L16 Derivative Tests

7/25/2019 L16 Derivative Tests

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Derivative Tests

Mathematics 100

Institute of Mathematics

Math 100 (Inst. of Mathematics) Derivative Tests 1 / 34


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Outline

1 Increasing and Decreasing functions

2 Relative and Absolute Extrema

3 Concavity and Points of Inflection

4 Curve Sketching



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Intuitive Definition of Increasing and Decreasing Functions

On the interval(, 0),as the graph goes from left to right, thegraph is rising.

On the interval(0,+),as the graph goes from left to right, the

graph is falling.

The function isincreasingat (,0)anddecreasingat(0,+).



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Increasing and Decreasing Functions

Definition

A functionf isincreasingon an interval Iif for anyx1 andx2 in the

intervalI,

x1


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Illustration

x1 x2

y1

y2

x1< x

2

y1 < y2

x1 x2

y2

y1

x1< x

2

y1 > y2



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Application of the Derivative

Consider the tangent lines at the intervals where the function is

increasing:

Remark

When the graph is increasing on an interval I, then for any x I,theslope of the tangent line at that point is positive.



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decreasing:

Remark

When the graph is decreasing on an interval I, then for any x I,theslope of the tangent line at that point is negative.



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constant.

a b

Remark

At the interval(a,b), where the graph is neither increasing nordecreasing, the tangent line is the horizontal line (slope is zero).



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Theorem

Theorem

Letfbe differentiable on an interval(a,b).

1. Iff(x)> 0 for allx (a,b)thenfis increasing on(a, b).

2. Iff(x)< 0 for allx (a,b)thenfis decreasing on(a,b).

3. Iff(x) =0 for all x (a,b)thenfis constant on(a,b).



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Critical Numbers

Remark

For a continuous functionf,f(x)can change sign only atx -values

wheref(x) =0 or is undefined.

We call thesex-values thecritical numbersof f.



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Examples

1. Find the intervals on whichf(x) =x3

3

2x2

is increasing ordecreasing.

Solution:

We need to determine the intervals when f(x)is positive and negative.

To determine these intervals, find the critical number(s). (values of xwhenf(x) =0.

f(x) = 3x2 3x

0 = 3x

2

3x0 = 3x(x 1)

x = 0, 1



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Examples

Solution (Contd):Determine the sign off(x)(notf(x)) for the intervals(,0),(0, 1),and(1,+).

(,0) (0,1) (1,+)x + +

x 1 +

f(x) =3x(x 1) + +

Hence,fis increasing at the interval(,0) (1,+)anddecreasing at the interval(0, 1).



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Relative and Absolute Extrema

Definition

Letfbe a function defined at c.

1. f(c)is called arelative maximumof fif there exists an interval(a,b)containingcsuch that for allx (a,b)

f(x) f(c)

2. f(c)is called arelative minimumof fif there exists an interval(a,b)containingcsuch that for allx (a,b)

f(x) f(c)

3. f(c)is anabsolute maximumoff iff(x) f(c)for everyx domf.f(c)is anabsolute minimumof f iff(x) f(c)for everyx domf.



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Illustration

relative minimum

relative maximum

relative minimum

absolute minimumno absolute maximum


FDT f R l i E


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FDT for Relative Extrema

First Derivative Test for Relative Extrema:

Letfbe continuous on the interval(a,b)in whichcis the only criticalnumber. Iffis differentiable on the interval (except possibly at c), then

1. Iff(x)> 0 forxc, thenf(c)

is a relative maximum.

2. Iff(x)< 0 forx 0 forx>c, thenf(c)is a relative minimum.

3. Iff(x)> 0 forx (a,b)such thatx =candf(x)


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Examples

1. Find all relative extrema of the function

f(x) =2x3 3x2 36x+14.

Solution:

Look for values of xwhere the sign of the first derivative changes.

Find the critical numbers: Set the derivative of f(x)equal to zero.

f(x) = 6x2 6x 36

= 6(x2 x 6)

= 6(x 3)(x+2) =0 x=3, 2

Sincef is defined for all real numbers, the only critical numbers of f arex= 2 andx=3.


f ( ) 2 3 3 2 36 14


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f(x) =2x3 3x2 36x+14

Solution (Cont.):

(, 2) 2 (2,3) 3 (3,+)

f(x) 58 67

x+2 0 + + +x 3 0 +

f(x) =6(x+2)(x 3) + 0 0 +Using the First Derivative Test, f(2) =58 is a relative maximum andf(3) = 67 is a relative minimum.


E l


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Examples

2. Find all relative extrema of the function f(x) =x4 4x3.

Solution:

f(x) =4x3 12x2 =4x2(x 3) =0 x=0,3

(,0) 0 (0,3) 3 (3,+)f(x) 0 27

x2 + 0 + + +

x 3 0 +

f

(x) =4x2

(x 3) 0 0 +Atx=0,f does not change sign from to+, vice versa. Hence theonly extremum isf(3) = 27 which is a relative minimum.


SDT for Relative Extrema


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SDT for Relative Extrema

Second Derivative Test for Relative Extrema

Supposef exists on some open interval containing candf(c) =0.

1. Iff(c)> 0, thenf(c)is a relative minimum.2. Iff(c)< 0, thenf(c)is a relative maximum.

3. Iff(c) =0, then the test fails.


Illustration


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Illustration

We use the second derivative test in the previous examples to

determine the relative extrema.

Example 1

Find the relative extrema off(x) =2x3 3x2 36x+14.

We need to find the values csuch thatf

(c) =0.(the criticalnumbers).x= 2 andx=3(Verify!)

Find the second derivative:

f(x) =12x 6

Evaluate the second derivative at the critical numbers:

f(2) =12(2) 6= 30< 0 relative maximumf(3) =12(3) 6= 30>0 relative minimum


Illustration


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Illustration

Example 2

Find the relative extrema off(x) =x4 4x3.

We need to find the values csuch thatf(c) =0.(the criticalnumbers).x=0 andx=3(Verify!)

Find the second derivative:f(x) =12x2 24x

Evaluate the second derivative at the critical numbers:

f(3) =12(3)2 24(3) =36> 0 relative minimum

f

(0) =12(0) 24(0) =0 test failsApply the FDT forx=0.


Definitions


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Definitions

Concavity

Letfbe differentiable on an open intervalI. The graph off is

1. concave upwardonI iff is increasing on the interval.

2. concave downwardonI iff is decreasing on the interval.


Definitions


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Definitions

concave upward

f is increasingconcave downward

f is decreasing

1. A curve that lies above its tangent lines is concave upward.

2. A curve that lies below its tangent lines is concave downward.


Definitions


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Definitions

Letfbe a function whose second derivative exists on an open interval I.

1. Iff(x)> 0 for allx I, then the graph of f is concave upward on I.

2. Iff(x)< 0 for allx I, then the graph of f is concave downward onI.


Definitions


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Definitions

For a continuous functionfwe can find the intervals on which thegraph is concave upward and concave downward through these steps:

a) Find the second derivative.

b) Locate thexvalues at whichf(x) =0 or undefined.

c) Use thexvalues to determine the test intervals.d) Test the sign of f(x)in each of the test intervals.

Point of Inflection

The point wherein the concavity changes is called apoint of

inflection.


Example


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Example

Determine the points of inflection and discuss the concavity of the

graph off(x) =x4 +x3 3x2 +1.

Solution:

Differentiating twice, we have

f(x) =12x2 +6x 6= 6(2x2 +x 1) =6(2x 1)(x+1)

By settingf(x) =0, we see the that the possible points of inflection

occur atx= 1 andx=1/2.


Example


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Example

Solution (Cont.):

(, 1) 1

1, 12

1

2

12, +

f(x) 2 7

16

2x 1 0 +

x+1 0 + + +f(x) =6(2x 1)(x+1) + 0 0 +

Hence the graph is concave up on the interval(, 1)

1

2, +

and is

concave down on

1,1

2


Curve Sketching (Examples)


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1. Sketch the graph off(x) =x3 9x2 +15x 12.

Solution:

Solve the first and second derivatives:

f

(x) =3x2 18x+ 15= 3(x2 6x+ 5) =3(x 1)(x 5) =0 x=1, 5

f(x) =6x 18= 6(x 3) =0 x=3

f(1) = 5 andf(5) = 37 hence(1, 5)and (5, 37)are the criticalpoints of the graph of f.

f(3) = 21, so(3, 21)is a possible point of inflection.




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Solution (Cont.):

(, 1) 1 (1, 3) 3 (3, 5) 5 (5,+)f(x) 5 21 37

x 1 0 + + + + +x 5 0 +

f(x) = 3(x1)(x5) + 0 0 +

f(x) = 6(x3) 0 + + +

graph of f(x) inc rel dec point of dec rel incconc down max conc down inflection conc up min conc up




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g ( p )

Solution (Cont.):

Graph:

3 2 1 1 2 3 4 5 6 7 8 9 10

40

35

30

25

20

15

10

5

5

10

0




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g ( p )

2. Sketch the graph off(x) = x3 +3x2 2.

Solution:

Derivatives:

f

(x) = 3x2 +6x=3x(x+2) =0 x=0, 2

f(x) = 6x+6= 6(x+1) =0 x=1

f(0) = 2 andf(2) =2, hence(0, 2)and (2, 2)are the critical points ofthe graph off.

Evaluating f atx=1, we get 0, so(1, 0)is a possible point of inflection.




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g ( p )

Solution (Cont.):

(,0) 0 (0, 1) 1 (1, 2) 2 (2,+)

f(x) 2 0 2

x 0 + + + + +

2 x + + + + + 0

f(x) =3x(2 x) 0 + + + 0

f(x) =6(1 x) + + + 0

graph of f(x) dec rel inc point of inc rel dec

conc up min conc up inflection conc down max conc down




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g ( p )

Solution (Cont.):

Graph:

1 1 2 3

2

1

1

2

3

0


Lecture Exercises


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1 Determine the interval/s over which the graph of the function

f(x) =e2x 3ex +x 2 is increasing/decreasing.

2 Sketch the graphs of the following functions.

a. f(x) =x3 3x2 +4b. f(x) =x3 6x2 +9x 3

c. f(x) = 3

4x4 4x3 +6x2

d. f(x) = x5 +5x


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