Concavity and the Second Derivative

Test Determine the intervals on which the graphs of functions are concave upward or concave downward.Find the points of inflection of the graphs of functions.Use the Second Derivative Test to find the relative extrema of functions.

Standard 4.5a

Concavity – the property of curving upward or downward

Concave upward

f’ is increasing

Concave downward

f’ is decreasing

Definition of Concavity

Let f be differentiable on an open interval I. The graph of f is

1. Concave upward on I if f’ is increasing on the interval.

2. Concave downward on I if f’ is decreasing on the interval.

Test for Concavity

Let f be a function whose second derivative exists on an open interval I.

1. If f ´´(x) > 0 for all x in I, then f is concave upward on I.

2. If f ´´(x) < 0 for all x in I, then f is concave downward on I.

Determine the intervals on which the graph is concave upward or concave downward.

1. Locate the x-values at which f ´´(x) = 0 or f ´´(x) is undefined.

2. Use these x-values to determine the test intervals.

3. Test the signs of f ´´(x) in each test interval.

Interval (-∞, -√3) (-√3, √3) (√3, ∞)Test Values x = -2 x = 0 x = 2

Sign of f ´´(x)

f ´´(-2) > 0 f ´´(0) < 0 f ´´(2) > 0

Conclusion Concave upward

Concave downward

Concave upward

Defintion of Point of Inflection

If the graph of a continuous function has a tangent line at point where the concavity changes from upward to downward (or vice versa) then the point is a point of inflection.

Property of Points of Inflection

If (c, f(c)) is a point of inflection of the graph of f, then either f ´´(c) = 0 or f ´´(c) is undefined at c.

Find the points of inflection of the graph.

Possible inflection point

(-∞, 4) (4, ∞)x = 0 x = 5

f ´´(0) < 0 f ´´(5) > 0Concave

downConcave up

Inflection point (4, 16)

It is possible for the second derivative to be zero at a point that is not a point of inflection.

* You must test to be certain that the concavity actually changes.

Find the points of inflection and discuss the concavity of the graph of the function.

Possible points of inflections: x = 0, x = 3

(0, 3) (3, ∞) x = 1 x = 4

f ´´(1) > 0 f ´´(4) < 0Concave upward Concave

downward

Inflection Point:

cIf f ´(c) = 0 and f ´´(c) >

0, f (c) is a relative minimum

Concave Upward

f ´´(c) > 0

cIf f ´(c) = 0 and f ´´(c) > 0, f (c) is a

relative minimum

Concave downward

f ´´(c) < 0

Second – Derivative Test

Let f ´(c) = 0 and let f ´´exist on an open interval containing c.

1. If f ´´(c) > 0, then f(c) is a relative minimum.2. If f ´´(c) < 0, then f(c) is a relative maximum.3. If f ´´(c) = 0 then the test fails. Use the First

Derivative Test.

Find the relative extrema using the Second-Derivative Test

1. Find the critical numbers.

Critical Numbers

3. Plug the critical numbers into the second derivative to determine relative extrema.

2. Find the second derivative.

Relative minimum (-1, 2)

Relative maximum (1, 2)Test fails

(-1, 0) (0, 1)f ´(-1/2)

>0f ´(1/2)

>0

Increasing

Increasing

(0,0) is neither a relative max or a

relative min

Find the relative extrema using the Second-Derivative Test.

Test fails

Relative min (3,-26 )

(-∞, 0) (0, 3)f ´(-1) < 0 f ´(1) < 0

Decreasing Decreasing

(0,1) is neither a relative max or a

relative min