solution

7
Q 1𝑗 1
3

1
4

1
5

1
6

1
7
153
140
.
𝑗 3
EX 78.3. Write in Σ notation: 6  9  12  15  18.
am Solution

We can solve this example in many ways, this depends on the way we look at the pattern of 6  9  12  15  18. For
example:
4
(1) 6  9  12  15  18 ˆ6  0 3  ˆ6  1 3  ˆ6  2 3  ˆ6  3 3  ˆ6  4 3 Q6  𝑖 3.
𝑖 0
6
Q𝑛
.R
(2) 6  9  12  15  18 ˆ2 3  ˆ3 3  ˆ4 3  ˆ5 3  ˆ6 3 3. Note that the letter of the index doesn’t
𝑛 2
matter at all, we can use any letter.
3 7 15 31
EX 78.4. Write in Σ notation: 4

8

16

32
.

Solution
A

5 𝑖
3
4

7
8

15
16

31
32 Q 2 2𝑖 1 . 

𝑖 2

79 Approximating area
EX: Estimate the area under the curve 𝑦 𝑥2 between 𝑥 0 and 𝑥 3 by approximating it with 6 rectangles. .
We will approximate the area using the following two methods:

Right Endpoints:

In this method we draw rectangles (the 6 rectangles) so the right side of each rectangle is as tall as the curve:

61
10

0.5 1 1.5 2 2.5 3

In this method it is obvious that base of each rectangle has size 0.5, and the height of each rectangle is given by the
value of our function 𝑦 𝑥2 on the right side of the rectangle. So, for example:
The area of the first rectangle 𝑅1 𝑏 ℎ 0.5 ˆ0.52 . Similarly, 𝑅2 0.5 ˆ12 , 𝑅3 0.5 ˆ1.52 , 𝑅4 0.5 ˆ22 ,
𝑅5 0.5 ˆ2.52 , and 𝑅6 0.5 ˆ32 . Hence, the area of all rectangles equal:
𝑅1  𝑅2  𝑅3  𝑅4  𝑅5  𝑅6 0.5 ˆ0.52  0.5 ˆ12  0.5 ˆ1.52  0.5 ˆ22  0.5 ˆ2.52  0.5 ˆ32 . This can
be written in sigma notation as:
6
Q ˆ0.5 ˆ0.5𝑖
2 91
11.375 A area under the curve.

zy
𝑖 1 8

Left Endpoints:

In this method we draw rectangles (the 6 rectangles) so the left side of each rectangle is as tall as the curve:
10

8
am
6

0.5 1 1.5 2 2.5 3

.R
In this method it is obvious that base of each rectangle has size 0.5, and the height of each rectangle is given by the
value of our function 𝑦 𝑥2 on the left side of the rectangle (notice that the left most rectangle in the above graph is
degenerate and has height 0). So, for example:
The area of the first rectangle 𝑅1 𝑏 ℎ 0.5 ˆ02 . Similarly, 𝑅2 0.5 ˆ0.52 , 𝑅3 0.5 ˆ12 , 𝑅4 0.5 ˆ1.52 ,
𝑅5 0.5 ˆ22 , and 𝑅6 0.5 ˆ2.52 . Hence, the area of all rectangles equal:
𝑅1  𝑅2  𝑅3  𝑅4  𝑅5  𝑅6 0.5 ˆ02  0.5 ˆ0.52  0.5 ˆ12  0.5 ˆ1.52  0.5 ˆ22  0.5 ˆ2.52 . This can
be written in sigma notation as:
A

6
Q ˆ0.5 ˆ0.5ˆ𝑖  1
2 55
8
6.875 @ area under the curve.
𝑖 1

Note. we can make better approximations by increasing the number of rectangles. Check the following example.
EX 79.1. Estimate the area under the curve 𝑦 𝑥2 between 𝑥 0 and 𝑥 3 by approximating it with 12 rectangles.

Solution

Right Endpoints method:

10

0.5 1 1.5 2 2.5 3

62
In this method it is obvious that base of each rectangle has size 0.25, and the height of each rectangle is given by the
value of our function 𝑦 𝑥2 on the right side of the rectangle. Hence, 𝑅𝑖 0.25 ˆ0.25 𝑖2 . Therefore, the area of
all rectangles equal:
12
Q 0.25 ˆ0.25 𝑖2 10.156.
𝑖 1

Left Endpoints method:

10

zy
0.5 1 1.5 2 2.5 3

In this method it is obvious that base of each rectangle has size 0.25, and the height of each rectangle is given by the
value of our function 𝑦 𝑥2 on the left side of the rectangle. Hence, 𝑅𝑖 0.25 ˆ0.25 ˆ𝑖  12 . Therefore, the area
of all rectangles equal:
12
Q 0.25 ˆ0.25 ˆ𝑖  1
2
7.906.
𝑖 1
am
Note. If we used 100 rectangles for example, then:
For Right Endpoints method:
100
Area of all rectangles should be Q𝑏 ℎ𝑖 . But we know that 𝑏 3
100
, and if we assume that the 𝑖-th right endpoint is
𝑖 1
3 3
labeled by 𝑥𝑖 , then 𝑥𝑖 100 𝑖. Hence, ℎ𝑖 ˆ𝑥𝑖 2 ˆ 100 𝑖2 . Therefore, area of all rectangles should equal:
100
Q 3
100
ˆ
3
100
𝑖2 9.1435.
.R
𝑖 1

For Left Endpoints method:

100
Area of all rectangles should be Q𝑏 ℎ𝑖 . But we know that 𝑏 3
100
, and if we assume that the 𝑖-th left endpoint is
𝑖 1
3 3
labeled by 𝑥𝑖 , then 𝑥𝑖 100 ˆ𝑖  1. Hence, ℎ𝑖 ˆ𝑥𝑖 2 ˆ 100 ˆ𝑖  12 . Therefore, area of all rectangles should
equal:
100
Q 3 3
A

2
ˆ ˆ𝑖  1 8.8654.
𝑖 1 100 100
EX 79.2. Estimate the area under the curve 𝑦 𝑥2 between 𝑥 0 and 𝑥 3 by approximating it with 𝑛 rectangles.

Solution

For Right Endpoints method:

𝑏 𝑛3 ∆𝑥, notice that we called the base ∆𝑥 because it is a little tiny bit of 𝑥. While the right endpoint 𝑥𝑖 3
𝑛
𝑖.
Hence, the height ℎ𝑖 ˆ𝑥𝑖 2 ˆ 𝑛3 𝑖2 . Therefore, area of all rectangles should equal:
𝑛
Q 3 3
ˆ 𝑖2 .
𝑖 1𝑛 𝑛
For Left Endpoints method:
𝑏 𝑛3 ∆𝑥. While the left endpoint 𝑥𝑖 3
𝑛
ˆ𝑖  1. Hence, the height ℎ𝑖 ˆ𝑥𝑖 2
3
ˆ 𝑛 ˆ𝑖  12 . Therefore, area
of all rectangles should equal:
𝑛
Q 3 3
𝑛
ˆ
𝑛
2
ˆ𝑖  1 .
𝑖 1

Note. The exact area under the curve is given by the limit:

63
 𝑛 𝑛
lim
𝑛 ª
Q 𝑛3 ˆ
3
𝑛
𝑖2 or lim
𝑛 ª
Q 𝑛3 ˆ
3
𝑛
ˆ𝑖  1
2
. These sums are called "Riemann Sums"
𝑖 1 𝑖 1
EX 79.3. Compute the area under the curve 𝑦 𝑥2 between 𝑥 0 and 𝑥 3.

Solution

𝑛
In order to compute the exact area we need to evaluate the limit lim
𝑛 ª
Q 𝑛3 ˆ
3
𝑛
𝑖2 :
𝑖 1
𝑛 𝑛
3 32 2 27 𝑛 𝑛
lim
𝑛
Q
3 3 2
ˆ 𝑖
𝑛
lim Q 2
𝑖 Q
lim 3 𝑖2 . At this point, we will use the fact that 𝑖2 Q 𝑛ˆ𝑛  1ˆ2𝑛  1
.
1𝑛 𝑛 𝑛 𝑖 1 6
𝑛 ª 𝑛 ª 𝑛 ª
𝑖 1 𝑖 𝑖 1
Hence:
𝑛
9 2𝑛3  3𝑛2  𝑛
lim
𝑛 ª
27
𝑛3 Q 𝑖2 lim
𝑛 ª
27 𝑛ˆ𝑛  1ˆ2𝑛  1
𝑛3 6
lim
𝑛 ª
9 𝑛ˆ𝑛  1ˆ2𝑛  1
2 𝑛3
lim
𝑛 ª 2 𝑛3
𝑖 1
9 3 1 9
lim ˆ2    2 9.
𝑛 2 ª 𝑛 𝑛2 2

zy
80 The fundamental theorem of calculus part 1
In the following example, the expression 𝑔 ˆ𝑥 1 𝑓 ˆ𝑡𝑑𝑡 means the net area between 1 and some value 𝑥 on the R𝑥
𝑥-axis. We will call 𝑔 ˆ𝑥 the accumulated area function because as 𝑥 increases 𝑔 ˆ𝑥 measures how much net area
has accumulated.
EX 80.1. Suppose 𝑓 ˆ𝑥 has the graph shown, and let 𝑔 ˆ𝑥 R1𝑥 𝑓 ˆ𝑡𝑑𝑡. Find:
am
4

3 𝑦 𝑓 ˆ𝑥

1 2 3 4 5 6 7 8
4
.R
3
2
1

1. 𝑔 ˆ1=𝑔 ˆ𝑥 R11 𝑓 ˆ𝑡𝑑𝑡 0.

A

2. 𝑔 ˆ2=𝑔 ˆ𝑥 R12 𝑓 ˆ𝑡𝑑𝑡 2.

4
3 𝑦 𝑓 ˆ𝑥

2
1

1 1 2 3 4 5 6 7 8

2
3
4

3. 𝑔 ˆ3=𝑔 ˆ𝑥 R13 𝑓 ˆ𝑡𝑑𝑡 5.

4
3 𝑦 𝑓 ˆ𝑥

2
1

1 1 2 3 4 5 6 7 8

2
3
4

64
4. 𝑔 ˆ4=𝑔 ˆ𝑥 R14 𝑓 ˆ𝑡𝑑𝑡 8.
4
3 𝑦 𝑓 ˆ𝑥

2
1

1 1 2 3 4 5 6 7 8

2
3
4

5. 𝑔 ˆ5=𝑔 ˆ𝑥 R15 𝑓 ˆ𝑡𝑑𝑡 9.

4
3 𝑦 𝑓 ˆ𝑥

2
1

1 1 2 3 4 5 6 7 8

2
3

zy
4

6. 𝑔 ˆ6=𝑔 ˆ𝑥 R16 𝑓 ˆ𝑡𝑑𝑡 8.

4
3 𝑦 𝑓 ˆ𝑥

2
1

1
2
3
1 2 3 4 5 6
am
7 8

4

7. 𝑔 ˆ7=𝑔 ˆ𝑥 R17 𝑓 ˆ𝑡𝑑𝑡 5.

4
3 𝑦 𝑓 ˆ𝑥

2
1
.R
1 1 2 3 4 5 6 7 8

2
3
4

8. 𝑔 ˆ0=𝑔 ˆ𝑥 R10 𝑓 ˆ𝑡𝑑𝑡  R01 𝑓 ˆ𝑡𝑑𝑡 2.


A

Note. If we plot all values of 𝑔 ˆ𝑥 on coordinate axes and connect the dots, we will notice the following:
10

4 𝑦 𝑔 ˆ𝑥
2

2 1 2 3 4 5 6 7 8

4
4
3 𝑦 𝑓 ˆ𝑥

2
1

1 1 2 3 4 5 6 7 8

2
3
4

(1) We know that the derivative 𝑔 ˆ𝑥 is positive where 𝑔 ˆ𝑥 is increasing. But 𝑔 ˆ𝑥 is increasing wherever we are
œ

adding on positive area, i.e. when 𝑓 ˆ𝑥 is positive. So, we have that:
𝑔 ˆ𝑥 A 0 where 𝑓 ˆ𝑥 A 0.
œ

65
(2) 𝑔 ˆ𝑥 is negative where 𝑔 ˆ𝑥 is decreasing. But 𝑔 ˆ𝑥 is decreasing wherever we are adding on negative
œ

area, i.e. when 𝑓 ˆ𝑥 is negative. So, we have that:

𝑔 ˆ𝑥 @ 0 where 𝑓 ˆ𝑥 @ 0.
œ

(3) 𝑔 ˆ𝑥 œ
0 where 𝑓 ˆ𝑥 0.

(4) If we look closer we will find that the rate at which 𝑔 ˆ𝑥 is increasing depends on the height of 𝑓 ˆ𝑥. So,
the rate of change of 𝑔 ˆ𝑥 which is 𝑔 ˆ𝑥 is behaving very much like the function 𝑓 ˆ𝑥 itself. In fact 𝑔 ˆ𝑥 𝑓 ˆ𝑥.
œ œ

Theorem 80.1. (Fundamental Theorem of Calculus, Part 1) If 𝑓 ˆ𝑥 is continuous on (︀𝑎, 𝑏⌋︀ then for 𝑎 B 𝑥 B 𝑏 the
function
𝑥
𝑔 ˆ𝑥 S𝑎 𝑓 ˆ𝑡𝑑𝑡

is continuous on (︀𝑎, 𝑏⌋︀ and differentiable on ˆ𝑎, 𝑏 and

œ
𝑔 ˆ𝑥 𝑓 ˆ𝑥.

zy
EX 80.2. ⌋︂
Find: ⌋︂
𝑑
1. 𝑑𝑥 R𝑥
5 ⌋︂𝑡  3𝑑𝑡=⌋︂𝑥  3.
2 2
𝑑
2. 𝑑𝑥 R𝑥
4 ⌋︂ 𝑡  3𝑑𝑡= 𝑥  3.⌋︂
2 2
⌋︂
𝑑
R sin 𝑥
4 𝑑 𝑥
3. 𝑑𝑥 𝑥 𝑡2  3𝑑𝑡=  𝑑𝑥 4 𝑡2  3𝑑𝑡
⌋︂
R  𝑥2  3.
𝑑
4. 𝑑𝑥 R4 ˆ
𝑡2  3𝑑𝑡:


𝑑
In this example we will use the chain rule which states that 𝑑𝑥
am
𝑓 ˆ𝑢ˆ𝑥 𝑑𝑓𝑑𝑢𝑢 𝑑𝑢𝑑𝑥𝑥 , knowing that 𝑢ˆ𝑥 ˆ  ˆ 
sinˆ𝑥.
So: ⌈︂
sin 𝑥 ⌋︂ 2 𝑢 ⌋︂ 2 ⌋︂
𝑑
𝑑𝑥 4 R ˆ
𝑡  3𝑑𝑡 𝑑𝑢
𝑑
4 𝑡  3𝑑𝑡 𝑑 sin
𝑑𝑥
𝑥
R
𝑢2  3 cosˆ𝑥 sin2 ˆ𝑥  3 cosˆ𝑥. ˆ 

81 The Fundamental Theorem of Calculus Part 2

Theorem 81.1. (Fundamental Theorem of Calculus, Part 2) If 𝑓 is continuous on (︀𝑎, 𝑏⌋︀, then
R
𝑏
S𝑎 𝑓 ˆ𝑥𝑑𝑥 𝐹 ˆ𝑏  𝐹 ˆ𝑎

œ
where 𝐹 is any antiderivative of 𝑓 , that is 𝐹 is any function such that 𝐹 𝑓.
Comments on "Fundamental Theorem of Calculus Part 2":
A.

𝑏
(1) If we think of 𝑓 as the derivative of 𝐹 (namely 𝑓 ˆ𝑥 𝐹 ˆ𝑥), then we get 𝑎 𝐹 ˆ𝑥𝑑𝑥 𝐹 ˆ𝑏  𝐹 ˆ𝑎. Which œ
R œ

means that: the integral of the derivative is the original function.

(2) The phrasing "any antiderivative" can be explained as follows: Suppose 𝐺ˆ𝑥 is another antiderivative for 𝑓 ˆ𝑥,
then 𝐺ˆ𝑥 𝐹 ˆ𝑥  𝑐. So, 𝐺ˆ𝑏  𝐺ˆ𝑎 𝐹 ˆ𝑏  𝑐  ˆ𝑓 ˆ𝑎  𝑐 𝐹 ˆ𝑏  𝐹 ˆ𝑎. Hence, this difference 𝐹 ˆ𝑏  𝐹 ˆ𝑎
is the same value no matter which antiderivative of 𝑓 we use, and this is why we can say "𝐹 is any antiderivative".
EX 81.1. Find R 15 3𝑥2




4
𝑥
𝑑𝑥

Solution

5


R1
5
3𝑥2


4
𝑥
𝑑𝑥 𝑥3  4 ln ⋃︀𝑥⋃︀⋁︀ ˆ53  4 ln ⋃︀5⋃︀  ˆˆ13  4 :0
⋃︀1⋃︀ 124  4 lnˆ5.
ln 

1


EX 81.2. Find R 1
4 𝑦 2 𝑦 1
𝑦2
1 𝑑𝑦

Solution

R R R
5 3 1
4 𝑦 2 𝑦 1 4 2 1 4 3 1 1 𝑦2 𝑦2 𝑦2
1 ˆ𝑦 𝑦  1𝑦 ⋁︀
 

1 1 𝑑𝑦  2 𝑑𝑦 1 𝑦
2  𝑦2  𝑦 2 𝑑𝑦 5  3  1
𝑦2 2 2 2
1
5 3 1 5 3 1
42 42 42 12 12 12 146
ˆ 5  3  1   ˆ 5  3  1  .......... 15
.
2 2 2 2 2 2

66
82 Proof of the Fundamental Theorem of Calculus
Theorem 82.1. (Fundamental Theorem of Calculus, Part 1) If 𝑓 ˆ𝑥 is continuous on (︀𝑎, 𝑏⌋︀ then for 𝑎 B 𝑥 B 𝑏 the
function
𝑥
𝑔 ˆ𝑥 S𝑎 𝑓 ˆ𝑡𝑑𝑡

is continuous on (︀𝑎, 𝑏⌋︀ and differentiable on ˆ𝑎, 𝑏 and

œ
𝑔 ˆ𝑥 𝑓 ˆ𝑥.

Proof. 𝑔 ˆ𝑥
œ
lim 𝑔 ˆ𝑥ℎ𝑔 ˆ𝑥
lim R
𝑎
𝑥ℎ
𝑓 ˆ𝑡𝑑𝑡R 𝑎
𝑥
𝑓 ˆ𝑡𝑑𝑡
. Now, by properties of integrals, the integral from (𝑎 to
ℎ 0 ℎ ℎ 0 ℎ
𝑥  ℎ) minus the integral from (𝑎 to 𝑥) is just the integral from (𝑥 to 𝑥  ℎ) (check the graph below).

zy
𝑎 𝑥 𝑥ℎ

R 𝑥ℎ
R R 𝑥 ℎ

R𝑥𝑥 ℎ 𝑓 ˆ𝑡𝑑𝑡 can be approximated by a skinny

𝑥
𝑓 ˆ𝑡𝑑𝑡 𝑓 ˆ𝑡𝑑𝑡 𝑓 ˆ𝑡𝑑𝑡 
Hence, lim 𝑎 𝑎
lim 𝑥
. Now informally,
ℎ 0 ℎ
am
ℎ 0
rectangle with height 𝑓 ˆ𝑥 and width ℎ (check the graph below).
ℎ

𝑓 ˆ𝑥
R
𝑎 𝑥𝑥ℎ

Therefore, lim R
𝑥ℎ
𝑓 ˆ𝑡𝑑𝑡
𝑥
ℎ
 lim
𝑓 ˆ𝑥. 𝑓 ˆ 𝑥

ℎ
ℎ 0
ℎ ℎ 0
But let’s make this argument a little more precise. Let 𝑀 be the maximum value that 𝑓 ˆ𝑥 achieves on the sub-interval
A.

(︀𝑥, 𝑥  ℎ⌋︀, and let 𝑚 be the minimum value (notice that 𝑓 ˆ𝑥 has to have a minimum value and a maximum value
because it is continuous).

𝑎 𝑥 𝑥ℎ

R𝑥𝑥 ℎ 𝑓 ˆ𝑡𝑑𝑡 B 𝑀 ℎ Ô𝑚 B R B 𝑀 . Hence, and from 𝐼𝑉 𝑇 R𝑥 ℎ

𝑥 ℎ 𝑥 ℎ
𝑓 ˆ𝑡𝑑𝑡 𝑓 𝑡 𝑑𝑡
Now, we know that 𝑚 ℎ B
 ˆ 
𝑥
ℎ
must
Therefore, lim R
𝑥 ℎ
𝑓 ˆ𝑡𝑑𝑡
equal 𝑓 ˆ𝑐 for some 𝑐 > (︀𝑥, 𝑥  ℎ⌋︀. 𝑥
ℎ
lim 𝑓 ˆ𝑐 such that 𝑐 between 𝑥 and 𝑥  ℎ. But as ℎ
ℎ 0 ℎ 0
approaches 0, 𝑐 will approach 𝑥. Consequentially:

lim 𝑓 ˆ𝑐 𝑓 ˆ𝑥.

ℎ 0

67
Theorem 82.2. (Fundamental Theorem of Calculus, Part 2) If 𝑓 is continuous on (︀𝑎, 𝑏⌋︀, then
𝑏
S𝑎 𝑓 ˆ𝑥𝑑𝑥 𝐹 ˆ𝑏  𝐹 ˆ𝑎

œ
where 𝐹 is any antiderivative of 𝑓 , that is 𝐹 is any function such that 𝐹 𝑓.
Proof. Let 𝐺ˆ𝑥 R 𝑥
𝑎 𝑓 ˆ𝑡𝑑𝑡, then (from the above theorem) 𝐺 ˆ𝑥 𝑓 ˆ𝑥, that is 𝐺ˆ𝑥 is antiderivative for 𝑓 ˆ𝑥.
0
œ

R𝑏 𝑎 
R
:

Now, 𝐺ˆ𝑏  𝐺ˆ𝑎 𝑎 𝑓 ˆ𝑡𝑑𝑡 𝑎 𝑓 ˆ𝑡𝑑𝑡 which means that the theorem is true if we used the antiderivative 𝐺ˆ𝑥.
But the theorem must be true for any antiderivative. That can be proved as follows:
Let 𝐹 ˆ𝑥 be any antiderivative of 𝑓 ˆ𝑥, then 𝐹 ˆ𝑥 𝐺ˆ𝑥  𝑐. Therefore, 𝐹 ˆ𝑏  𝐹 ˆ𝑎 ˆ𝐺ˆ𝑏  𝑐  ˆ𝐺ˆ𝑎  𝑐
𝑏
𝐺ˆ𝑏  𝐺ˆ𝑎 𝑎 𝑓 ˆ𝑡𝑑𝑡. R
83 The Substitution Method
EX 83.1. Find R 2𝑥 sinˆ𝑥2 𝑑𝑥

zy
Solution

Let 𝑢 Ô 𝑑𝑢 2𝑥𝑑𝑥. Hence R 2𝑥 sinˆ𝑥2𝑑𝑥 R sinˆ𝑢𝑑𝑢

𝑥2  cosˆ𝑢  𝑐  cosˆ𝑥2   𝑐.
EX 83.2. Find R 1 𝑥3𝑥 𝑑𝑥. 
2
am Solution

Let 𝑢 1  3𝑥2 Ô 𝑑𝑢 6𝑥𝑑𝑥 Ô 16 𝑑𝑢 𝑥𝑑𝑥. So:

R R R
1
𝑥 6 𝑑𝑢 1 1 1 1
13𝑥2
𝑑𝑥 𝑢 6 𝑢
𝑑𝑢 6
ln ⋃︀𝑢⋃︀  𝑐 6
ln ⋂︀1  3𝑥2 ⋂︀  𝑐.
EX 83.3. Find R𝑒 7𝑥
𝑑𝑥.

Solution
.R
Let 𝑢 Ô 𝑑𝑢 7𝑑𝑥 Ô 71 𝑑𝑢
7𝑥 𝑑𝑥. Hence, R 𝑒7𝑥 𝑑𝑥 R 𝑒𝑢 1
7
𝑑𝑢 1 𝑢
7
𝑒  𝑐 1 7𝑥
7
𝑒  𝑐.
EX 83.4. Find R𝑒 ln𝑥𝑥 𝑑𝑥.
2
𝑒 ˆ 

Solution

Ô
A

Let 𝑢 lnˆ𝑥 𝑑𝑢 𝑥1 𝑑𝑥. Note that we should consider the bounds of integration while solving this example,
but we can worry about the bounds now or later. We will solve the example using the two methods:
(1) Worry about bounds of integration now:
When 𝑥 𝑒, then 𝑢 lnˆ𝑒 1. Also when 𝑥 𝑒2 , then 𝑢 lnˆ𝑒2  2. So:
2

R𝑒𝑒 R12 𝑢𝑑𝑢

2
lnˆ𝑥 𝑢2 22 12 1
𝑥
𝑑𝑥 2
⋁︀ 2

2 2
.
1
(2) Worry about bounds of integration later:
𝑒2

R lnˆ𝑥
𝑥
𝑑𝑥 R 𝑢𝑑𝑢 𝑢2
2
ˆ lnˆ𝑥2
2
⋁︀
ˆ lnˆ𝑒2 2
2

ˆ lnˆ𝑒2
2
22
2

12
2
1
2
.
𝑒

84 Why U-Substitution Works

𝑈 -Substitution is based on the chain rule, recall that the chain rule says:
𝑑
𝑑𝑥
𝑓 ˆ𝑔 ˆ𝑥 𝑓 ˆ𝑔 ˆ𝑥𝑔 ˆ𝑥. Now if we take the integral of both sides of this equation with respect to 𝑥, we get:
œ œ

R 𝑑
𝑑𝑥
𝑓 ˆ𝑔 ˆ𝑥𝑑𝑥 R
𝑓 ˆ𝑔 ˆ𝑥𝑔 ˆ𝑥𝑑𝑥. But on the left side of the previous equation we are taking the integral of a
œ œ

derivative, and we knew that the integral is an antiderivative. Hence, the integral (antiderivative) of a derivative is
the function itself, namely:
R 𝑑
𝑑𝑥
𝑓 ˆ𝑔 ˆ𝑥𝑑𝑥 𝑓 ˆ𝑔 ˆ𝑥. Now when we do use substitution we actually writing:

68
𝑓 ˆ𝑔 ˆ𝑥  𝑐 R
𝑓 ˆ𝑔 ˆ𝑥  𝑔 ˆ𝑥𝑑𝑥. Because then we are seeing an expression of the form 𝑓 ˆ𝑔 ˆ𝑥𝑔 ˆ𝑥𝑑𝑥 and
œ œ
R œ œ

recognize 𝑢 as 𝑔 ˆ𝑥 and 𝑑𝑢 as 𝑔 ˆ𝑥𝑑𝑥. So, we are rewriting this expression 𝑓 ˆ𝑔 ˆ𝑥𝑔 ˆ𝑥𝑑𝑥 as:
œ
R œ œ

R 𝑓 ˆ𝑢𝑑𝑢 𝑓 ˆ𝑢  𝑐 𝑓 ˆ𝑔 ˆ𝑥  𝑐.

œ

85 Average Value of a Function

The average of a list of numbers 𝑞1 , 𝑞2 , 𝑞3 , ...., 𝑞𝑛 is:
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑞1 𝑞2 𝑞𝑛3 .... 𝑞𝑛
 
𝑛
𝑖 1 𝑞𝑖
 
.
P
𝑛
While for a continuous function 𝑓 ˆ𝑥 on an interval (︀𝑎, 𝑏⌋︀ we could estimate the average value of the function by
sampling it at a bunch of 𝑥-values 𝑥1 , 𝑥2 , 𝑥3 , ...., 𝑥𝑛 spaced a distance of ∆𝑥 apart. Hence, the approximated average

P
here is: 𝑛
𝑖 1 𝑓 ˆ𝑥𝑖 
𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑒𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑓 𝑥1 𝑓 𝑥2 𝑓𝑛 𝑥3 .... 𝑓 𝑥𝑛 ˆ  ˆ  ˆ
.   ˆ 

zy
Δ𝑥

𝑎 𝑥1𝑥2
am
𝑥𝑛 𝑏
But the approximation gets better as the number of sample points gets larger, so we can define the exact average as:

𝑎𝑣𝑒𝑟𝑎𝑔𝑒 lim
P𝑛𝑖 1 𝑓 ˆ𝑥𝑖  .
𝑛 ª 𝑛
Δ𝑥
But we can make the above sum look more like a Riemann Sum by multiplying by Δ𝑥
. So we get:

P𝑛𝑖 1 𝑓 ˆ𝑥𝑖  P𝑛𝑖 1 𝑓 ˆ𝑥𝑖 

.R
lim ∆𝑥
ÐÐÐÐÐÐÐÐ
lim 𝑛 Δ𝑥 𝑏𝑎
∆𝑥 𝑛 ª Δ𝑥 0
𝑎𝑣𝑒𝑟𝑎𝑔𝑒 lim 𝑎𝑣𝑒𝑟𝑎𝑔𝑒
𝑛 ª 𝑛 ∆𝑥 𝑏𝑎
lim P 𝑛
𝑓 ˆ𝑥𝑖  Δ𝑥 R𝑏
𝑓 ˆ𝑥𝑑𝑥
R𝑎𝑏 𝑓 ˆ𝑥𝑑𝑥 .
ÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐÐ
𝑖 1 𝑎
Δ𝑥 0
𝑎𝑣𝑒𝑟𝑎𝑔𝑒
𝑏𝑎
1
EX 85.1. Find the average value of the function 𝑔 ˆ𝑥 15𝑥
on the interval (︀2, 5⌋︀.
A

Solution

𝑔𝑎𝑣𝑒 R 5 1
2 15𝑥
. Let 𝑢 1  5𝑥 Ô 𝑑𝑢  5𝑑𝑥 Ô 
1
𝑑𝑢 𝑑𝑥. Note that at 𝑥 2𝑢  9 and at 𝑥 5𝑢 24. So:

52 5
24
R R

5 1 24 1 1
𝑔𝑎𝑣𝑒 2 15𝑥 9 𝑢 ˆ 5 𝑑𝑢 1 1
ˆ 5  ln ⋃︀𝑢⋃︀⋁︀ 
1
ˆlnˆ24  lnˆ9 
1
lnˆ 24  
1
lnˆ 83   0.065.
52 3 3 15 15 9 15
 9

Question. Is there a number 𝑐 in the interval (︀𝑎, 𝑏⌋︀ for which 𝑔 ˆ𝑐 equals its average value? If so, find all
such numbers 𝑐. If not, explain why not .
In order to answer this question we may try to solve this the following equation for 𝑐:
15 1
𝑔 ˆ𝑐 𝑔𝑎𝑣𝑒 1
Ô
1 5𝑐

1
15
lnˆ 83 

1  5𝑐 ln 158
3
5𝑐Ô 15
ln 38
 1


ˆ 
Ô 

ˆ 
Ô 𝑐 lnˆ 8 
3
5

 3.25. Hence, the
answer is Yes. This actually follows from "Mean Value Theorem for Integrals".

86 Mean Value Theorem for Integrals

This part of the video is the last part, and it shows two proofs of "Mean Value Theorem for Integrals". The same two
proofs was shown before in the video from 09  09  10 to 09  14  50, and we stated them in Section 67.

69
Acknowledgements
Although I never met or emailed the greet Dr. Linda Green, I would like to thank her for this greet course. Also i
would like to thank freeCodeCamp.org channel for preparing the video of the course.

Department of Mathematics, Faculty of Education, Al-Azhar University, Cairo, Egypt.

E-mail address: hosam7101996@gmail.com

zy
am
.R
A

70