CHAPTER 3SECTION 3.4

CONCAVITY AND THE SECOND DERIVATIVE

TEST

Definition of Concavity and Figure 3.24

Sketch 4 graphs a)1 decreasing and concave upb)1 increasing and concave up, c)1 decreasing and concave down,

d)1 increasing and concave down

x

y

x

y

x

y

x

y

a b

cd

x x

x x

y y

y y

Concave upward

• Look at these two graphs. Each is concave upward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph. For each of the graphs above sketch the tangent lines to the graph at a number of different points.

x

y

x

y

x x

y y

Concave upward

• As we move from left to right the slopes of the tangent lines are getting less negative. That is they are increasing.x

y

x

y

Concave upward

• As we move from left to right the slopes of the tangent lines are getting larger. That is they are increasing.x

y

When a graph is concave upward

The slope of the tangent lines are increasing.

Concave downward

• Look at these two graphs. Each is concave downward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph. For each of the graphs above sketch the tangent lines to the graph at a number of different points.

x

y

x

y

x x

y y

Concave downward

• As we move from left to right, the slopes of the tangent lines are getting more negative.

• They are decreasing.x

y

Concave downward

• As we move from left to right the slopes of the tangent lines are getting smaller. That is they are decreasing.

x

y

When a graph is concave downward

The slopes of the tangent lines are decreasing.

Putting it all together

• For a function f that is differentiable on an interval I, the graph of f is

• (i) Concave up on I, if the slope of the tangent line is increasing on I or

• (ii) Concave down on I, if the slope of the tangent line is decreasing on I

Linking knowledge

• (i) Concave up on I, if the slope of the tangent line is increasing on I.

• If the slope of the tangent line is increasing and the slope of the tangent line is represented by the first derivative and to determine when something is increasing we had to take the derivative, then to find where the slope of the tangent line (f ‘(x)) is increasing we will need to take the derivative of f ‘(x) or find the second derivative f “(x)

I know, this is a very large run on sentence.

Linking knowledge

• (ii) Concave down on I, if the slope of the tangent line is decreasing on I

• If the slope of the tangent line is decreasing and the slope of the tangent line is represented by the first derivative and to determine when something is decreasing we had to take the derivative, then to find where the slope of the tangent line (f ‘(x)) is decreasing we will need to take the derivative of

• f ‘(x) or find the second derivative f “(x)

Definition of concavity

• For a function f that is differentiable on an interval I, the graph of f is

• (i) Concave up on I, if f’ is increasing on I or• (ii) Concave down on I, if f’ is decreasing on I

Theorem 3.7Test for concavity

I.in downward concave is ofgraph e th then I,in allfor 0)( If 2.

I.in upward concave is ofgraph e th then I,in allfor 0)( If 1.

I. intervalopen an on exists derivative second hosefunction w a be fLet

fxxf

fxxf

Putting it all together

• Given the function f(x)• f(x) = 0 x-intercepts• f(x) undefined vertical asymptote• f(x)>0 Q-1 or Q-2• f(x)<0 Q-3 or Q-4

Putting it all together

decreasing is 0)(increasing is 0)(

asymptote) tical ver orpt ner number(cor critical undefined )(

number critical 0)()( Find

fxffxf

xfxf

xf

Putting it all together

downward concave is decreasing is 0)(

upward concave is ;increasing is 0)(

??????? undefined )(number cal"hypercriti" 0)(

)( Find

ffxfffxf

xfxf

xf

Determining concavity

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

• Concavity find second derivative.

• Find hypercritical numbers.

• Set up a chart• Find concavity

36)( 2

x

xf

32

2

42

222

22

22

12

3136)(

3)12)(2)(3(2312)(

312)(

)2()3(6)(

36)(

xxxf

xxxxxxf

xxxf

xxxf

xxf

c = 1; c = -1 and f” is defined on the entire # line

32

2

3136)(

xxxf Setting up the chart

interval Test points

Sign of f” f ‘ concave

(-∞, -1) -2 + inc upward

(-1,1) 0 - dec downward

(1,∞) 2 + inc upward

Points of inflection

• A point of inflection for the graph of f is that point where the concavity changes.

Theorem 3.7 Test for Concavity

Definition of Point of Inflection and Figure 3.28

Theorem 3.8 Points of Inflection

Theorem 3.9 Second Derivative Test and Figure 3.31

• Example 1: Graph the function f given by

• and find the relative extrema.• 1st find graph the function.

f (x) x3 3x2 9x 13,

• Example 1 (continued): • 2nd solve f (x) = 0.

• Thus, x = –3 and x = 1 are critical values.

3x2 6x 9 0x2 2x 3 0

(x 3)(x 1) 0

x 3 0x 3

orx 1 0

x 1

• Example 1 (continued): • 3rd use the Second Derivative Test with –3 and 1.

• Lastly, find the values of f (x) at –3 and 1.

• So, (–3, 14) is a relative maximum and (1, –18) is a • relative minimum.f ( 3) ( 3)3 3( 3)2 9( 3) 13 14

f (1) (1)3 3(1)2 9(1) 13 18

f ( 3) 6( 3) 6 18 6 12 0 : Relative maximum

f (1) 6(1) 6 6 6 12 0 : Relative minimum

Second Derivative Testa. If f’’(c) > 0 then ________________________

If c is a critical number of f’(x) and…

b. If f’’(c) < 0 then ________________________

c. If f’’(c) = 0 or undefined then __________________________________

Second Derivative Testa. If f’’(c) > 0 then ________________________ (c, f(c)) is a relative min

If c is a critical number of f’(x) and…

b. If f’’(c) < 0 then ________________________

c. If f’’(c) = 0 or undefined then __________________________________

(c, f(c)) is a relative max

the test fails (use 1st Derivative test)

The second derivative gives the same information about the first derivative that the first derivative gives about the original function.

If f’’(x) > 0 ______________ If f’’(x) < 0 ______________ If f’’(x) = 0 ______________

Concave upward Concave downward

Inflection Points

If f’(x) > 0 ______________ If f’(x) < 0 ______________ If f’(x) = 0 ______________

______________ ______________ ________________

For f(x) to increase, _____________

For f’(x) to increase, _____________

For f(x) to decrease, _____________

For f’(x) to decrease, _____________

The second derivative gives the same information about the first derivative that the first derivative gives about the original function.

If f’’(x) > 0 ______________ If f’’(x) < 0 ______________ If f’’(x) = 0 ______________

Concave upwardSlopes increase

Concave downwardSlopes decrease

f(x) increases f(x) decreases f(x) is constant

Inflection Points Where concavity changes

Occur at critical numbers of f”(x)

If f’(x) > 0 ______________ If f’(x) < 0 ______________ If f’(x) = 0 ______________

f’(x) decreasesf’(x) increases

f(x) is conc up f(x) is conc down

f’(x) is constant

f(x) is a straight line______________ ______________ ________________

For f(x) to increase, _____________

For f’(x) to increase, _____________

For f(x) to decrease, _____________

For f’(x) to decrease, _____________

f’(x) > 0

f’’(x) > 0

f’(x) < 0

f’’(x) < 0

Sketch 126 3f x x

Include extrema, inflection points, and intervals of concavity.

Sketch 126 3f x x

22' 6 3 2f x x x

22

12

3

x

x

Critical numbers: 12 0x 2 3 0x 0x

No VA’s

smooth

None

Include extrema, inflection points, and intervals of concavity.

22 2

42

3 12 12 2 3 2''

3

x x x xf x

x

2 2 2

42

12 3 3 4

3

x x x

x

2

32

12 3 3

3

x

x

2

32

36 1

3

x

x

Critical numbers: 2 1 0x 2 3 0x 1x None

1. Find the extrema of 2sin cos2 on 0,2f x x x

1. Find the extrema of 2sin cos2 on 0,2f x x x

' 2cos 2sin2f x x x Crit numbers: 0 2cos 2sin2x x

'' 2sin 4cos2f x x x

0 2cos 2 2sin cosx x x 0 2cos 1 2sinx x

0 2cos 0 1 2sinx x 120 cos sinx x

3 52 2 6 6, ,x x

2'' 2f

32'' 6f

56'' 3f

6'' 3f 326f

2 1f

32 3f

25 36f

rel min at

rel max at

rel min at

rel max at

36 2,

2 ,1

5 36 2,

32 , 3

2nd Derivative Test

2. Sketch 22sin on 0,4xf x

2. Sketch 22sin on 0,4xf x

2' cos xf x Crit numbers:

,3 ,5 ,7x

12'' 3f

12''f 2f

23f rel min at

rel max at ,2

3 , 2

2nd Derivative Test

20 cos x3

2 2 2,x 5 72 2,

12 2'' sin xf x Crit numbers: 20 sin x

2 ,2 ,3 ,4x 2 ,4 ,6 ,8x Intervals: 0, 2 ,4

Test values: 3

Inf pt

2x

2 0f

2 ,0

f ’’(test pt) f(x) down up

2sin xf x

2

-2

24

Find a Function

Describe the function at the point x=3 based on the following:

4)3( f

0)3( f

6)3( f

3

(3, 4)

Find a Function

Describe the function at the point x=5 based on the following:

0)5( f

0)5( f

0)5( f 5

Find a Function

Given the function is continuous at the point x=2, sketch a graph based on the following:

3)2( fDNEf )2(

20)(20)(20)(

xallforxfxforxfxforxf

2

(2,3)

WHY? BECAUSE f’(x) is POSITVE!!!!!!!!!!!!!!!