4.1 maximum and minimum values has an absolute minimum at · has an absolute minimum at = 0. 0 = 0...

1a 1b

§4.1 Maximum and Minimum Values

A function 𝑓 has an absolute maximum at a no. 𝑐 in

its domain 𝒟 if

𝑓 𝑐 ≥ 𝑓(𝑥) for all 𝑥 ∈ 𝒟.

A function 𝑓 has an absolute minimum at a no. 𝑐 in

its domain 𝒟 if

𝑓 𝑐 ≤ 𝑓(𝑥) for all 𝑥 ∈ 𝒟.

The no. 𝑓(𝑐) is the maximum value or minimum

value, respectively, of 𝑓 on 𝒟.

Example. 𝑦 = 𝑓 𝑥 = 𝑥2, 𝒟 = [−1,2)

−2…0…2…𝑥-, 0…4…𝑦-, 𝑓

𝑓 has an absolute minimum at 𝑥 = 0.

𝑓 0 = 0 is the absolute minimum value.

𝑓 has no absolute maximum.■

The maximum and minimum values of 𝑓 are the

extreme values of 𝑓.

Extreme Value Theorem

If 𝑓 is continuous on a closed interval [𝑎, 𝑏] then 𝑓

attains an absolute maximum and an absolute

minimum on [𝑎, 𝑏].

Example. 𝑦 = 𝑓 𝑥 = 𝑥2, 𝒟 = [−1,2]

−2…0…2…𝑥-, 0…4…𝑦-, 𝑓

2a 2b

𝑓 has an absolute minimum at 𝑥 = 0.

𝑓 has an absolute maximum at 𝑥 = 2.

A function has a local maximum at a no. 𝑐 if there is

an open interval 𝐼 containing 𝑐 such that

𝑓 𝑐 ≥ 𝑓(𝑥) for all 𝑥 ∈ 𝐼

A function has a local minimum at a no. 𝑐 if there is

an open interval 𝐼 containing 𝑐 such that

𝑓 𝑐 ≤ 𝑓(𝑥) for all 𝑥 ∈ 𝐼

Example.

…𝑎…𝑐…𝑑…𝑏…𝑥-, 𝑓

𝑓 has a local maximum at 𝑐.

𝑓 has a local minimum at 𝑑.

𝑓 has an absolute maximum at 𝑏.■

?? Practice with maxima and minima

Fermat’s Theorem

If 𝑓 has a local maximum or minimum at a no. 𝑐, and

if 𝑓 ′(𝑐) exists, then 𝑓 ′ 𝑐 = 0.

Why. …𝑐… ,𝑓, tangent at local max.

3a 3b

A critical number of a function 𝑓 is a no. 𝑐 in the

domain of 𝑓 such that either

𝑓 ′ 𝑐 = 0 or 𝑓 ′(𝑐) does not exist.

Rephrase Fermat’s Theorem.

If 𝑓 has a local maximum or minimum at a no. 𝑐, then

𝑐 is a critical no.

Example. Find the critical numbers of

𝑓 𝑥 = 4𝑥3 − 9𝑥2 − 12𝑥 + 3

Since 𝑓 is a polynomial, 𝑓′ exists at all real numbers.

𝑓 ′ = 12𝑥2 − 18𝑥 − 12

= 6 2𝑥2 − 3𝑥 − 2

= 6(2𝑥 + 1)(𝑥 − 2)

Critical numbers are 𝑥 = −1

2, 𝑥 = 2■

Example. Find the critical numbers of

𝐺 𝑥 = 𝑥2 − 𝑥3

𝐺 ′ 𝑥 =1

3 𝑥2 − 𝑥 −

2

3(2𝑥 − 1)

=1

3

2𝑥 − 1

𝑥2 − 𝑥 2

3

𝐺 ′ = 0 at 𝑥 =1

2.

𝐺′ does not exist for 𝑥 = 0 or 𝑥 = 1.

… 0…1…𝑥-, …0…𝑦-, 𝐺

■

4a 4b

Closed Interval Method

To find absolute maximum and minimum values of a

continuous function 𝑓 on a closed interval [𝑎, 𝑏].

1. Find the values of 𝑓 at the critical numbers in

[𝑎, 𝑏].

2. Find 𝑓(𝑎) and 𝑓(𝑏).

3. The largest of these is the absolute maximum

and the smallest of these is the absolute min.

Example. Find the absolute maximum and minimum

values of

𝑓 𝑥 = 1 − 2𝑥 − 𝑥2 on [−4,1].

1. Find the critical numbers

𝑓 ′ 𝑥 = −2 − 2𝑥 = −2(1 + 𝑥)

𝑥 = −1 is a critical number.

𝑓 −1 = 1 + 2 − 1 = 2

2. 𝑓 −4 = 1 + 8 − 16 = −7

𝑓 1 = 1 − 2 − 1 = −2

3. The absolute max. value is 2 at 𝑥 = −1.

The absolute min. value is −7 at 𝑥 = −4. ■

Example. Find the absolute maximum and minimum

values of

𝑓 𝑥 =ln 𝑥

𝑥 on [1, 3]

1. 𝑓(𝑥) is differentiable on 1,3].

𝑓 ′ 𝑥 = ln 𝑥 ′ 𝑥−(ln 𝑥)𝑥 ′

𝑥2

=1

𝑥 𝑥−ln 𝑥

𝑥2

=1−ln 𝑥

𝑥2

𝑥 = 𝑒 makes the numerator zero

𝑓 𝑒 =ln 𝑒

𝑒=

1

𝑒= 0.367…

2. 𝑓 1 =ln 1

1= 0

𝑓 3 =ln 3

3= 0.366…

3. 𝑓 has an absolute maximum value of

1

𝑒= 0.367… at 𝑥 = 𝑒

𝑓 has an absolute minimum value of 0 at 𝑥 = 1

■

5a 5b

4.2 The Mean Value Theorem

Rolle’s Theorem Let 𝑓 satisfy

1. 𝑓 is continuous on the closed interval [𝑎, 𝑏].

2. 𝑓 is differentiable on (𝑎, 𝑏)

3. 𝑓 𝑎 = 𝑓(𝑏)

Then there is a no. 𝑐 on (𝑎, 𝑏) such that 𝑓 ′ 𝑐 = 0.

Why? …𝑎…𝑏…, … ,𝑓, 𝑓 𝑎 = 𝑓 𝑏 , tangent

Example. Verify that

𝑓 𝑥 = sin 𝑥 + cos 𝑥, 0 ≤ 𝑥 ≤ 2𝜋

satisfies the hypotheses of Rolle’s theorem. Find all

numbers 𝑐 satisfying the conclusions of Rolle’s

theorm.

1. 𝑓 is continuous on [0,2𝜋].

2. 𝑓 is differentiable on (0,2𝜋).

3. 𝑓 0 = 0 + 1 = 𝑓(2𝜋)

𝑓 does satisfy the hypotheses. Look for 𝑐 such that

𝑓 ′ 𝑐 = cos 𝑐 − sin 𝑐 = 0.

axes, unit circle, radius at angle 𝑐, pt. (cos 𝑐 , sin 𝑐)

At 𝑐 = 𝜋/4and 𝑐 = 5𝜋/4 we have cos 𝑐 = sin 𝑐.■

6a 6b

The Mean Value Theorem

Let 𝑓 satisfy

1. 𝑓 is continuous on the closed interval 𝑎, 𝑏 .

2. 𝑓 is differentiable on (𝑎, 𝑏).

Then there is a no. 𝑐 on (𝑎, 𝑏) such that

𝑓 ′ 𝑐 =𝑓 𝑏 − 𝑓(𝑎)

𝑏 − 𝑎

Picture …𝑎…𝑏… ,… ,𝑓,𝑓 𝑎 ,𝑓 𝑏 , line segment 𝑔,

c, tangent to 𝑓 at 𝑐

?? Estimate numbers 𝑐 satisfying the Mean Value

Theorem.

Proof. Let 𝑕 = 𝑓 − 𝑕…𝑎…𝑏…, … , 𝑕

By Rolle’s theorem there is a 𝑐 ∈ (𝑎, 𝑏) such that

𝑕′ 𝑐 = 0or

𝑓 ′ 𝑐 − 𝑔′ 𝑐 = 0

Let 𝑚 be the slope of line segment 𝑔.

𝑚 =𝑓 𝑏 − 𝑓(𝑎)

𝑏 − 𝑎

Then 𝑓 ′ 𝑐 = 𝑚■

7a 7b

Example. Verify that 𝑓 ′ 𝑥 = 𝑥, 1 ≤ 𝑥 ≤ 4

satisfies the hypotheses of the Mean Value Theorem.

Find all numbers 𝑐 that satisfy the conclusion.

1. 𝑓(𝑥) is continuous on [1,4]

2. 𝑓(𝑥) is differentiable on (1,4)

𝑓 does satisfy the hypotheses. Find a no. 𝑐 that

satisfies the conclusions. We want

𝑓 ′ 𝑐 =𝑓 𝑏 − 𝑓(𝑎)

𝑏 − 𝑎

= 4 − 1

4 − 1=

1

3

𝑓 𝑥 = 𝑥1

2

𝑓 ′ 𝑥 =1

2𝑥−

1

2 =1

3

𝑥−1

2 =2

3

𝑥1

2 =3

2

𝑥 =9

4= 2.25 ∈ (1,4)■

Example. At 2pm you leave Pullman for the Spokane

airport. After driving north 85 miles, you arrive at

3: 30pm.

Let 𝑠(𝑡) be your position as a function of time 𝑡.

Assume 𝑠(𝑡) is continuous and differentiable. Recall

that

𝑠′ 𝑡 = 𝑣 𝑡 ,

your instantaneous velocity.

Notation of Mean Value theorem

𝑎 = 0 hours 𝑠 𝑎 = 0 miles

𝑏 = 11

2hours𝑠 𝑏 = 85 miles

By the Mean Value theorem there is a no. 𝑐 such that

𝑠′ 𝑐 =𝑠 𝑏 − 𝑠(𝑎)

𝑏 − 𝑎

=85 − 0

11

2− 0

miles

hour≈ 57

miles

hour

At 𝑡 = 𝑐 instantaneous velocity = average velocity.

■

8a 8b

Theorem. If 𝑓 ′ 𝑥 = 0 for all 𝑥 ∈ (𝑎, 𝑏), then 𝑓 is

constant on (𝑎, 𝑏).

Picture …𝑎…𝑏…𝑥-, …, 𝑓

Proof. Let 𝑥1, 𝑥2 be any two numbers such that

𝑎 < 𝑥1 < 𝑥2 < 𝑏

By the Mean Value theorem, there is a no. 𝑐 in

(𝑥1 , 𝑥2) such that

𝑓 ′ 𝑐 =𝑓 𝑥2 − 𝑓(𝑥1)

𝑥2 − 𝑥1

By hypothesis, 𝑓 ′ 𝑐 = 0.

Therefore 𝑓 𝑥1 = 𝑓(𝑥2)■

Corollary. If 𝑓 ′ 𝑥 = 𝑔′(𝑥) for all 𝑥 ∈ (𝑎, 𝑏) then

𝑓 𝑥 = 𝑔 𝑥 + 𝐶 on (𝑎, 𝑏).

Picture …𝑎…𝑏…𝑥-, …, 𝑓,𝑔

Proof. Let 𝑕 = 𝑓 − 𝑔. Then

𝑕′ 𝑥 = 𝑓 ′ 𝑥 − 𝑔′ 𝑥 = 0 for 𝑥 ∈ (𝑎, 𝑏)

Then by the previous theorem

𝑕 is constant on (𝑎, 𝑏).

𝑕 𝑥 = 𝐶

𝑓 𝑥 − 𝑔 𝑥 = 𝐶

𝑓 𝑥 = 𝑔 𝑥 + 𝐶. ■

9a 9b

§4.3 Derivatives and the Shapes of Graphs

Increasing/Decreasing Test

(a) If 𝑓 ′ 𝑥 > 0 on an interval, then 𝑓 is

increasing on that interval.

(b) If 𝑓 ′ 𝑥 < 0 on an interval, then 𝑓 is

decreasing on that interval.

Picture …𝑥-,…,𝑓,𝑓 ′ > 0, 𝑓 ′ < 0

Proof. Follows from the Mean Value Theorem.

Example. Find the intervals where the function

𝑓 𝑥 = 𝑥3 − 2𝑥2 + 𝑥

is increasing or decreasing.

Solution. Find the critical numbers.

Since 𝑓 is differentiable at every real no., need only

find where 𝑓 ′ = 0.

𝑓 ′ 𝑥 = 3𝑥2 − 4𝑥 + 1

= 3𝑥 − 1 (𝑥 − 1)

= 3 𝑥 −1

3 𝑥 − 1

The critical numbers are 𝑥 = 1/3 and 𝑥 = 1.

Construct a sign chart for 𝑓′:

Interval 𝑥 − 13 𝑥 − 1 𝑓′ 𝑓

𝑥 < 1/3 − − + increasing 1

3< 𝑥 < 1

+ − − decreasing

𝑥 > 1 + + + increasing

■

10a 10b

The First Derivative Test

Suppose 𝑐 is a critical number of a continuous

function 𝑓.

a. If 𝑓′ changes from positive to negative at 𝑐,

then 𝑓 has a local maximum at 𝑐.

b. If 𝑓′ changes from negative to positive at 𝑐,

then 𝑓 has a local minimum at 𝑐.

c. If 𝑓′ does not change sign at 𝑐, then 𝑓 has no

local extremum at 𝑐.

Picture

… ,…, curve with 𝑓 ′ > 0,𝑓 ′ < 0,𝑓 ′ > 0,𝑓 ′ > 0

Previous Example (Continued). Find the 𝑥-

coordinates of the local maxima and minima of

𝑓 𝑥 = 𝑥3 − 2𝑥2 + 𝑥

By the sign chart and first derivative test

𝑓 has a local maximum at 𝑥 = 1/3

𝑓 has a local minimum at 𝑥 = 1

Graph 𝑓. Find the local maximum and minimum

values.

𝑓 1

3 =

1

27−

2

9+

1

3=

1 − 6 + 9

27=

4

27≈ 0.15

𝑓 1 = 1 − 2 + 1 = 0

Note also 𝑓 0 = 0.

… 0…1

3… 1…𝑥-,…0… 0.15…, curve

■

11a 11b

Example (Alternative Sign Chart) Find the local

maximum and minimum values of

𝑔 𝑥 = 𝑥 − 2 sin(𝑥)

Solution. Find the critical numbers of 𝑔. Since 𝑔 is

differentiable at every real number, need only find

where 𝑔′ = 0.

𝑔′ 𝑥 = 1 − 2 cos(𝑥) cannot be factored

Then 𝑔′ 𝑥 = 0 when cos 𝑥 = 1/ 2.

45°-45°-90° triangle

recall cos 𝜋

4 =

1

2. See also cos −

𝜋

4 =

1

2.

The critical numbers are

𝑥 =−𝜋

4 and 𝑥 =

𝜋

4

These divide the real line into 3 subintervals.

Evaluate 𝑔′ at test points in each subinterval.

𝑔′ −𝜋

2 = 1 > 0,

𝑔′ 0 = 1 − 2 < 0,

𝑔′ 𝜋

2 = 1 > 0

Alternate sign chart.

−𝜋…𝜋…𝑥-, 𝑔′ > 0, 𝑔′ < 0, 𝑔′ > 0 , segments

with corresponding slopes

Since 𝑔′ is continuous, it cannot change sign within a

subinterval.

12a 12b

By the first derivative test

a. 𝑔 has a local minimum at 𝑥 = −𝜋/4

b. 𝑔 has a local maximum at 𝑥 = 𝜋/4

This can be read directly off the sign chart! ■

Concavity

… ,… ,𝑓 CU

curve above tangents

𝑓′ increasing

𝑓 is concave up (CU)

… ,… ,𝑓,CD

curve below tangents

𝑓′ decreasing

𝑓 is concave down (CD)

Definitions.

If the graph of 𝑓 lies above its tangents on an interval

𝐼, then 𝑓 is concave up on 𝐼.

If the graph of 𝑓 lies below its tangents on an interval

𝐼, then 𝑓 is concave down on 𝐼.

?? graph of 𝑔where is 𝑔 CU and CD?

Concavity Test

a. If 𝑓 ′′ 𝑥 > 0 on 𝐼, then 𝑓 is CU on 𝐼.

b. If 𝑓 ′′ 𝑥 < 0 on 𝐼, then 𝑓 is CD on 𝐼.

Why?

a. 𝑓 ′′ > 0 ⇒ 𝑓 ′ increasing ⇒ 𝑓 CU

b. 𝑓 ′′ < 0 ⇒ 𝑓′ decreasing ⇒ 𝑓 CD ■

13a 13b

Example. Find the intervals where the function

𝑓 𝑥 = 𝑥3 − 2𝑥2 + 𝑥

is CU and CD

Solution.

𝑓 ′ 𝑥 = 3𝑥2 − 4𝑥 + 1

𝑓 ′′ 𝑥 = 6𝑥 − 4

= 6 𝑥 −2

3

sign chart for 𝑓′′

interval 𝑥 − 2/3 𝑓′′ 𝑓 𝑥 < 2/3 − − CD 𝑥 > 2/3 + + CU

■

A point 𝑃 on a curve is an inflection point if the curve

changes from concave up to concave down, or from

concave down to concave up at 𝑃.

Example (continued).

𝑓 𝑥 = 𝑥3 − 2𝑥2 + 𝑥

𝑓 ′′ 𝑥 = 6 𝑥 −2

3

𝑓 has an inflection point at 𝑥 =2

3.

𝑓 2

3 =

2

3

3

− 2 2

3

2

+2

3

=8

27−

8

9+

2

3

=8 − 24 + 18

27

=2

27

The inflection point is 2

3,

2

27 ■

?? graph of 𝑔 where are the inflection points?

14a 14b

The Second Derivative Test

Suppose 𝑓′′ is continuous on an open interval

containing 𝑐.

a) If 𝑓 ′ 𝑐 = 0 and 𝑓 ′′ 𝑐 > 0, then 𝑓 has a local

minimum at 𝑐.

b) If 𝑓 ′ 𝑐 = 0 and 𝑓 ′′ 𝑐 < 0, then 𝑓 has a local

maximum at 𝑐.

Picture

…𝑐… ,… ,𝑓 CU, horizontal tangent at 𝑐

𝑓 ′ 𝑐 = 0

𝑓 ′′ (𝑐) > 0

the slope is increasing

𝑓 has a local minimum

…𝑐… ,… ,𝑓 CD, horizontal tangent at 𝑐

𝑓 ′ 𝑐 = 0

𝑓 ′′ 𝑐 < 0

the slope is decreasing

𝑓 has a local maximum

15a 15b

Example. Apply the second derivative test to find the

𝑥-coordinates of the maxima and minima of

𝑓 𝑥 = 𝑥3 − 2𝑥2 + 𝑥

Solution. Start by finding the critical numbers

𝑓 ′ 𝑥 = 3𝑥2 − 4𝑥 + 1

= 3𝑥 − 1 (𝑥 − 1)

= 3 𝑥 −1

3 (𝑥 − 1)

𝑓 ′ 𝑥 = 0 for 𝑥 = 1/3 and 𝑥 = 1

Find the sign of 𝑓′′ at each critical no. where

𝑓 ′ 𝑐 = 0

𝑓 ′′ 𝑥 = 6𝑥 − 4

= 6 𝑥 −2

3

𝑓 ′′ 1

3 < 0 a local max. by the 2nd derivative test

𝑓 ′′ 1 > 0 a local min. by the 2nd derivative test

graph

𝑓 0 = 0

𝑓 1

3 =

4

27≈ 0.15

𝑓 1 = 0

recall the inflection point 2

3,

2

27

… 0…1

3…

2

3… .1…𝑥-,…0…0.15…𝑦-, 𝑓

Note the local max, the local min & inflection point.■

16a 16b

Example. Find the local maximum and minimum

values of 𝑓 using the first and second derivative tests.

𝑓 𝑥 =𝑥

𝑥2 + 4

Solution. Start by finding the critical numbers (this is

common to both tests).

𝑓 ′ 𝑥 = 1 𝑥2 + 4 − 𝑥(2𝑥)

𝑥2 + 4 2

=4 − 𝑥2

𝑥2 + 4 2

= 2 − 𝑥 (2 + 𝑥)

𝑥2 + 4 2

critical numbers are 𝑥 = −2 and 𝑥 = 2

Apply the first derivative test

Construct a sign chart for 𝑓′

Interval 2 − 𝑥 2 + 𝑥 𝑓′ 𝑓

𝑥 < −2 + − − decreasing −2 < 𝑥 < 2 + + + increasing

𝑥 > 2 − + − decreasing


𝑓 has a local min at 𝑥 = −2 and a local max at 𝑥 = 2

Apply the second derivative test

Must calculate the second derivative

𝑓 ′′ 𝑥 =−2𝑥 𝑥2 + 4 2 − 4 − 𝑥2 2 𝑥2 + 4 (2𝑥)

𝑥2 + 4 4

=−2𝑥 𝑥2 + 4 − 4𝑥(4 − 𝑥2)

𝑥2 + 4 3

=−2𝑥3 − 8𝑥 − 16𝑥 + 4𝑥3

𝑥2 + 4 3

=2𝑥3 − 24𝑥

𝑥2 + 4 3

=2𝑥(𝑥2 − 12)

𝑥2 + 4 3

Then

𝑓 ′′ −2 =−4(4−12)

4+4 3> 0 local min at 𝑥 = −2

𝑓 ′′ 2 =4(8−12)

4+4 3< 0 local max at 𝑥 = 2■

17a 17b

Compare the first and second derivative tests

1. The first derivative test may be easier to apply.

2. The first derivative tests handles more cases.

You can apply it to a function that has a corner

or cusp at 𝑥 = 𝑐, where the 2nd derivative DNE.

axes, curve, tangent 1st& 2nd deriv. tests OK

function w/ corner 1st derivative test only

3. The second derivative tests generalizes to

calculus of several variables.

§4.4 Curve Sketching

Curve Sketching Checklist

A. Domain

B. Intercepts: 𝑥 and 𝑦

C. Symmetry: odd, even or periodic

D. Asymptotes: horizontal and vertical

E. Intervals of increase and decrease

F. Local maxima and minima

G. Concavity and points of inflection

H. Sketch the curve!

18a 18b

Example. Sketch the graph of 𝑓 𝑥 = 𝑥3 + 6𝑥2 + 9𝑥

A. Domain

𝑓 is a polynomial. The domain is all real numbers.

B. Intercepts

What is 𝑥 when 𝑦 = 0?

𝑥3 + 6𝑥2 + 9𝑥 = 0

𝑥 𝑥2 + 6𝑥 + 9 = 0

𝑥 𝑥 + 3 2 = 0

𝑥 intercept 𝑥 = 0 or 𝑥 = −3 ***

What is 𝑦 when 𝑥 = 0? 𝑓 0 = 0

𝑦 intercept 𝑦 = 0 ***

C. Symmetry. The polynomial involves both even

and odd powers of 𝑥. No symmetry.

D. Asymptotes.No horizontal or vertical asymptotes.

E. Intervals of increase or decrease?

𝑓 ′ 𝑥 = 3𝑥2 + 12𝑥 + 9

= 3(𝑥2 + 4𝑥 + 3)

= 3 𝑥 + 1 (𝑥 + 3)

The sign of 𝑓′ changes when 𝑓 ′ = 0

Critical numbers at 𝑥 = −1 and 𝑥 = −3

Sign chart for 𝑓′

…− 3…− 1…𝑥-,𝑓 ′ > 0,𝑓 ′ < 0,𝑓 ′ > 0, strokes

𝑓 increases on (−∞,−3) and (−1,∞) ***

𝑓 decreases on (−3,−1) ***

F. Local max and min values


𝑓 has a local max. at 𝑥 = −3.

𝑓 −3 = −3 3 + 6 −3 2 + 9 −3

= −27 + 54 − 27 = 0

Local max at −3,0 ***

𝑓 has a local min at 𝑥 = −1

𝑓 −1 = −1 3 + 6 −1 2 + 9 −1

= −1 + 6 − 9 = −4

Local min at (−1,−4) ***

19a 19b

G. Concavity and inflection points

𝑓 ′′ 𝑥 = 6𝑥 + 12 = 6(𝑥 + 2)


…− 2…𝑥-, CD 𝑓 ′′ < 0, CU 𝑓 ′′ > 0 ***

𝑓 −2 = −2 3 + 6 −2 2 + 9(−2)

= −8 + 6 4 − 18 = −2

Inflection point (−2,−2) ***

Sketch 𝑓

−4…1…𝑥-,−5…0…𝑦-,𝑓

Whew! ■

Example. Sketch the graph of 𝑔 𝑥 =𝑥

𝑥2−9

A. Domain 𝑥 ∈ ℝ 𝑥 ≠ −3 or 3}

= −8,−3 ∪ −3,3 ∪ (3,∞)

B. 𝑥-intercept: 𝑥 = 0

𝑦-intercept: 𝑦 = 0

C. Symmetry: 𝑓 𝑥 =odd function

even function= odd function

D. Horizontal asymptote: a value approached as 𝑥 → ±∞?

indeterminate form type ∞/∞

lim𝑥→±∞

𝑥

𝑥2 − 9= lim

𝑥→±∞

1

2𝑥= 0

Vertical asymptotes: denom= 0 and num≠ 0 at 𝑥 = −3 and 𝑥 = 3

lim𝑥→3+

𝑥

𝑥2 − 9= ∞ lim

𝑥→3−

𝑥

𝑥2 − 9= −∞


𝑓 ′ 𝑥 = 1 (𝑥2 − 9) − 𝑥(2𝑥)

𝑥2 − 9 2=

−𝑥2 − 9

𝑥2 − 9 2< 0

20a 20b

𝑓′(𝑥) is always decreasing in its domain; no max or

min values


𝑓 ′′ 𝑥

= −𝑥2 − 9 ′ 𝑥2 − 9 2 − −𝑥2 − 9 𝑥2 − 9 2 ′

𝑥2 − 9 4

= −2𝑥 𝑥2 − 9 2 − (−𝑥2 − 9)2(𝑥2 − 9)(2𝑥)

𝑥2 − 9 4

= −2𝑥 (𝑥2 − 9) − (−𝑥2 − 9)(4𝑥)

𝑥2 − 9 3

=−2𝑥3 + 18𝑥 + 4𝑥3 + 36𝑥

𝑥2 − 9 3

=2𝑥3 + 54𝑥

𝑥2 − 9 3

=2𝑥(𝑥2 + 27)

𝑥2 − 9 3


interval 2𝑥 𝑥2 − 9 𝑓′′ 𝑓 𝑥 < −3 − + − CD

−3 < 𝑥 < 0 − − + CU 0 > 𝑥 > 3 + − − CD 𝑥 > 3 + + + CU

Inflection point at (0,0)

Sketch graph

−4…− 2…0…2…4…𝑥-,−2… 0…2…𝑦-,𝑓

where 𝑓 2 =2

4−9=

−2

5 and 𝑓 4 =

4

16−9=

4

7

■

21a 21b

Example Graph 𝑦 = 𝑓 𝑥 = ln(cos(𝑥))

A. Domain ln 𝑧 is defined for 𝑧 > 0

Sketch cos 𝑥

−2𝜋…0… 2𝜋…𝑥-,−1…0… 1…𝑦-, cos 𝑥

cos 𝑥 > 0 for 𝒟 = ⋯∪ –𝜋

2,𝜋

2 ∪

3𝜋

2,

5𝜋

2 ∪ …

𝒟 is the domain of 𝑓 ***

𝑓(𝑥) will have period 2𝜋

restrict this discussion to −𝜋

2< 𝑥 <

𝜋

2

B. 𝑥-intercept when 𝑦 = ln(cos 𝑥) = 0, cos 𝑥 = 1

𝑥 = ⋯ ,−2𝜋, 0, 2𝜋,… ***

𝑦- intercept 𝑦 = 𝑓 0 = ln(cos 0) = ln 1 = 0 ***

C. Symmetry cos(𝑥) is periodic and even

𝑦 = ln(cos𝑥) is periodic and even too ***

D. Horizontal asymptotes: none

Vertical asymptotes:

ln 𝑧 has a vertical asymptote at 𝑧 = 0

ln(cos(𝑥)) ………………………….. cos 𝑥 = 0

occurs for 𝑥 = ⋯ ,−3𝜋

2,−

𝜋

2,𝜋

2,

3𝜋

2,… ***


𝑓 𝑥 = ln(cos𝑥)

𝑓 ′ 𝑥 =−sin 𝑥

cos 𝑥= − tan 𝑥

−𝜋/2… 0…𝜋/2…𝑥-,… 0…, tan 𝑥 ,− tan 𝑥

𝑓 ′ > 0 on (−𝜋

2, 0) and 𝑓 ′ < 0 on 0,

𝜋

2

22a 22b

F. Local max and min values

Critical number at 𝑥 = 0.

By the first derivative test, a local max at 𝑥 = 0

𝑓 has a local max at (0, 0) ***


From the graph of 𝑓′ we see 𝑓 ′′ = 𝑓 ′ ′ < 0

𝑓 is CD ***

Graph 𝑦 = ln(cos 𝑥)

−2𝜋…0… 2𝜋…𝑥-,−1…0… 1…𝑦-, ln( cos 𝑥)

■

§4.5Optimization Problems

Problem Solving Strategy

1. Read problem carefully.

2. Draw a picture if relevant.

3. Assign notation to the given information and

unknown quantities.

4. Express quantity to be optimized in terms of

one variable.

5. Find the domain of the variable.

6. Find the absolute maximum or minimum (give

a clear argument that it is an abs. max or min)

and answer the question.

23a 23b

Example. A farmer want to fence an area of 1.5 million square feet in a rectangular field and then divide it into half with a fence parallel to one of the sides of the rectangle. What are the dimensions of the field that minimize the length of the fence?

2. fence, 𝑥, 𝑥, 𝑥, 𝑥,𝑦, 𝑦,𝑦

3. Area of field 𝐴 = 1.5 106 square feet.

Length of fence 𝐿

𝐴 = 2𝑥𝑦

𝐿 = 4𝑥 + 3𝑦

4. Express 𝐿 in terms of one variable.

𝑦 = 𝐴/(2𝑥)

𝐿 𝑥 = 4𝑥 + 3𝐴/(2𝑥)

5. 0 < 𝑥 < ∞

6. Find a critical number

𝐿′ 𝑥 = 4 +3

2𝐴 −𝑥−2 = 4 −

3

2𝐴

1

𝑥2

At the critical no. 𝑐, 𝐿′ 𝑐 = 0:

4 =3

2𝐴

1

𝑐2 or 𝑐2 =

3

8𝐴

𝑐 = 3

8𝐴

1

2=

3

8⋅

3

2106

1

2=

3

4103 = 750 feet

Is this an absolute minimum? Write

𝐿′ 𝑥 = 1

𝑥2(4𝑥2 −

3

2𝐴) =

4

𝑥2 𝑥2 −

3

8𝐴

=4

𝑥2 𝑥2 − 7502

For 𝑥 < 750 ft, 𝐿′ < 0. For 𝑥 > 750 ft, 𝐿′ > 0.

… 750…𝑥-, slopes

𝐿 has an absolute minimum at 𝑥 = 750 feet.

𝑦 =𝐴

2𝑥=

1.5 106

1.5 103= 1000 feet. ■

24a 24b

First Derivative Test For Absolute Extreme Values

Suppose 𝑐 is a critical no. of a continuous function 𝑓

defined on an interval.

a) If 𝑓 ′ 𝑥 > 0 for all 𝑥 < 𝑐 and 𝑓 ′ 𝑥 < 0 for all

𝑥 > 𝑐, then 𝑓(𝑐) is an absolute max. value of

𝑓.

b) If 𝑓 ′ 𝑥 < 0 for all 𝑥 < 𝑐 and 𝑓 ′ 𝑥 > 0 for all

𝑥 > 𝑐, then 𝑓(𝑐) is an absolute min. value of 𝑓.

Sign chart for part (a) …𝑐…, 𝑓′ values, strokes

Sign chart for part (b) …𝑐…, 𝑓′ values, strokes

Example. From a piece of cardboard of side 1 foot,

cut out square corners of side 𝑥 and fold up flaps to

make an open top box. What 𝑥 gives the maximum

volume?

2.square, corners of side 𝑥, middles 12 − 2𝑥

3. 𝑉 = volume of box

𝑥 = size of corners

4. 𝑉 𝑥 = 12 − 2𝑥 12 − 2𝑥 𝑥

= 4𝑥 6 − 𝑥 2

5. 0 ≤ 𝑥 ≤ 6 inches

25a 25b

6. Find the critical numbers

𝑉 ′ 𝑥 = 4 6 − 𝑥 2 + 4𝑥 2 6 − 𝑥 (−1)

= 4 6 − 𝑥 [ 6 − 𝑥 − 2𝑥]

= 4(6 − 𝑥)(6 − 3𝑥)

= 12(6 − 𝑥)(2 − 𝑥)

𝑥 = 2 and 𝑥 = 6 are critical numbers.

Use the closed interval method.

a) Evaluate 𝑉 at the endpoints of the domain.

𝑉 0 = 0.

𝑉 6 = 0.

b) Evaluate 𝑉 at the critical point

𝑉 2 = 8 ⋅ 16 = 128

c) The largest value gives the absolute max. of 𝑉.

𝑥 = 2 inches ■

Example. A woman at a point A on the shore of a

circular lake wants to arrive at a point C opposite A

on the other side in the shortest possible time. She

can walk at a rate of 4 miles per hour and row at a

rate of 2 miles per hour. How should she proceed?

2. circle, 𝐴…0…𝐶, radius 𝑟, 𝑟, 𝐵, segment 𝑥,

∠0𝐴𝐵 = 𝜃, arc 𝑦 3. 𝑥(𝜃) distance rowing, 𝜃 = 0 row only

𝑦(𝜃) distance walking, 𝜃 = 𝜋/2 walk only

𝑇 𝜃 =1

2𝑥 𝜃 +

1

4𝑦(𝜃)

4. Express quantity to be optimized in terms of 1

variable

add ∠0𝐵𝐴 = 𝜃,∠𝐴𝑂𝐵 = 𝜋 − 2𝜃,∠𝐵0𝐶 = 2𝜃

26a 26b

𝑦 𝜃 = 𝑟(2𝜃)

add segment BC Note ∠𝐴𝐵𝐶 is a right angle. Let

∠0𝐵𝐶 = ∠0𝐶𝐵 = 𝛼… Obtain 𝜃 + 𝛼 = 𝜋/2 𝑥

2𝑟= cos(𝜃)

𝑥 𝜃 = 2𝑟 cos(𝜃)

time spent traveling

𝑇 𝜃 =1

2𝑥 𝜃 +

1

4𝑦(𝜃)

5. domain: 0 ≤ 𝜃 ≤ 𝜋/2

6. Find the absolute min. value

Use the closed interval method.

𝑇 0 =1

2𝑥(0) =

1

2(2𝑟) = 𝑟

𝑇 𝜋

2 =

1

4𝑦

𝜋

2 =

1

4𝜋𝑟 ≈ 0.79𝑟

Find the critical numbers

𝑇 ′ 𝜃 =1

2𝑥 ′ 𝜃 +

1

4𝑦′ 𝜃

=1

2(−2𝑟 sin 𝜃 +

1

4 2𝑟 = 𝑟

1

2− sin 𝜃

Then the critical value of 𝜃 gives

sin 𝜃 =1

2⟹ 𝜃 = 𝜋/6

𝑇 𝜋

6 =

1

2𝑥

𝜋

6 +

1

4𝑦

𝜋

6

=1

2 2𝑟 cos

𝜋

6 +

1

4 𝑟 2 ⋅

𝜋

6

= 𝑟 3

2+ 𝑟

𝜋

12≈ 1.13𝑟

The shortest time corresponds to 𝜃 = 𝜋/2, or

walking all the way.■

STOP §4.7 Antiderivatives

A function 𝐹 is called an antiderivative of 𝑓 on an

interval 𝐼 if 𝐹′ 𝑥 = 𝑓(𝑥) for all 𝑥 in 𝐼

Example. Find an antiderivative of 𝑓 𝑥 = 𝑥

27a 27b

𝐹 𝑥 =1

2𝑥2

?? 𝐺 𝑥 =■

Recall: If 𝐹′ = 𝐺′ for all 𝑥 on an interval 𝐼, then

𝐹 𝑥 = 𝐺 𝑥 + 𝐶on 𝐼.

Theorem. If 𝐹 is an antiderivative of 𝑓 on an interval

𝐼, then the most general antiderivative of 𝑓 on 𝐼 is

𝐹 𝑥 + 𝐶,

where 𝐶 is an arbitrary constant.

Example. Find the most general antiderivative of

𝑓 𝑥 = 𝑥.

𝐹 𝑥 =1

2𝑥2 + 𝐶■


𝑓 𝑥 = 𝑥𝑛 .

𝑑

𝑑𝑥

1

𝑛 + 1

𝑑

𝑑𝑥𝑥𝑛+1 =

1

𝑛 + 1 𝑛 + 1 𝑥𝑛 = 𝑥𝑛

The most general antiderivative is

𝐹 𝑥 =1

𝑛+1𝑥𝑛+1 + 𝐶■

Table of Antiderivatives

Function

Antiderivative

𝑥𝑛 , 𝑛 ≠ 1 1

𝑛 + 1𝑥𝑛+1 + 𝐶

1

𝑥

ln 𝑥 + 𝐶

cos(𝑥) sin 𝑥 + 𝐶 sin(𝑥) − cos 𝑥 + 𝐶 sec2(𝑥) tan 𝑥 + 𝐶

1

1 − 𝑥2

sin−1 𝑥 + 𝐶

1

1 + 𝑥2

tan−1 𝑥 + 𝐶

𝑎 𝑓(𝑥) 𝑎𝐹 𝑥 + 𝐶 𝑓 𝑥 + 𝑔(𝑥) 𝐹 𝑥 + 𝐺 𝑥 + 𝐶

where 𝐹 and 𝐺 are functions satisfying 𝐹′ 𝑥 = 𝑓(𝑥)

and 𝐺 ′ 𝑥 = 𝑔(𝑥).


𝑓 𝑥 =𝑥2 + 𝑥 + 1

𝑥

28a 28b

Solution. First simplify algebraically

𝑓 𝑥 = 𝑥 + 1 + 𝑥−1

then

𝐹 𝑥 =1

2𝑥2 + 𝑥 + ln 𝑥 + 𝐶■


𝑓 𝑡 = sin 𝑡 − 2 𝑡

Solution

𝐹 𝑡 = − cos 𝑡 − 2 2

3𝑡

3

2 + 𝐶

= − cos 𝑡 −4

3𝑡

3

2 + 𝐶■

Geometrical Picture of Antiderivatives

Consider

𝑓 𝑥 = 𝑥

general form of antiderivative

𝐹 𝑥 =1

2𝑥2 + 𝐶

get a family of antiderivatives corresponding to

different values of 𝐶

−2…0…2…𝑥-,−1…0…3…𝑦-, several curves

Differential Equations

A differential equation is an equation involving one

or more derivatives of an unknown function

Example. Solve

𝑑𝑦

𝑑𝑥= 𝑥

with the extra condition that 𝑦 = 1 when 𝑥 = 0.

General solution

𝑦 =1

2𝑥2 + 𝐶

particular solution

𝑦 =1

2𝑥2 + 1■

Example. Solve for 𝑓(𝑥)

𝑓 ′ 𝑥 = 4 − 3 1 + 𝑥2 −1

29a 29b

with 𝑓 1 = 0.

General solution.

𝑓 𝑥 = 4𝑥 − 3 tan−1 𝑥 + 𝐶

find the particular solution

0 = 𝑓 1 = 4 − 3 tan−1 1 + 𝐶

= 4 − 3 𝜋

4 + 𝐶

then 𝐶 =3𝜋

4− 4.

𝑓 𝑥 = 4𝑥 − 3 tan−1 𝑥 +3𝜋

4− 4■

Example. Solve for 𝑓(𝑥)

𝑓 ′′ 𝑥 = 20𝑥3 − 10

with 𝑓 1 = 1 and 𝑓 ′ 1 = −5.

General antiderivative of 𝑓′′

𝑓 ′ 𝑥 = 20 1

4𝑥4 − 10𝑥 + 𝐶

use 𝑓 ′(1) = −5

𝑓 ′ 1 = 5 − 10 + 𝐶 = −5

we see 𝐶 = 0. Thus

𝑓 ′ 𝑥 = 5𝑥4 − 10𝑥

general antiderivative of 𝑓′

𝑓 𝑥 = 𝑥5 − 5𝑥2 + 𝐷

use 𝑓 1 = 1

𝑓 1 = 1 − 5 + 𝐷 = 1

we see 𝐷 = 5

conclude

𝑓 𝑥 = 𝑥5 − 5𝑥2 + 5■

Position, Velocity and Acceleration

Consider motion along a straight line

𝑠(𝑡) position at time 𝑡 units are feet

30a 30b

𝑣(𝑡) velocity at time 𝑡 units are feet/sec

𝑎(𝑡)acceleration at time 𝑡 units are feet/sec2

where

𝑠′ 𝑡 = 𝑣(𝑡)

𝑣′ 𝑡 = 𝑎(𝑡)

Example. You are standing on the edge of a cliff 96

feet above the Snake river.

cliff, water, 0…96…𝑠- with origin at water

(a) At 𝑡 = 0, throw a rock straight up at 80 ft/sec.

Find 𝑣(𝑡) and 𝑠(𝑡) given 𝑎 = −32 ft/sec2.

First consider

𝑣′ 𝑡 = 𝑎 = −32

general solution

𝑣 𝑡 = −32𝑡 + 𝐶

we have the initial condition that 𝑣 0 = 80 ft/sec

𝑣 𝑡 = −32𝑡 + 80

Next consider

𝑠′ 𝑡 = 𝑣 𝑡 = −32𝑡 + 80

general solution

𝑠 𝑡 = −16𝑡2 + 80𝑡 + 𝐷

we have a second initial condition 𝑠 0 = 96 ft.

𝑠 𝑡 = −16𝑡2 + 80𝑡 + 96

(b) When does the rock hit the water?

Find 𝑡 such that 𝑠 𝑡 = 0

−16𝑡2 + 80𝑡 + 96 = 0

divide by 16

−𝑡2 + 5𝑡 + 6 = 0

31a 31b

factor

𝑡2 − 5𝑡 − 6 = 0

𝑡 + 1 𝑡 − 6 = 0

The answer we want corresponds to the positive root

𝑡 = 6 seconds. ■

4.1 maximum and minimum values has an absolute minimum at · has an absolute minimum at = 0. 0 = 0...

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