3.4 Concavity and the Second Derivative Test

In the past, one of the important uses of derivatives was as an aid in curve sketching. We usually use a calculator or a computer to draw complicated graphs, it is still important to understand the relationships between derivatives and graphs.

First derivative:

y is positive Curve is rising.

y is negative Curve is falling.

y is zero Possible local maximum or minimum.

Concavity of a Function

As you look at the graph of a function …  

… if the function CURVES UP, like a cup,

we say the function is _______________.

   …if the function CURVES DOWN, like a frown,

we say the function is _______________. 

CONCAVE UP

CONCAVE DOWN

++ – –

What do their eyes mean???

Second derivative:

y is positive Curve is concave up.

y is negative Curve is concave down.

y is zero Possible inflection point(where concavity changes).

++ – –

y is positive y is negative

Example 1Graph 23 23 4 1 2y x x x x

There are roots at and .1x 2x

23 6y x x

0ySet

20 3 6x x

20 2x x

0 2x x

0, 2x

First derivative test:

y0 2

0 0

21 3 1 6 1 3y negative

21 3 1 6 1 9y positive

23 3 3 6 3 9y positive

We can use a chart to organize our thoughts.

Possible local extrema at x = 0, 2.

Example 1Graph 23 23 4 1 2y x x x x

There are roots at and .1x 2x

23 6y x x

0ySet

20 3 6x x

20 2x x

0 2x x

0, 2x

First derivative test:

y0 2

0 0

maximum at 0x

minimum at 2x

Possible local extrema at x = 0, 2.

Local Maximum at x = 0

Local Minimum at x = 2

f ’’(x) = 6x – 6 = 6(x – 1)

f ’’(2) = 6(2 – 1)= 6> 0

f ’’(0) = 6(0 – 1) = –6<0

Theorem 3.7 Test for Concavity

Example 2Graph 23 23 4 1 2y x x x x

23 6y x x First derivative test:

y0 2

0 0

NOTE: On the AP Exam, it is not sufficient to simply draw the chart and write the answer. You must give a written explanation!

There is a local maximum at (0,4) because for all x in and for all x in (0,2) .

0y( ,0) 0y

There is a local minimum at (2,0) because for all x in(0,2) and for all x in .

0y(2, )0y

Because the second derivative atx = 0 is negative, the graph is concave down and therefore (0,4) is a local maximum.

Example 2Graph 23 23 4 1 2y x x x x

There are roots at and .1x 2x

23 6y x x

Or you could use the second derivative test:

6 6y x

0 6 0 6 6y

2 6 2 6 6y Because the second derivative atx = 2 is positive, the graph is concave up and therefore (2,0) is a local minimum.

Possible local extrema at x = 0, 2.

Example 3Graph

4

1)(

2

2

x

xxf

2222

22

)4(

10

)4(

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

x

x

x

xxxxxf

42

222

)4(

)2)(1)(2)(10()4)(10()(''

x

xxxxxf

There are one zero (x = 0) for f ’(x) = 0, and there are no the zeros for f ’’(x) = 0, but f(x) is not continuous at x = ±2.

Interval –∞ < x <–2 –2< x < 2 2 < x < +∞

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

0)4(

)43(1032

2

x

x

4

51lim

4

1lim

222

2

2 xx

xxx

4

51lim

4

1lim

222

2

2 xx

xxx

4

51lim

4

1lim

222

2

2 xx

xxx

4

51lim

4

1lim

222

2

2 xx

xxx

,4

1lim

2

2

2

x

xx 4

1lim

2

2

2

x

xx

do not exist.

Definition of Point of Inflection

Theorem 3.8 Points of Inflection

Can you give an example (or draw a sketch of a graph) for why the point of inflection could occur where f ’’(c) does not exist?

inflection point at 1x There is an inflection point at x = 1 because the second derivative changes from negative to positive.

Example 4Graph 23 23 4 1 2y x x x x

6 6y x

We then look for inflection points by setting the second derivative equal to zero.

0 6 6x

6 6x

1 x

Possible inflection point at .1x

y1

0

0 6 0 6 6y negative

2 6 2 6 6y positive

Make a summary table:

x y y y

1 0 9 12 rising, concave down

0 4 0 6 local max

1 2 3 0 falling, inflection point

2 0 0 6 local min

3 4 9 12 rising, concave up

Example 5 Determine the points of inflection and discuss the concavity of the graph of

34 4)( xxxf

Solution

)2(122412)('' 2 xxxxxf

Taking the 1st and 2nd derivative:

Setting f ’’(x) = 0 to find the zeros of f ’’(x) is x = 0 and x = 2:

Interval –∞ < x < 0 0< x < 2 2 < x < +∞

Test Value x = –1 x = 1 x = 3

Sign of f ’ ’(x)

f ’’(–1)> 0 f ’’(1) < 0 f ’’(3)> 0

Conclusion

Concave Up Concave down Concave Up

Points of inflection

(x^4-4x^3)/15

Theorem 3.9 Second Derivative Test

The second derivative can be used to perform a simple test for the relative min. and max. The test is based on f ’(c) = 0.

Example 6 Find the relative extrema for

Solution

Taking the 1st and 2nd derivative.

22 )4()2(8

1)( xxxf

)]4()2(2)4)(2(2[8

1)(' 22 xxxxxf

)4)(1)(2(2

1)]2()4)[(4)(2(

4

1 xxxxxxx

)45)(2(2

1 2 xxx

The zeros of 1st derivative are:

4 and ,1 ,2 xxx

Example 6

Point (–2, 0) (1, –81/8) (4, 0)

Sign of f ’’(x)

f ’’(–2)=–9 < 0 f ’’(1) = 9/2 > 0 f ’’(3)=–9 < 0

Conclusion

Relative Maximum

Relative Minimum Relative Maximum

)22(2

3)]52)(2(45[

2

1)('' 22 xxxxxxxf

]3)1[(2

3 2 xRelative Maximum

Relative Minimum

Setting f ’’(x) = 0 to find the zeros of f ’’(x):

31x 31 and x

Example 6

Find the points of inflection

Points of Inflection

Interval –∞< x < 1– 31/2 1– 31/2 < x < 1+ 31/2 1+ 31/2 < x < +∞

Sign of f ’’(x)

f ’’(x) < 0 f ’’(x) > 0 f ’’(x) < 0

Conclusion

Concave Down Concave Up Concave Down

Homework

Pg. 195 11-25 odds, 29, 35, 37, 48, 49-55 odds, 69