﻿ ﻿Big Idea 2: Derivatives . Graphs of Functions and Derivatives - Review the Knowledge You Need to Score High - 5 Steps to a 5: AP Calculus AB 2017 (2016)

## 5 Steps to a 5: AP Calculus AB 2017 (2016)

### Graphs of Functions and Derivatives

IN THIS CHAPTER

Summary: Many questions on the AP Calculus AB exam involve working with graphs of a function and its derivatives. In this chapter, you will learn how to use derivatives both algebraically and graphically to determine the behavior of a function. Applications of Rolle’s Theorem, the Mean Value Theorem, and the Extreme Value Theorem are also shown.

Key Ideas

Rolle’s Theorem, Mean Value Theorem, and Extreme Value Theorem

Test for Increasing and Decreasing Functions

First and Second Derivative Tests for Relative Extrema

Test for Concavity and Point of Inflection

Curve Sketching

Graphs of Derivatives

8.1 Rolle’s Theorem, Mean Value Theorem, and Extreme Value Theorem

Main Concepts: Rolle’s Theorem, Mean Value Theorem, Extreme Value Theorem

• Set your calculator to Radians and change it to Degrees if/when you need to. Do not forget to change it back to Radians after you have finished using it in Degrees.

Rolle’s Theorem

If f is a function that satisfies the following three conditions:

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

2. f is differentiable on the open interval (a , b )

3. f (a ) = f (b ) = 0

then there exists a number c in (a , b ) such that f ′(c ) = 0. (See Figure 8.1-1 .)

Figure 8.1-1

Note that if you change condition 3 from f (a ) = f (b ) = 0 to f (a ) = f (b ), the conclusion of Rolle’s Theorem is still valid.

Mean Value Theorem

If f is a function that satisfies the following conditions:

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

2. f is differentiable on the open interval (a , b )

then there exists a number c in (a , b ) such that . (See Figure 8.1-2 .)

Figure 8.1-2

Example 1

If f (x ) = x 2 + 4x – 5, show that the hypotheses of Rolle’s Theorem are satisfied on the interval [–4, 0] and find all values of c that satisfy the conclusion of the theorem. Check the three conditions in the hypotheses of Rolle’s Theorem:

(1) f (x ) = x 2 + 4x – 5 is continuous everywhere since it is polynomial.

(2) The derivative f ′(x ) = 2x + 4 is defined for all numbers and thus is differentiable on (–4, 0).

(3) f (0) = f (–4) = –5. Therefore, there exists a c in (–4, 0) such that f ′(c ) = 0. To find c , set f ′(x ) = 0. Thus, 2x + 4 = 0 ⇒ x = –2, i.e., f ′(–2) = 0. (See Figure 8.1-3 .)

Figure 8.1-3

Example 2

Let . Using Rolle’s Theorem, show that there exists a number c in the domain of f such that f ′(c ) = 0. Find all values of c .

Note f (x ) is a polynomial and thus f (x ) is continuous and differentiable everywhere.

Enter . The zeros of y 1 are approximately –2.3, 0.9, and 2.9 i.e., f (–2.3) = f (0.9) = f (2.9) = 0. Therefore, there exists at least one c in the interval (–2.3, 0.9) and at least one c in the interval (0.9, 2.9) such that f ′(c ) = 0. Use d [Differentiate ] to find f ′(x ): f ′(x ) = x 2x – 2. Set f ′(x ) = 0 ⇒ x 2x – 2 = 0 or (x – 2)(x + 1) = 0.

Thus, x = 2 or x = –1, which implies f ′(2) = 0 and f ′(–1) = 0. Therefore, the values of c are –1 and 2. (See Figure 8.1-4 .)

Figure 8.1-4

Example 3

The points P (1, 1) and Q (3, 27) are on the curve f (x ) = x 3 . Using the Mean Value Theorem, find c in the interval (1, 3) such that f ′(c ) is equal to the slope of the secant .

The slope of secant is . Since f (x ) is defined for all real numbers, f (x ) is continuous on [1, 3]. Also f ′(x ) = 3x 2 is defined for all real numbers. Thus, f (x ) is differentiable on (1, 3). Therefore, there exists a number c in (1, 3) such that f ′(c ) = 13.

Set f ′(c ) = 13 ⇒ 3(c )2 = 13 or . Since only is in the interval (1, 3), . (See Figure 8.1-5 .)

Figure 8.1-5

Example 4

Let f be the function f (x ) = (x – 1)2/3 . Determine if the hypotheses of the Mean Value Theorem are satisfied on the interval [0, 2], and if so, find all values of c that satisfy the conclusion of the theorem.

Enter y 1 = (x – 1)2/3 . The graph y 1 shows that there is a cusp at x = 1. Thus, f (x ) is not differentiable on (0, 2) which implies there may or may not exist a c in (0, 2) such that . The derivative and . Set . Note that f is not differentiable (a + x = 1). Therefore, c does not exist. (See Figure 8.1-6 .)

Figure 8.1-6

• The formula for finding the area of an equilateral triangle is where s is the length of a side. You might need this to find the volume of a solid whose cross sections are equilateral triangles.

Extreme Value Theorem

If f is a continuous function on a closed interval [a , b ], then f has both a maximum and a minimum value on the interval.

Example 1

If f (x ) = x 3 + 3x 2 – 1, find the maximum and minimum values of f on [– 2, 2]. Since f (x ) is a polynomial, it is a continuous function everywhere. Enter y 1 = x 3 + 3x 2 – 1. The graph of y 1 indicates that f has a minimum of –1 at x= 0 and a maximum value of 19 at x = 2. (See Figure 8.1-7 .)

Figure 8.1-7

Example 2

If , find any maximum and minimum values of f on [0, 3]. Since f (x ) is a rational function, it is continuous everywhere except at values where the denominator is 0. In this case, at x = 0, f (x ) is undefined. Since f (x ) is not continuous on [0, 3], the Extreme Value Theorem may not be applicable. Enter . The graph of y 1 shows that as x → 0+ , f (x ) increases without bound (i.e., f (x ) goes to infinity). Thus, f has no maximum value. The minimum value occurs at the endpoint x = 3 and the minimum value is . (See Figure 8.1-8 .)

Figure 8.1-8

8.2 Determining the Behavior of Functions

Main Concepts: Test for Increasing and Decreasing Functions, First Derivative Test and Second Derivative Test for Relative Extrema, Test for Concavity and Points of Inflection

Test for Increasing and Decreasing Functions

Let f be a continuous function on the closed interval [a , b ] and differentiable on the open interval (a , b ).

1. If f ′(x ) > 0 on (a , b ), then f is increasing on [a , b ].

2. If f ′(x ) < 0 on (a , b ), then f is decreasing on [a , b ].

3. If f ′(x ) = 0 on (a , b ), then f is constant on [a , b ].

Definition: Let f be a function defined at a number c . Then c is a critical number of f if either f ′(c ) = 0 or f ′ (c ) does not exist. (See Figure 8.2-1 .)

Figure 8.2-1

Example 1

Find the critical numbers of f (x ) = 4x 3 + 2x 2 .

To find the critical numbers of f (x ), you have to determine where f ′(x ) = 0 and where f ′(x ) does not exist. Note f ′(x ) = 12x 2 + 4x , and f ′(x ) is defined for all real numbers. Let f ′(x ) = 0 and thus 12x 2 + 4x = 0, which implies 4x(3x + 1) = 0 ⇒ x = –1/3 or x = 0. Therefore, the critical numbers of f are 0 and –1/3. (See Figure 8.2-2 .)

Figure 8.2-2

Example 2

Find the critical numbers of f (x ) = (x – 3)2/5 .

. Note that f ′(x ) is undefined at x = 3 and that f ′(x ) ≠ 0. Therefore, 3 is the only critical number of f . (See Figure 8.2-3 .)

Figure 8.2-3

Example 3

The graph of f ′ on (1, 6) is shown in Figure 8.2-4 . Find the intervals on which f is increasing or decreasing.

Figure 8.2-4

Solution: (See Figure 8.2-5 .)

Figure 8.2-5

Thus, f is decreasing on [1, 2] and [5, 6] and increasing on [2, 5].

Example 4

Find the open intervals on which f (x ) = (x 2 – 9)2/3 is increasing or decreasing.

Step 1: Find the critical numbers of f .

Set f ′(x ) = 0 ⇒ 4x = 0 or x = 0.

Since f ′(x ) is a rational function, f ′(x ) is undefined at values where the denominator is 0. Thus, set x 2 – 9 = 0 ⇒ x = 3 or x = –3. Therefore, the critical numbers are –3, 0, and 3.

Step 2: Determine intervals.

Intervals are (–∞, –3), (–3, 0), (0, 3), and (3, ∞).

Step 3: Set up a table.

Step 4: Write a conclusion. Therefore, f (x ) is increasing on [–3, 0] and [3, ∞) and decreasing on (– ∞, –3] and [0, 3]. (See Figure 8.2-6 .)

Figure 8.2-6

Example 5

The derivative of a function f is given as f ′(x ) = cos(x 2 ). Using a calculator, find the values of x on such that f is increasing. (See Figure 8.2-7 .)

Figure 8.2-7

Using the [Zero ] function of the calculator, you obtain x = 1.25331 is a zero of f ′ on . Since f ′(x ) = cos(x 2 ) is an even function, x = –1.25331 is also a zero on . (See Figure 8.2-8 .)

Figure 8.2-8

Thus, f is increasing on [–1.2533, 1.2533].

• Be sure to bubble in the right grid. You have to be careful in filling in the bubbles especially when you skip a question.

First Derivative Test and Second Derivative Test for Relative Extrema

First Derivative Test for Relative Extrema

Let f be a continuous function and c be a critical number of f . (Figure 8.2-9 .)

Figure 8.2-9

1. If f ′(x ) changes from positive to negative at x = c (f ′ > 0 for x< c and f ′ < 0 for x > c ), then f has a relative maximum at c .

2. If f ′(x ) changes from negative to positive at x = c (f ′ < 0 for x < c and f ′ > 0 for x > c ), then f has a relative minimum at c .

Second Derivative Test for Relative Extrema

Let f be a continuous function at a number c .

1. If f ′(c ) = 0 and f ″(c ) < 0, then f (c ) is a relative maximum.

2. If f ′(c ) = 0 and f ″(c ) > 0, then f (c ) is a relative minimum.

3. If f ′(c ) = 0 and f ″(c ) = 0, then the test is inconclusive. Use the First Derivative Test.

Example 1

The graph of f ′, the derivative of a function f , is shown in Figure 8.2-10 . Find the relative extrema of f .

Figure 8.2-10

Solution: (See Figure 8.2-11 .)

Figure 8.2-11

Thus, f has a relative maximum at x = –2, and a relative minimum at x = 3.

Example 2

Find the relative extrema for the function .

Step 1: Find f ′(x ).

f ′(x ) = x 2 – 2x – 3

Step 2: Find all critical numbers of f (x ).

Note that f ′(x ) is defined for all real numbers.

Set f ″(x ) = 0: x 2 – 2x – 3 = 0 ⇒ (x – 3)(x + 1) = 0 ⇒ x = 3 or x = –1.

Step 3: Find f ″(x ): f ″ (x ) = 2x – 2.

Step 4: Apply the Second Derivative Test.

f ″(3) = 2(3) – 2 = 4 ⇒ f (3) is a relative minimum.

f ″(–1) = 2(–1) – 2 = –4 ⇒ f (–1) is a relative maximum.

and .

Therefore, –9 is a relative minimum value of f and is a relative maximum value. (See Figure 8.2-12 .)

Figure 8.2-12

Example 3

Find the relative extrema for the function f (x ) = (x 2 – 1)2/3 .

Using the First Derivative Test

Step 1: Find f ′(x ).

Step 2: Find all critical numbers of f .

Set f ′(x ) = 0. Thus, 4x = 0 or x = 0.

Set x 2 – 1 = 0. Thus, f ′(x ) is undefined at x = 1 and x = –1. Therefore, the critical numbers are –1, 0, and 1.

Step 3: Determine intervals.

The intervals are (–∞, –1), (–1, 0), (0, 1), and (1, ∞).

Step 4: Set up a table.

Step 5: Write a conclusion.

Using the First Derivative Test, note that f (x ) has a relative maximum at x = 0 and relative minimums at x = –1 and x = 1.

Note that f (–1) = 0, f (0) = 1, and f (1) = 0. Therefore, 1 is a relative maximum value and 0 is a relative minimum value. (See Figure 8.2-13 .)

Figure 8.2-13

• Do not forget the constant, C , when you write the antiderivative after evaluating an indefinite integral, e.g., cos xdx = sin x + C .

Test for Concavity and Points of Inflection

Test for Concavity

Let f be a differentiable function.

1. If f ″ > 0 on an interval I, then f is concave upward on I.

2. If f ″ < 0 on an interval I, then f is concave downward on I.

(See Figures 8.2-14 and 8.2-15 .)

Figure 8.2-14

Figure 8.2-15

Points of Inflection

A point P on a curve is a point of inflection if:

1. the curve has a tangent line at P, and

2. the curve changes concavity at P (from concave upward to downward or from concave downward to upward).

(See Figures 8.2-168.2-18 .)

Figure 8.2-16

Figure 8.2-17

Figure 8.2-18

Note that if a point (a , f (a )) is a point of inflection, then f ″(c ) = 0 or f ″(c ) does not exist. (The converse of the statement is not necessarily true.)

Note: There are some textbooks that define a point of inflection as a point where the concavity changes and do not require the existence of a tangent at the point of inflection. In that case, the point at the cusp in Figure 8.2-18would be a point of inflection.

Example 1

The graph of f ′, the derivative of a function f , is shown in Figure 8.2-19 . Find the points of inflection of f and determine where the function f is concave upward and where it is concave downward on [–3, 5].

Figure 8.2-19

Solution: (See Figure 8.2-20 .)

Figure 8.2-20

Thus, f is concave upward on [–3, 0) and (3, 5], and is concave downward on (0, 3).

There are two points of inflection: one at x = 0 and the other at x = 3.

Example 2

Using a calculator, find the values of x at which the graph of y = x 2 e x changes concavity.

Enter y 1 = x 2 ∗ e x and y 2 = d (y 1(x ), x , 2). The graph of y 2, the second derivative of y , is shown in Figure 8.2-21 . Using the [Zero ] function, you obtain x = –3.41421 and x = –0.585786. (See Figures 8.2-21 and 8.2-22 .)

Figure 8.2-21

Figure 8.2-22

Thus, f changes concavity at x = –3.41421 and x = –0.585786.

Example 3

Find the points of inflection of f (x ) = x 3 – 6x 2 + 12x – 8 and determine the intervals where the function f is concave upward and where it is concave downward.

Step 1: Find f ′(x ) and f ″(x ).

f ′(x ) = 3x 2 – 12x + 12

f ″(x ) = 6x – 12

Step 2: Set f ″(x ) = 0.

6x – 12 = 0

x = 2

Note that f ″(x ) is defined for all real numbers.

Step 3: Determine intervals.

The intervals are (–∞, 2) and (2, ∞).

Step 4: Set up a table.

Since f (x ) has change of concavity at x = 2, the point (2, f (2)) is a point of inflection. f (2) = (2)3 – 6(2)2 + 12(2) – 8 = 0.

Step 5: Write a conclusion.

Thus, f (x ) is concave downward on (–∞, 2), concave upward on (2, ∞), and f (x ) has a point of inflection at (2, 0). (See Figure 8.2-23 .)

Figure 8.2-23

Example 4

Find the points of inflection of f (x ) = (x – 1)2/3 and determine the intervals where the function f is concave upward and where it is concave downward.

Step 1: Find f ′(x ) and f ″(x ).

Step 2: Find all values of x where f ″(x ) = 0 or f ″(x ) is undefined.

Note that f ″(x ) ≠ 0 and that f ″(1) is undefined.

Step 3: Determine intervals.

The intervals are (–∞, 1) and (1, ∞).

Step 4: Set up a table.

Note that since f (x ) has no change of concavity at x = 1, f does not have a point of inflection.

Step 5: Write a conclusion.

Therefore, f (x ) is concave downward on (–∞, ∞) and has no point of inflection. (See Figure 8.2-24 .)

Figure 8.2-24

Example 5

The graph of f is shown in Figure 8.2-25 and f is twice differentiable. Which of the following statements is true?

Figure 8.2-25

(A) f (5) < f ′(5) < f ″(5)

(B) f ″(5) < f ′(5) < f (5)

(C) f ′(5) < f (5) < f ″(5)

(D) f ′(5) < f ″(5) < f (5)

The graph indicates that (1) f (5) = 0; (2) f ′(5) < 0, since f is decreasing; and (3) f ″(5) > 0, since f is concave upward. Thus, f ′(5) < f (5) < f ″(5), choice (C).

• Move on. Do not linger on a problem too long. Make an educated guess. You can earn many more points from other problems.

8.3 Sketching the Graphs of Functions

Main Concepts: Graphing without Calculators, Graphing with Calculators

Graphing without Calculators

General Procedure for Sketching the Graph of a Function

Steps:

1. Determine the domain and if possible the range of the function f (x ).

2. Determine if the function has any symmetry, i.e., if the function is even (f (x ) = f (–x )), odd (f (x ) = –f (–x )), or periodic (f (x + p ) = f (x )).

3. Find f ′(x ) and f ″(x ).

4. Find all critical numbers (f ′(x ) = 0 or f ′(x ) is undefined) and possible points of inflection (f ″(x ) = 0 or f ″(x ) is undefined).

5. Using the numbers in Step 4, determine the intervals on which to analyze f (x ).

6. Set up a table using the intervals, to

(a) determine where f (x ) is increasing or decreasing.

(b) find relative and absolute extrema.

(c) find points of inflection.

(d) determine the concavity of f (x ) on each interval.

7. Find any horizontal, vertical, or slant asymptotes.

8. If necessary, find the x -intercepts, the y -intercepts, and a few selected points.

9. Sketch the graph.

Example 1

Sketch the graph of .

Step 1: Domain: all real numbers x ≠ ± 5.

Step 2: Symmetry: f (x ) is an even function (f (x ) = f (–x )); symmetrical with respect to the y -axis.

Step 3:

Step 4: Critical numbers:

f ′(x ) = 0 ⇒ –42x = 0 or x = 0

f ′(x ) is undefined at x = ±5 which are not in the domain.

Possible points of inflection:

f ″(x ) ≠ 0 and f ″(x ) is undefined at x = ±5 which are not in the domain.

Step 5: Determine intervals:

Intervals are (–∞, –5), (–5, 0), (0, 5), and (5, ∞).

Step 6: Set up a table:

Step 7: Vertical asymptote: x = 5 and x = –5

Horizontal asymptote: y = 1

Step 8:

x -intercept: (–2, 0) and (2, 0)

(See Figure 8.3-1 .)

Figure 8.3-1

Graphing with Calculators

Example 1

Using a calculator, sketch the graph of f (x ) = –x 5/3 + 3x 2/3 indicating all relative extrema, points of inflection, horizontal and vertical asymptotes, intervals where f (x ) is increasing or decreasing, and intervals where f (x ) is concave upward or downward.

1. Domain: all real numbers; Range: all real numbers

2. No symmetry

3. Relative maximum: (1.2, 2.03)

Relative minimum: (0, 0)

Points of inflection: (–0.6, 2.56)

4. No asymptote

5. f (x ) is decreasing on (–∞, 0], [1.2, ∞) and increasing on (0, 1.2).

6. Evaluating f ″(x ) on either side of the point of inflection (–0.6, 2.56)

f (x ) is concave upward on (–∞, –0.6) and concave downward on (–0.6, ∞). (See Figure 8.3-2 .)

Figure 8.3-2

Example 2

Using a calculator, sketch the graph of f (x ) = e x 2 /2 , indicating all relative minimum and maximum points; points of inflection; vertical and horizontal asymptotes; and intervals on which f (x ) is increasing, decreasing, concave upward, or concave downward.

1. Domain: all real numbers; Range (0, 1]

2. Symmetry: f (x ) is an even function, and thus is symmetrical with respect to the y -axis.

3. Relative maximum: (0, 1)

No relative minimum

Points of inflection: (–1, 0.6) and (1, 0.6)

4. y = 0 is a horizontal asymptote; no vertical asymptote.

5. f (x ) is increasing on (–∞, 0] and decreasing on [0, ∞).

6. f (x ) is concave upward on (–∞, –1) and (1, ∞); and concave downward on (–1, 1).

(See Figure 8.3-3 .)

Figure 8.3-3

• When evaluating a definite integral, you do not have to write a constant C , e.g., . Notice, no C .

8.4 Graphs of Derivatives

The functions f , f ′, and f ″ are interrelated, and so are their graphs. Therefore, you can usually infer from the graph of one of the three functions (f , f ′, or f ″) and obtain information about the other two. Here are some examples.

Example 1

The graph of a function f is shown in Figure 8.4-1 . Which of the following is true for f on (a , b )?

Figure 8.4-1

I. f ′ ≥ 0 on (a , b )

II. f ″ > 0 on (a , b )

Solution:

I. Since f is strictly increasing, f ′ ≥ 0 on (a , b ) is true.

II. The graph is concave downward on (a , 0) and upward on (0, b ). Thus, f ″ > 0 on (0, b ) only. Therefore, only statement I is true.

Example 2

Given the graph of f ′ in Figure 8.4-2 , find where the function f : (a) has its relative maximum(s) or relative minimums, (b) is increasing or decreasing, (c) has its point(s) of inflection, (d) is concave upward or downward, and (e) if f(–2) = f (2) = 1 and f (0) = –3, draw a sketch of f .

Figure 8.4-2

(a) Summarize the information of f ′ on a number line:

The function f has a relative maximum at x = –4 and at x = 4, and a relative minimum at x = 0.

(b) The function f is increasing on interval (–∞, –4] and [0, 4], and f is decreasing on [–4, 0] and [4, ∞).

(c) Summarize the information of f ″ on a number line:

A change of concavity occurs at x = –2 and at x = 2 and f ′ exists at x = –2 and at x = 2, which implies that there is a tangent line to the graph of f at x = –2 and at x = 2. Therefore, f has a point of inflection at x = –2 and at x = 2.

(d) The graph of f is concave upward on the interval (–2, 2) and concave downward on (–∞, –2) and (2, ∞).

(e) A sketch of the graph of f is shown in Figure 8.4-3 .

Figure 8.4-3

Example 3

Given the graph of f ′ in Figure 8.4-4 , find where the function f (a) has a horizontal tangent, (b) has its relative extrema, (c) is increasing or decreasing, (d) has a point of inflection, and (e) is concave upward or downward.

Figure 8.4-4

(a) f ′(x ) = 0 at x = –4, 2, 4, 8. Thus, f has a horizontal tangent at these values.

(b) Summarize the information of f ′ on a number line:

The First Derivative Test indicates that f has relative maximums at x = –4 and 4; and f has relative minimums at x = 2 and 8.

(c) The function f is increasing on (–∞, –4], [2, 4], and [8, ∞) and is decreasing on [–4, 2] and [4, 8].

(d) Summarize the information of f ″ on a number line:

A change of concavity occurs at x = –1, 3, and 6. Since f ′(x ) exists, f has a tangent at every point. Therefore, f has a point of inflection at x = –1, 3, and 6.

(e) The function f is concave upward on (–1, 3) and (6, ∞) and concave downward on (–∞, –1) and (3, 6).

Example 4

A function f is continuous on the interval [–4, 3] with f (–4) = 6 and f (3) = 2 and the following properties:

(a) Find the intervals on which f is increasing or decreasing.

(b) Find where f has its absolute extrema.

(c) Find where f has the points of inflection.

(d) Find the intervals where f is concave upward or downward.

(e) Sketch a possible graph of f .

Solution:

(a) The graph of f is increasing on [1, 3] since f ′ > 0 and decreasing on [–4, –2] and [–2, 1] since f ′ < 0.

(b) At x = –4, f (x ) = 6. The function decreases until x = 1 and increases back to 2 at x = 3. Thus, f has its absolute maximum at x = –4 and its absolute minimum at x = 1.

(c) A change of concavity occurs at x = –2, and since f ′(–2) = 0, which implies a tangent line exists at x = –2, f has a point of inflection at x = –2.

(d) The graph of f is concave upward on (–4, –2) and concave downward on (–2, 1) and (1, 3).

(e) A possible sketch of f is shown in Figure 8.4-5 .

Figure 8.4-5

Example 5

If f (x ) = |ln(x + 1)|, find . (See Figure 8.4-6 .)

Figure 8.4-6

The domain of f is (–1, ∞).

f (0) = |ln(0 + 1)| = |ln(1)| = 0

Therefore, .

8.5 Rapid Review

1. If f ′(x ) = x 2 – 4, find the intervals where f is decreasing. (See Figure 8.5-1 .)

Figure 8.5-1

Answer: Since f ′(x ) < 0 if –2 < x < 2, f is decreasing on (–2, 2).

2. If f ″(x ) = 2x – 6 and f ′ is continuous, find the values of x where f has a point of inflection. (See Figure 8.5-2 .)

Figure 8.5-2

Answer: Thus, f has a point of inflection at x = 3.

3. (See Figure 8.5-3 .) Find the values of x where f has change of concavity.

Figure 8.5-3

Answer: f has a change of concavity at x = 0. (See Figure 8.5-4 .)

Figure 8.5-4

4. (See Figure 8.5-5 .) Find the values of x where f has a relative minimum.

Figure 8.5-5

Answer: f has a relative minimum at x = –2. (See Figure 8.5-6 .)

Figure 8.5-6

5. (See Figure 8.5-7 .) Given f is twice differentiable, arrange f (10), f ′(10), f ″(10) from smallest to largest.

Figure 8.5-7

Answer: f (10) = 0, f ′(10) > 0 since f is increasing, and f ″(10) < 0 since f is concave downward. Thus, the order is f ″(10), f (10), f ′(10).

6. (See Figure 8.5-8 .) Find the values of x where f ′ is concave up.

Figure 8.5-8

Answer: f ′ is concave upward on (–∞, 0). (See Figure 8.5-9 .)

Figure 8.5-9

8.6 Practice Problems

Part A The use of a calculator is not allowed.

1 . If f (x ) = x 3x 2 – 2x , show that the hypotheses of Rolle’s Theorem are satisfied on the interval [–1, 2] and find all values of c that satisfy the conclusion of the theorem.

2 . Let f (x ) = e x . Show that the hypotheses of the Mean Value Theorem are satisfied on [0, 1] and find all values of c that satisfy the conclusion of the theorem.

3 . Determine the intervals in which the graph of is concave upward or downward.

4 . Given f (x ) = x + sin x 0 ≤ x ≤ 2π , find all points of inflection of f .

5 . Show that the absolute minimum of on [– 5, 5] is 0 and the absolute maximum is 5.

6 . Given the function f in Figure 8.6-1 , identify the points where:

(a) f ′ < 0 and f ″ > 0,

(b) f ′ < 0 and f ″ < 0,

(c) f ′ = 0,

(d) f ″ does not exist.

Figure 8.6-1

7 . Given the graph of f ″ in Figure 8.6-2 , determine the values of x at which the function f has a point of inflection.

Figure 8.6-2

8 . If f ″(x ) = x 2 (x + 3)(x – 5), find the values of x at which the graph of f has a change of concavity.

9 . The graph of f ′ on [–3, 3] is shown in Figure 8.6-3 . Find the values of x on [–3, 3] such that (a) f is increasing and (b) f is concave downward.

Figure 8.6-3

10 . The graph of f is shown in Figure 8.6-4 and f is twice differentiable. Which of the following has the largest value:

(a) f (–1)

(b) f ′(–1)

(c) f ″(–1)

(d) f (–1) and f ′(–1)

Figure 8.6-4

Sketch the graphs of the following functions indicating any relative and absolute extrema, points of inflection, intervals on which the function is increasing, decreasing, concave upward, or concave downward.

11 . f (x ) = x 4x 2

12 .

Part B Calculators are allowed.

13 . Given the graph of f ′ in Figure 8.6-5 , determine at which of the four values of x (x 1 , x 2 , x 3 , x 4 ) f has:

(a) the largest value,

(b) the smallest value,

(c) a point of inflection,

(d) and at which of the four values of x does f ″ have the largest value.

Figure 8.6-5

14 . Given the graph of f in Figure 8.6-6 , determine at which values of x is

(a) f ′(x ) = 0

(b) f ″(x ) = 0

(c) f ′ a decreasing function

Figure 8.6-6

15 . A function f is continuous on the interval [–2, 5] with f (–2) = 10 and f (5) = 6 and the following properties:

(a) Find the intervals on which f is increasing or decreasing.

(b) Find where f has its absolute extrema.

(c) Find where f has points of inflection.

(d) Find the intervals where f is concave upward or downward.

(e) Sketch a possible graph of f .

16 . Given the graph of f ′ in Figure 8.6-7 , find where the function f

(a) has its relative extrema.

(b) is increasing or decreasing.

(c) has its point(s) of inflection.

(d) is concave upward or downward.

(e) if f (0) = 1 and f (6) = 5, draw a sketch of f .

Figure 8.6-7

17 . If f (x ) = |x 2 – 6x – 7|, which of the following statements about f are true?

I. f has a relative maximum at x = 3.

II. f is differentiable at x = 7.

III. f has a point of inflection at x = –1.

18 . How many points of inflection does the graph of y = cos(x 2 ) have on the interval [–π , π ]?

Sketch the graphs of the following functions indicating any relative extrema, points of inflection, asymptotes, and intervals where the function is increasing, decreasing, concave upward, or concave downward.

19 . f (x ) = 3e x 2 /2

20 . f (x ) = cos x sin2 x [0, 2π ]

8.7 Cumulative Review Problems

(Calculator) indicates that calculators are permitted.

21 . Find if (x 2 + y 2 )2 = 10xy .

22 . Evaluate

23 . Find if y = cos(2x ) + 3x 2 – 1.

24 . (Calculator) Determine the value of k such that the function

is continuous

for all real numbers.

25 . A function f is continuous on the interval [–1, 4] with f (–1) = 0 and f (4) = 2 and the following properties:

(a) Find the intervals on which f is increasing or decreasing.

(b) Find where f has its absolute extrema.

(c) Find where f has points of inflection.

(d) Find intervals on which f is concave upward or downward.

(e) Sketch a possible graph of f .

8.8 Solutions to Practice Problems

Part A—The use of a calculator is not allowed.

1 . Condition 1: Since f (x ) is a polynomial, it is continuous on [–1, 2].

Condition 2: Also, f (x ) is differentiable on (–1, 2) because f ′(x ) = 3x 2 – 2x – 2 is defined for all numbers in [–1, 2].

Condition 3: f (–1) = f (2) = 0. Thus, f (x ) satisfies the hypotheses of Rolle’s Theorem which means there exists a c in [–1, 2] such that f ′(c ) = 0. Set f ′(x ) = 3x 2 – 2x – 2 = 0. Solve 3x 2 – 2x – 2 = 0, using the quadratic formula and obtain . Thus, x ≈ 1.215 or –0.549 and both values are in the interval (–1, 2). Therefore, .

2 . Condition 1: f (x ) = ex is continuous on [0, 1].

Condition 2: f (x ) is differentiable on (0, 1) since f ′(x ) = ex is defined for all numbers in [0, 1].

Thus, there exists a number c in [0, 1] such that . Set f ′(x ) = e x = (e – 1). Thus, e x = (e – 1). Take ln of both sides. ln(e x ) = ln(e – 1) ⇒ x = ln(e – 1).

Thus, x ≈ 0.541 which is in the interval (0, 1). Therefore, c = ln(e – 1).

3 .

Set f ″ > 0. Since (3x 2 + 25) > 0, ⇒ (x 2 – 25)3 > 0 ⇒ x 2 – 25 > 0, x < –5 or x > 5. Thus, f (x ) is concave upward on (–∞, –5) and (5, ∞) and concave downward on (–5, 5).

4 . Step 1: f (x ) = x + sin x ,
f ′(x ) = 1 + cos x ,
f ″ = – sin x .

Step 2: Set f ″(x ) = 0 ⇒ – sin x = 0 or
x = 0, π , 2π .

Step 3: Check intervals.

Step 4: Check for tangent line: At x = π , f ′(x ) = 1 + (–1) ⇒ 0 there is a
tangent line at x = π .

Step 5: Thus, (π , π ) is a point of inflection.

5 . Step 1: Rewrite f (x ) as
f (x ) = (25 – x 2 )1/2 .

Step 2:

Step 3: Find critical numbers. f ′(x ) = 0; at x = 0; and f ′(x ) is undefined at x = ±5.

Step 4:

f ′(0) = 0 and
(and f (0) = 5) ⇒ (0, 5) is a relative maximum. Since f (x ) is continuous on [–5, 5], f (x ) has both a maximum and a minimum value on [–5, 5] by the Extreme Value Theorem. And since the point (0,5) is the only relative extremum, it is an absolute extremum. Thus, (0,5) is an absolute maximum point and 5 is the maximum value. Now we check the endpoints, f (–5) = 0 and f (5) = 0. Therefore, (–5, 0) and (5, 0) are the lowest points for fon [–5, 5]. Thus, 0 is the absolute minimum value.

6 . (a) Point A f ′ < 0 ⇒ decreasing and f ″ > 0 ⇒ concave upward.

(b) Point E f ′ < 0 ⇒ decreasing and f ″ < 0 ⇒ concave downward.

(c) Points B and D f ′ = 0 ⇒ horizontal tangent.

(d) Point C f ″ does not exist ⇒ vertical tangent.

7 . A change in concavity ⇒ a point of inflection. At x = a , there is a change of concavity; f ″ goes from positive to negative ⇒ concavity changes from upward to downward. At x = c , there is a change of concavity; f ″ goes from negative to positive ⇒ concavity changes from downward to upward. Therefore, f has two points of inflection, one at x = a and the other at x = c .

8 . Set f ″(x ) = 0. Thus, x 2 (x + 3)(x – 5) = 0 ⇒ x = 0, x = –3 or x = 5. (See Figure 8.8-1 .)

Thus, f has a change of concavity at x = –3 and at x = 5.

Figure 8.8-1

9 . (See Figure 8.8-2 .)

Thus, f is increasing on [2, 3] and concave downward on (0, 1).

Figure 8.8-2

10 . The correct answer is (A).
f (–1) = 0; f ′(0) < 0 since f is decreasing and f ″(–1) < 0 since f is concave downward. Thus, f (–1) has the largest value.

11 . Step 1: Domain: all real numbers.

Step 2: Symmetry: Even function (f (x ) = f (– x )); symmetrical with respect to the y -axis.

Step 3: f ′(x ) = 4x 3 – 2x and f ″(x ) = 12x 2 – 2.

Step 4: Critical numbers: f ′(x ) is defined for all real numbers. Set f ′(x ) = 4x 3 – 2x = 0 ⇒ 2x (2x 2 – 1) = 0 ⇒ x = 0 or x = ± . Possible points of inflection: f ″(x ) is defined for all real numbers. Set f ″(x ) = 12x 2 – 2 = 0
x = ± .

Step 5: Determine intervals:

Intervals are:

,

, and

.

Since f ′(x ) is symmetrical with respect to the y -axis, you only need to examine half of the intervals.

Step 6: Set up a table (Table 8.8-1 ). The function has an absolute minimum value of (–1/4) and no absolute maximum value.

Table 8.8-1

Step 7: Sketch the graph. (See Figure 8.8-3 .)

Figure 8.8-3

12 . Step 1: Domain: all real numbers x ≠ 4.

Step 2: Symmetry: none.

Step 3: Find f ′ and f ″.

Step 4: Critical numbers: f ′(x ) ≠ 0 and f ′(x ) is undefined at x = 4.

Step 5: Determine intervals.

Intervals are (–∞, 4) and (4, ∞).

Step 6: Set up table as below:

Step 7: Horizontal asymptote:

. Thus, y = 1 is a horizontal asymptote. Vertical asymptote:

and

; Thus, x = 4 is a vertical asymptote.

Step 8: x -intercept: Set f ′(x ) = 0 ⇒ x + 4 = 0; x = –4. y -intercept: Set x = 0 ⇒ f (x ) = –1.

Step 9: Sketch the graph. (See Figure 8.8-4 .)

Figure 8.8-4

13 . (a)

The function f has the largest value (of the four choices) at x = x 1 . (See Figure 8.8-5 .)

Figure 8.8-5

(b) And f has the smallest value at x = x 4 .

(c)

A change of concavity occurs at x = x 3 , and f ′(x 3 ) exists which implies there is a tangent to f at x = x 3 . Thus, at x = x 3 , f has a point of inflection.

(d) The function f ″ represents the slope of the tangent to f ′. The slope of the tangent to f ′ is the largest at x = x 4 .

14 . (a) Since f ′(x ) represents the slope of the tangent, f ′(x ) = 0 at x = 0, and x = 5.

(b) At x = 2, f has a point of inflection which implies that if f ″(x ) exists, f ″(x ) = 0. Since f ′(x ) is differentiable for all numbers in the domain, f ″(x ) exists, and f ″(x ) = 0 at x = 2.

(c) Since the function f is concave downward on (2, ∞), f ″ < 0 on (2, ∞) which implies f ′ is decreasing on (2, ∞).

15 . (a) The function f is increasing on the intervals (–2, 1) and (3, 5) and decreasing on (1, 3).

(b) The absolute maximum occurs at x = 1, since it is a relative maximum, f (1) > f (–2) and f (5) < f (–2). Similarly, the absolute minimum occurs at x = 3, since it is a relative minimum, and f (3) < f (5) < f (–2).

(c) No point of inflection. (Note that at x = 3, f has a cusp.)

Note: Some textbooks define a point of inflection as a point where the concavity changes and do not require the existence of a tangent. In that case, at x = 3, f has a point of inflection.

(d) Concave upward on (3, 5) and concave downward on (–2, 3).

(e) A possible graph is shown in Figure 8.8-6 .

Figure 8.8-6

16 . (a)

The function f has its relative minimum at x = 0 and its relative maximum at x = 6.

(b) The function f is increasing on [0, 6] and decreasing on (– ∞, 0] and [6, ∞).

(c)

Since f ′(3) exists and a change of concavity occurs at x = 3, f has a point of inflection at x = 3.

(d) Concave upward on (–∞, 3) and downward on (3, ∞).

(e) Sketch a graph. (See Figure 8.8-7 .)

Figure 8.8-7

17 . (See Figure 8.8-8 .)

Figure 8.8-8

The graph of f indicates that a relative maximum occurs at x = 3, f is not differentiable at x = 7, since there is a cusp at x = 7 and f does not have a point of inflection at x = –1, since there is no tangent line at x = –1. Thus, only statement I is true.

18 . (See Figure 8.8-9 .)

Figure 8.8-9

Enter y 1 = cos(x 2 )

Using the [Inflection ] function of your calculator, you obtain three points of inflection on [0, π ]. The points of inflection occur at x = 1.35521, 2.1945, and 2.81373. Since y 1 = cos(x 2 ) is an even function, there is a total of 6 points of inflection on [–π , π ]. An alternate solution is to enter . (y 1 (x ), x , 2).

The graph of y 2 indicates that there are 6 zeros on [–π , π ].

19 . Enter y 1 = 3 ∗ e ∧ (–x ∧ 2/2). Note that the graph has a symmetry about the y -axis. Using the functions of the calculator, you will find:

(a) a relative maximum point at (0, 3), which is also the absolute maximum point;

(b) points of inflection at (–1, 1.819) and (1, 1.819);

(c) y = 0 (the x -axis) a horizontal asymptote;

(d) y 1 increasing on (–∞, 0] and decreasing on [0, ∞); and

(e) y 1 concave upward on (–∞, –1) and (1, ∞) and concave downward on (–1, 1). (See Figure 8.8-10 .)

Figure 8.8-10

20 . (See Figure 8.8-11 .) Enter y 1 = cos(x ) ∗ (sin(x )) ∧ 2. A fundamental domain of y 1 is [0, 2 π ]. Using the functions of the calculator, you will find:

Figure 8.8-11

(a) relative maximum points at (0.955, 0.385), (π , 0), and (5.328, 0.385), and relative minimum points at (2.186, – 0.385) and (4.097, – 0.385);

(b) points of inflection at (0.491, 0.196), , (2.651, – 0.196), (3.632, – 0.196), , and (5.792, 0.196);

(c) no asymptote;

(d) function is increasing on intervals (0, 0.955), (2.186, π ), and (4.097, 5.328), and decreasing on intervals (0.955, 2.186), (π , 4.097), and (5.328, 2π );

(e) function is concave upward on intervals (0, 0.491), , and (5.792, 2π ), and concave downward on the intervals , (2.651,3.632), and .

8.9 Solutions to Cumulative Review Problems

21 .

22 . Substituting x = 0 in the expression leads to , ana indeterminant form. Apply L’Hôpital’s Rule and you have or . Alternatively,

23 . y = cos(2x ) + 3x 2 – 1

24 . (Calculator) The function f is continuous everywhere for all values of k except possibly at x = 1. Checking with the three conditions of continuity at x = 1:

(1) f (1) = (1)2 – 1 = 0

(2) thus, 2 + k 0 ⇒ k = – 2. Since , therefore, .

(2) . Thus, k = –2.

25 . (a) Since f> 0 on (-1, 0) and (0, 2), the function f is increasing on the intervals [-1, 0] and [0, 2]. Since f< 0 on (2, 4), f is decreasing on [2, 4].

(b) The absolute maximum occurs at x = 2, since it is a relative maximum and it is the only relative extremum on (–1, 4). The absolute minimum occurs at x = –1, since f (–1) < f (4) and the function has no relative minimum on [–1, 4].

(c) A change of concavity occurs at x = 0. However, f ′(0) is undefined, which implies f may or may not have a tangent at x = 0. Thus, f may or may not have a point of inflection at x = 0.

(d) Concave upward on (–1, 0) and concave downward on (0, 4).

(e) A possible graph is shown in Figure 8.9-1 .

Figure 8.9-1

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