The slope of the slope is how fast the slope is changing.

3.4 Concavity and the 2nd Derivative Test

We learned how to locate intervals where a function is increasing or decreasing with the first derivative test. Locating intervals where f ’ increases or decreases determines where graph is curving upward or downward.

To find open intervals where the function is concave upward or downward, we need to find intervals on which f ‘ is increasing or decreasing.

So to use this test for concavity, we need to locate x-values where f “(x) = 0 or is undefined. These values separate the intervals. Finally, test the sign of the test values in these intervals.

Ex 1 p. 191 Determining Concavity

Determine the open intervals on which the graph of

is concave upward or downward. 2

3( )

6f x

x

Solution: The function is continuous on the real numbers (no domain problems!) Next, find the second derivative.

2 2'( ) 3( 6) (2 )f x x x 2 2

6

( 6)

x

x

2 2 2 1

2 4

( 6) ( 6) ( 6 )(2)( 6) (2 )''( )

( 6)

x x x xf x

x

2 2 2

2 4

( 6)( 6) ( 6) 4

( 6)

x x x

x

2 13( 6)x

2

2 3

( 6) 3 6

( 6)

x

x

Because f “ (x) = 0 at now we know interval borders

2x

2

2 3

(18) 2"( )

( 6)

xf x

x

Interval

Test Value

x = -3 x = 0 x =3

Sign of f ”(x)

f ”(-3) > 0

f ”(0) < 0 f ”(3) > 0

Conclusion

Concave up

Concave down

Concave up

( , 2) ( 2, 2) ( 2, )

1

0.5

5

(- 2, .375) ( 2, .375)

f x = 3

x2+6

Ex 2 p 192 Determining ConcavityDetermine intervals on which the graph of

is concave upward or downward.

2

2

1( )

4

xf x

x

Solution. Differentiate twice, find zeros or undefined values2 2

2 2 2 2

( 4)(2 ) ( 1)(2 ) (2 )( 5)'( )

( 4) ( 4)

x x x x xf x

x x

2 2

10

( 4)

x

x

2 2 2 1

2 4

( 4) ( 10) ( 10 )(2)( 4) (2 )"( )

( 4)

x x x xf x

x

2

2 3

10(3 4)

( 4)

x

x

"( ) 0 can't happen. The function is not continuous at x = 2f x

Use x = -2 and 2 to split intervals

2

2 3

10(3 4)"( )

( 4)

xf x

x

Intervals (-∞, -2) (-2, 2) (2, ∞)

Test value x = -3 x = 0 x = 3

Sign of f “ f “(-3) > 0 f “(0) < 0 f “(3) > 0

Conclusion Concave up

Concave down

Concave up

4

2

-2

-4

-5 5

concavedown

concave upconcave up

f x = x2+1

x2-4

Our last problem had two points of inflection, which can be thought of as points where the graph changes from concave upward to concave downward or vice versa.

Ex 3 p. 193 Points of Inflection

Determine points of inflection and discuss concavity of 4 3( ) 4f x x x

3 2'( ) 4 12f x x x 2"( ) 12 24 12 ( 2)f x x x x x

12 ( 2) 0x x when x=0, 2, so possible points of inflection are when x = 0, or 2

Interval (-∞, 0) (0, 2) (2, ∞)

Test value x =-1 x = 1 x = 3

Sign of f “ f “(-1) > 0 f “(1) < 0 f “(3) > 0

Conclusion Concave up

Concave down

Concave up

Points of inflection happen when x = 0, and 2

Ex 4. p195 Using the Second Derivative Test to find relative max or mins

Find the relative extrema of f(x) = -3x5 + 5x3

Find critical numbers first (what makes first derivative = 0) 4 2 2 2'( ) 15 15 15 ( 1)f x x x x x

0, 1,1x So critical numbers are

Using f “(x) = -60x3 + 30x, apply 2nd Derivative test.

Point on f(x)

(-1, -2) (1, 2) (0, 0)

Sign of f”(x)

f ” (-1) > 0 f “(1) < 0 f “(0) = 0

Conclusion Concave up so relative min

Concave down so relative max

Test fails

Looking on either side of x = 0, the first derivative is positive, so at x = 0 is neither a max or min.

2

2

1

-1

-2

-3

rel max at x = 1

test failedat x=0

Rel min at x = -1

f x = -3x5+5x3

To summarize:

Concavity •look for zeros of f” and see where it is zero or function is not continuous.•Set up intervals with those values•Test intervals – if f”>0 it is concave up. If f” < 0 it is concave down in interval

2nd Derivative test•Look for critical numbers of FIRST derivative•Evaluate SECOND derivative at those values•If f” > 0 then that critical number is relative min•If f” < 0 then that critical number is relative max•If f” = 0, then test fails and you’ll have to look at 1st Derivative test to determine max, min or neither.

You’ll need to get in your head the 1st der. Test and 2nd der. Test and how they are used!

3.4 Assign. P. 195/ 1-69 every other odd, 79-82