3.4 Concavity and the Second Derivative Test. In the past, one of the important uses of derivatives...

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