Copyright © 2016, 2012 Pearson Education, Inc. 2.2 - 1.

Copyright © 2016, 2012 Pearson Education, Inc. 2.2 - 1

2.2


Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs

OBJECTIVE• Find the relative extrema of a function using the Second-Derivative Test.

• Graph a continuous function in a manner that shows concavity.


DEFINITION:

Suppose that f is a function whose derivative f exists at every point in an open interval I. Then

f is concave up on I if f is concave down on I

f is increasing over I. if f is decreasing over I.

2.2 Using Second Derivatives to Find Maximum and Minimum Values and Sketch Graphs


THEOREM 4: A Test for Concavity

1. If f (x) > 0 on an interval I, then the graph of f is concave up. ( f is increasing, so f is turning upon I.)

2. If f (x) < 0 on an interval I, then the graph of f is concave down. ( f is decreasing, so f is turning down on I.)



THEOREM 5: The Second Derivative Test for Relative Extrema

Suppose that f is differentiable for every x in an open interval (a, b) and that there is a critical value c in (a, b) for which f (c) = 0. Then:

1. f (c) is a relative minimum if f (c) > 0.2. f (c) is a relative maximum if f (c) < 0.

For f (c) = 0, the First-Derivative Test can be used to determine whether f (c) is a relative extremum.



Example 1: Graph the function f given by

and find the relative extrema.

1st find f (x) and f (x).


2( ) 3 6 9,f x x x ( ) 6 6.f x x


Example 1 (continued): 2nd solve f (x) = 0.

Thus, x = –3 and x = 1 are critical values.


3 0 1 0or

3 1

x x

x x

( 3)( 1) 0x x

23 6 9 0x x


Example 1 (continued): 3rd use the Second Derivative Test with –3 and 1.

Lastly, find the values of f (x) at –3 and 1.

So, (–3, 14) is a relative maximum and (1, –18) is a relative minimum.


3 2(1) (1) 3(1) 9(1) 13 18f

3 2( 3) ( 3) 3( 3) 9( 3) 13 14f


Example 1 (concluded): Then, by calculating and plotting a few more points, we can make a sketch of f (x), as shown below.




Quick Check 1

Find the relative extrema of the function given by , and sketch the graph.

First find and :

g 3 5( ) 10 6g x x x

g x g x

2 430 30g x x x

360 120g x x x



Quick Check 1 Continued

Second, solve :

Thus , , and .

Therefore, there are critical values at , , and .

0g x

2 430 30 0g x x x 2 230 (1 ) 0x x

0x 1x 1x

0,0 1,4 1, 4




Third, use the Second Derivative Test with , , and :

, no extrema here, based on first derivative test.

, relative maximum here.

, relative minimum here.

Therefore there is a relative maximum at and a relative minimum at .

0 1 1

30 60(0) 120(0) 0g

3(1) 60(1) 120(1) 0g 3( 1) 60( 1) 120( 1) 0g

1,4 1, 4



Quick Check 1 Concluded

Using the information we gathered, we can plot : g x



THEOREM 6: Finding Points of Inflection

If f has a point of inflection, it must occur at a point x0, in the domain of f, where

or does not exist.0( ) 0f x 0( )f x



Quick Check 2

Determine the points of inflection for the function given by

We will use Theorem 6 (Finding Points of Inflection):

If a function has a point of inflection, it must occur at a point , where:

or does not exist.

3 5( ) 10 6g x x x

0x

0( ) 0f x 0( )f x




So, to find the point of inflection, we need to find where .

First, we need to find what is, then find where it equals 0.

If then,

Now we know that the points of inflection are when and . We can now solve for both, which gives us

, , and as the points of inflection.

( ) 0g x

( )g x3 5( ) 10 6g x x x ( ) ( ( ))g x g x

2 4(30 30 )x x 360 120x x 260 1 2x x

60 0x 21 2 0x

(0,0) 2,2.475

2

2, 2.475

2


Strategy for Sketching Graphs:

a) Derivatives and Domain. Find f (x) and f (x). Note the domain of f.

b) Critical values of f. Find the critical values by solving f (x) = 0 and finding where f (x) does not exist. These numbers yield candidates for relative maxima or minima. Find the function values at these points.

c) Increasing and/or decreasing; relative extrema. Substitute each critical value, x0, from step (b) into f (x). If f (x0) < 0, then f (x0) is a relative maximum



Strategy for Sketching Graphs (continued):

and f is increasing on the left of x0 and decreasing on the right. If f (x0) > 0, then f (x0) is a relative minimum and f is decreasing on the left of x0 and increasing on the right.

d) Inflection Points. Determine candidates for inflection points by finding where f (x) = 0 or where f (x) does not exist. Find the function values at these points.



Strategy for Sketching Graphs (concluded):

e) Concavity. Use the candidates for inflection points from step (d) to define intervals. Substitute test values into f (x) to determine where the graph is concave up ( f (x) > 0 ) and where it is concave down ( f (x) < 0 ).

f) Sketch the graph. Sketch the graph using the information from steps (a) – (e), calculating and plotting extra points as needed.



Example 2: Find the relative extrema of the function f given by

and sketch the graph.

a) Derivatives and Domain.

The domain of f is all real numbers.


( ) 6 .f x x

2( ) 3 3,f x x


Example 2 (continued): b) Critical values of f.

And we have f (–1) = 4 and f (1) = 0.


23 3x 23x

0

32x 1

x 1


Example 2 (continued): c) Increasing and/or Decreasing; relative extrema.

So (–1, 4) is a relative maximum, and f (x) is increasing on (–∞, –1] and decreasing on [–1, 1]. The graph is also concave down at the point (–1, 4).

So (1, 0) is a relative minimum, and f (x) is decreasing on [–1, 1] and increasing on [1, ∞). The graph is also concave up at the point (1, 0).



Example 2 (continued): d) Inflection Points.

And we have f (0) = 2.

e) Concavity. From step (c), we can conclude that f is concave down on the interval (–∞, 0) and concave up on (0, ∞).


x 0


Example 2 (concluded)

f) Sketch the graph. Using the points from steps (a) – (e), the graph follows.




Quick Check 3

Find the relative maxima and minima of the function given by and sketch the graph.

a.) Derivatives and Domain

The Domain of is all real numbers.

f

2 41 8f x x x

316 4f x x x

216 12f x x

f




b.) Find the critical values:

And we have , , and .

316 4 0x x

24 4 0x x 24 0x

2 4x 2 4x

2x

0 1f 2 17f 2 17f




c.) Increasing and/or decreasing; relative extrema.

So is a relative minimum, and is decreasing on and increasing on . Also, the graph is concave up at .

So is a relative maximum, and is increasing on and decreasing on . Also, the graph is concave down at .

So is a relative maximum, and is increasing on and decreasing on . Also, the graph is concave down at .

0 16 0f

0,1 2,0 f x 0,2 0,1

22 16 12(2) 16 48 0f

2,17 f x 0,2[2, ) 2,17

22 16 12( 2) 16 48 0f

2,17 f x ( , 2] 2,0 2,17




d.) Inflection Points.

e.) Concavity: From step c.) we can conclude that is concave down on the intervals and , and is concave up on the interval .

216 12 0x 212 16x

2 4

3x

12

3x

, 2 2, 2,2

f




We can graph from the points we have gathered: f x


Example 3: Graph the function f given by

List the coordinates of any extreme points and points of inflection. State where the function is increasing or decreasing, as well as where it is concave up or concave down.



Example 3 (continued)a) Derivatives and Domain.

The domain of f is all real numbers.



Example 3 (continued)b) Critical values. Since f (x) is never 0, the only critical value is where f (x) does not exist. Thus, we set its denominator equal to zero.

And, we have


2 33(2 5)x 2 3(2 5)x

2 5x 2x

x

0

5

5

2

00

5

2f

1 35

2 5 12

5

2f

0 1

5

2f

1


Example 3 (continued)c) Increasing and/or decreasing; relative extrema.

Since f (x) does not exist, the Second Derivative Test fails. Instead, we use the First Derivative Test.


5

2f

5 3

8

59 2 5

2

5

2f

8

9 08

0

5

2f


Example 3 (continued)c) Increasing and/or decreasing; relative extrema

(continued). Selecting 2 and 3 as test values on either side of

Since f’(x) is positive on both sides of is not an extremum.


2 3 2 3

2 2 2 2(3) 0

3(2 3 5) 3(1) 3 1 3f

2 3 2 3

2 2 2 2(2) 0

3(2 2 5) 3( 1) 3 1 3f


Example 3 (continued)d) Inflection points. Since f (x) is never 0, we only need to find where f (x) does not exist. And, since f (x) cannot exist where f (x) does not exist, we

know from step (b) that a possible inflection point is ( 1).



Example 3 (continued)e) Concavity. Again, using 2 and 3 as test points on

either side of

Thus, is a point of inflection.


5

3

8 8 8(3) 0

9 1 99(2 3 5)

f

5

3

8 8 8(2) 0

9 1 99(2 2 5)

f


Example 3 (concluded)f) Sketch the graph. Using the information in steps (a) – (e), the graph follows.




Section Summary

• The second derivative f is used to determine the concavity of the graph of function f.

• If f (x) > 0 for all x in an open interval I, then the graph of f is concave up over I.

• If f (x) < 0 for all x in an open interval I, then the graph of f is concave down over I.

• If c is a critical value and f (c) > 0, then f (c) is a relative minimum.

• If c is a critical value and f (c) < 0, then f (c) is a relative maximum.



Section Summary Concluded

• If c is a critical value and f (c) = 0, then the First-Derivative Test must be used to determine if f (c) is a relative extrema.

• If f (x0) = 0 or f (x0) does not exist, and there is a change in concavity to the left and to the right of x0, then the point( x0, f (x0) ) is called a point of inflection.

• Finding relative extrema, intervals over which a function is increasing or decreasing, intervals of upward or downward concavity, and points of inflection is all part of a strategy for accurate curve sketching.

Copyright © 2016, 2012 Pearson Education, Inc. 2.2 - 1.

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