Section 3.4

You have already seen that locating the intervals in which a function f increases or decreases helps to describe the graph. In Section 3.4 you will see how locating the intervals in which f ꞌ increases or decreases can be used to determine where the graph of f is curving upward (concave up) or curving downward (concave down)

Definition of ConcavityLet f be differentiable on an open interval I. The graph of f is concave upward on I if f ꞌ is increasing on the interval and concave downward on I if f ꞌ is deceasing on the interval.

f ꞌ

f ꞌꞌ

f is increasing from −∞ to −1.f ꞌ on this interval is positive and decreasing. By definition f is concave down.f ꞌꞌ on this interval is negative.

f ꞌ

f ꞌꞌ

f is decreasing from −1 to 1.f ꞌ on this interval is negative and decreasing until x = 0 where it is increasing.By definition f is concave down until x = 0 and then it becomes concave up.f ꞌꞌ on this interval is negative until x = 0 where it becomes positive.

f ꞌ

f ꞌꞌ

f is increasing from 1 to ∞.f ꞌ on this interval is positive and increasing.By definition f is concave up.f ꞌꞌ on this interval is positive.

This graphic representation may help when deciding on concavity.

Test for ConcavityLet f be a function whose second derivative exists on an open interval I.1. If f ꞌꞌ(x) > 0 for all x in I, then the graph of f

is concave upward in I.2. If f ꞌꞌ(x) < 0 for all x in I, then the graph of f

is concave downward in I.A linear equation has no concavity since its 2nd derivative is always 0.

Example 1Determine the open intervals on which the graph of

is concave upward or downward.f is continuous since the denominator can never be equal to 0.

2

63

f xx

126 3

f x x

22' 6 3 2

f x x x

22' 12 3

f x x x

22' 12 3

f x x x

2 32 2'' 12 3 12 2 3 2

f x x x x x

2 32 2 212 3 48 3

x x x

32 2 212 3 3 4

x x x

32 212 3 3 3

x x

2

32

36 1

3

x

x

2

32

36 10

3

x

x

236 1 0 x

2 1 0 x

1x

Use the intervals (-∞, -1), (-1, 1), and (1, ∞) to find where the graph is concave up or concave down.

-1 1

2

32

36 1''

3

xf x

x

+ − +

f is concave up on the intervals (-∞, -1); (1, ∞).f is concave down on the interval (-1, 1).

f ꞌꞌ

Concaveupward

Concavedownward

Concaveupward

Example 2Determine the open intervals on which the graph of

is concave upward or downward.

2

2

14

xf xx

2

2

14

xf xx

2 2

22

2 4 2 1'

4

x x x xf x

x

3 3

22

2 8 2 2'4

x x x xf xx

22

10

4

x

x

22

10'4

xf xx

22 2

42

10 4 10 2 4 2''

4

x x x xf x

x

2 2 2

42

4 10 4 4''

4

x x xf x

x

2

32

10 3 4''

4

xf x

x

2

32

10 3 40

4

x

x

There are no values of x that will make the 2nd derivative equal to 0 but at x = -2 and x = 2 the function is not continuous. Therefore the intervals will be (-∞, -2), (-2, 2), and (2, ∞).

-2 2

+ − +

f is concave up on the intervals (-∞, -2); (2, ∞).f is concave down on the interval (-2, 2).

2

32

10 3 4''

4

xf x

x

Concaveupward

Concavedownward

Concaveupward

Definition of Point of InflectionLet f be a function that is continuous on an open interval, let c be a point in the interval. If the graph of f has a tangent line at this point (c, f(c)), then this point is a point of inflection of the graph if the concavity of f changes from upward to downward (or downward to upward) at the point.In example 1 there were points of inflection at x = −1 and x = 1.

The concavity of f changes at a point of inflection. Note that a graph crosses its tangent line at a point of inflection.On the AP exam to change concavity f ꞌꞌ must change signs.

Points of InflectionIf (c, f(c)) is a point of inflection of the graph of f, then either f ꞌꞌ(c) = 0 or f ꞌꞌ does not exist at x = c.

Example 3Determine the points of inflection and discuss the concavity of the graph of

4 34 . f x x x

4 34 f x x x

3 2' 4 12 f x x x

2'' 12 24 12 2 f x x x x x

So the intervals are (-∞, 0), (0, 2), and (2, ∞).

0 2

+ − +

f is concave up on the intervals (-∞, 0); (2, ∞).f is concave down on the interval (0, 2).The points of inflection are (0, 0) and (2, −16)

2'' 12 24 12 2 f x x x x x

f ꞌꞌ

x

y

Concaveupward

Concavedownward

Concaveupward

The points of inflection are at (0, 0) and (2, -16).

It is possible for the 2nd derivative to be 0 at a point that is not a point of inflection.

For example f (x) = x4.f ꞌ(x) = 4x3

f ꞌꞌ(x) = 12x2

The critical # for f ꞌꞌ is 0.

The point (0, 0) is not a point of inflection.The graph is concave up on the intervals (−∞, 0) and (0, ∞).

+ +

0

f ꞌꞌ

Making Connections

f

f ꞌ

From x < −3.5 f ꞌ is positive and decreasing sof is increasing on this interval and concave down.At x = −3.5 there is a relative maximum.

f

f ꞌ

From −3.5 < x < −2.1 f ꞌ is negative and decreasing sof is decreasing on this interval and concave down.At x = −2.1 there is a point of inflection.

f

f ꞌ

From −2.1 < x < −1.3 f ꞌ is negative and increasing sof is decreasing on this interval and concave up.At x = −2.1 there is a relative minimum.

f

f ꞌ

From −1.3 < x < −0.3 f ꞌ is positive and increasing sof is increasing on this interval and concave up.At x = −0.3 there is a point of inflection.

f

f ꞌ

From −0.3 < x < 0f ꞌ is positive and decreasing sof is increasing on this interval and concave down.At x = 0 there is a relative maximum.

f

f ꞌ

From 0 < x < 0.7f ꞌ is negative and decreasing sof is decreasing on this interval and concave down.At x = 0.7 there is a point of inflection.

f

f ꞌ

From x > 1.5 f ꞌ is positive and increasing sof is increasing on this interval and concave up.

Given the graph of a derivative find where…a. f is increasing: (−∞, −3.5); (−1.3, 0); (1,5, ∞)b. f is decreasing: (−3.5, −1.3); (0, 1.5)c. relative maxima at: x = −3.5, 0d. relative minima at: x = −1.3, 1.5e. f is concave up: (−2.1, −0.3); (1.5, ∞)f. f is concave down: (−∞, −2.1); (−0.3, 1.5)g. f has points of inflection at: x = −2.1, −0.3, 1.5

f ꞌ

f ꞌꞌ

From x < −2.1 f ꞌ is decreasing sof ꞌꞌ is negative on this interval.f ꞌꞌ = 0 at x = −2.1.

f ꞌ

f ꞌꞌ

From −2.1 < x < −0.3 f ꞌ is increasing sof ꞌꞌ is positive on this interval.f ꞌꞌ = 0 at x = −0.3.

f ꞌ

f ꞌꞌ

From −0.3 < x < 0.7 f ꞌ is decreasing sof ꞌꞌ is negative on this interval.f ꞌꞌ = 0 at x = 0.7.

f ꞌ

f ꞌꞌ

From x > 0.7 f ꞌ is increasing sof ꞌꞌ is positive on this interval.

Each of the following columns is headed by the graph of a function f. Underneath this graph are three others that are, in random order, the graphs of f ꞌ, f ꞌꞌ, and a function F whose derivative is f, i.e., Fꞌ = f. Label each graph with the correct function.

f ꞌ

f is increasing 0 < x < 0.3 andx > 1f is decreasing 0.3 < x < 1f has a relative maximum at 0.3 and a relative minimum at 1

f ꞌ is positive 0 < x < 0.3 and x > 1f ꞌ is negative 0.3 < x < 1f ꞌ = 0 at x = 0.3, 1

f

f ꞌ

f ꞌ is decreasing 0 < x < 0.7f ꞌ is increasing x > 0.7f ꞌ has a relative minimum at x = 0.7

f ꞌꞌ is negative 0 < x < 0.7f ꞌꞌ is positive x > 0.7

f

f ꞌꞌ

f ꞌ

f is positive x > 0

F is increasing x > 0f

f ꞌꞌ

F

f ꞌ

f is decreasing 0.7 < x < 3.5

f ꞌ is negative 0.7 < x < 3.5

f

f ꞌ

f ꞌ is increasing 0.7 < x < 2f ꞌ is decreasing 2 < x < 3.5f ꞌ has a relative maximum at x = 2

f ꞌꞌ is positive 0.7 < x < 2f ꞌꞌ is negative 2 < x < 3.5

f

f ꞌꞌ

f ꞌ

f is positive 0.7 < x < 2f is negative 2 < x < 3.5

F is increasing 0.7 < x < 2F is decreasing 2 < x < 3.5

f

f ꞌꞌ

F

f ꞌ

f is increasing x < 4f is decreasing x > 4f has a relative maximum at x = 4

f ꞌ is positive x < 4f ꞌ is negative x > 4

f

f ꞌ

f ꞌ is increasing x < 3 and x > 5f ꞌ is decreasing 3 < x < 5f ꞌ has a relative maximum at x = 3f ꞌ has a relative minimum at x = 5

f ꞌꞌ is positive x < 3 and x > 5f ꞌꞌ is negative 3 < x < 5

f

f ꞌꞌ

f ꞌ

f is positive

F is increasing f

f ꞌꞌ

F

f ꞌ

f is increasing and linear

f ꞌ is positive and horizontal

f

f ꞌ

f ꞌ is horizontal with a slope of 0.

f ꞌꞌ is the x-axis.

f

f ꞌꞌ

f ꞌ

f is positive x > 2f is negative x < 2

F is increasing x > 2F is decreasing x < 2

f

f ꞌꞌ

F

In addition to testing for concavity, the 2nd derivative test can be used to perform a simple test for relative maxima and minima. The test is based on the fact that if the graph of a function f is concave upward on an open interval containing c, and f ꞌ(c) = 0, f (c) must be a relative minimum of f. Similarly, if the graph of a function f is concave downward on an open interval containing c, and f ꞌ(c) = 0, f (c) must be a relative maximum of f.

Second Derivative TestLet f be a function such that f ꞌ(c) = 0 and the second derivative of f exists on an open interval containing c.1. If f ꞌꞌ(c) > 0, then f has a relative minimum at

(c, f(c)).2. If f ꞌꞌ(c) < 0, then f has a relative maximum at

(c, f(c)).If f ꞌꞌ(c) = 0, the test fails. That is, f may have a relative minimum, relative maximum, or neither. In such cases, you can use the First Derivative Test.

Example 4Find the relative extrema for f(x) = -3x5 + 5x3.Then find the intervals that the graph is concave up and concave down. Finally find the points of inflection.

1. Find the critical numbers for f by finding the 1st derivative.f ꞌ(x) = -15x4 + 15x2

x = -1, 0, 12. Use the 2nd Derivative Test to find the relative

minimum or relative maximum.f ꞌꞌ(x) = -60x3 + 30x

0 1

+ = −

At x = -1 there is a relative minimum.At x = 0 we need to use the 1st Derivative Test.At x = 1 there is a relative maximum.

3'' 60 30( ) f x x x

-1

3'' 60 30( ) f x x x

2'' 30 2) 1( f x x x

3. Find the critical #s for the second derivative

10,2

x

0

+ −

3 2'' 60 3( 0) 30 2 1 f x x x x x

12

12

+ −

1 is concave up on the intervals , and 2

1 10, and concave down on the intevals , 02 2

1 and , .2

f