Module 2 Examples

8/12/2019 Module 2 Examples

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CS 546: Module 2

Spring 2014


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Outline

Horizontal and Vertical Asymptotes

Local minimum/maximum

Inflection points Antiderivative/Integral

Definite Integral


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definition: horizontal asymptote


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example

Here we see that as x

approaches negative

infinity and positive

infinity y approaches

0.


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recipe for finding

horizontal asymptotes

(1) Find the limit of the function as x approaches infinity

(2) Find the limit of the function as x approaches negative infinity


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recipe for finding the limit of afractional polynomial as thevariable approaches infinity


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recipe for finding the horizontalasymptote of a fractional

polynomial as the variableapproaches infinity


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example


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recipe for finding the limit atx approaches negative infinity

(1) Replace x with (-x)

(2) Find the limit as x goes to positive infinity instead


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1

2limlim

2

x

xy xx

0

example

1)(

2)(lim 2

x

xx

1

21lim

2

x

xx


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)1(

25limlim

3

3

x

xy xx

515

example

125lim

3

3

xxx


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)1(

25limlim

3

3

x

xy xx

5

example

)1)((2)(5lim 3

3

xxx

)1(

25lim

3

3

x

x

x

1

25lim

3

3

x

xx


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definition: vertical asymptote


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example

1x


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recipe for findinghorizontal asymptotes

Find the points where the function has a restriction on

the domain


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example

1x

When x = 1, the function does notexist, thus there is a restriction on

the domain when x = 1.


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Derivatives

Last weeks module covered derivatives. What

can we use derivatives for?

graphing aid

finding critical points

Local and global maxima

Local and global minima

Points of inflection


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example

xxh 2)(' 2)( xxh

x

)(xh

002)0(' xh

212)1(' xh

422)2(' xh

x h(x) h(x)

-2 4 -4

-1 1 -2

0 0 0

1 1 2

2 4 4

By looking at the derivative, h(x), we can see that h(x) is decreasing when x is less

than 0 because the derivative is negative. We can also see that h(x) is increasing when

x is greater than 0 because the derivative is positive.


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0x

5.1x

2.2x

8.2x

maximum

minimum

maximum

inflection point

In this plot we see that

there are two local maxima

and one local minimum.

Notice that right before the

local maximum the function

is increasing (the derivative

is positive), and right after

the local maximum thefunction is decreasing (the

derivative is negative).

Notice that right before the

local minimum the function

is decreasing (the derivativeis negative), and right after

the local minimum the

function is increasing (the

derivative is positive).

There is also one inflectionpoint.


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definition: inflection point


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finding points of inflection

If a function has an inflection point, then its second derivative

equals 0 at this point. Thus, inflection points are among the points

at which the second derivative equals 0. So, to find inflection points

we first find the second derivative of a function, set it equal to zero

and then find the roots of the equation.


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definition: second derivative


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example23)(' xxh

3)( xxh ?)('' xh

xxh 6)(''


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recipe for findingcritical points (local minima,

maxima and inflection points)(1) Find the derivative of the function

Set the derivative equal to 0 and solve for x.

Plug in values between critical points to determine if the identified critical pointsare local minima, local maxima, or inflection points.

(2) Find the second derivative of the function Set the second derivative equal to 0 to find any points where a change in

concavity occurs.

Any additional critical points identified here will be inflection points.

E pl fi di g l l


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Example: finding localminimum/maximum

x

h(x) = ex(x+3)

1. Find the derivativeh(x) = ex(x+4)

Set the derivative equal to 0 and solve for x.

h(x) = ex(x+4) = 0

The function will equal 0 when ex= 0 (this never happens)or when x+4 = 0 (this happens at -4).

x=-4 is either a minimum or a maximum.

2. Plug in values around this pointh(x=-3.9) = 0.002, h(x=-4.1) = -0.002

Since the derivative is positive before x=-4 (the function isincreasing) and negative after x=-4 (the function is

decreasing), x=-4 is a maximum.

E pl : fi di g i fl tio


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Example: finding inflectionpoint

x

h(x) = ex(x+3), h(x) = ex(x+4)

1. Find the second derivative

h(x) = ex(x+5)

Set the second derivative equal to 0 and solve for x.

h(x) = ex(x+5) = 0

The function will equal 0 when ex= 0 (this never

happens) or when x+5 = 0 (this happens at -5).The inflection point occurs at x=-5.


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example

a. Horizontal or vertical asymptotes, if they exist.

b. Points of maximum or minimum, and the functions value at these points.

c. Inflection points and the value of the function at these points.

d. Describe the behavior of the graph of this function.

35

53 xxy


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35

53 xxy a. Horizontal or vertical asymptotes, if they exist.

example


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If there is a restriction on the domain, then there will be a vertical asymptote on

the edge of that restriction. If , then there will be a vertical

asymptote at .

If the limits in (1) or (2) = (a constant, not infinity), then there is a horizontal

asymptote at .

recipe for findinghorizontal asymptotes

(1) Find the limit of the function as x approaches infinity(2) Find the limit of the function as x approaches negative infinity

recipe for findingvertical asymptotes

Find the points where the function has a restriction on the domain

),(),( ccxcx

cy

c


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Horizontal asymptotes:


35

53lim xxx


35 53lim xxx

example

35

53 xxy


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recipe for finding the limit atx approaches negative infinity

(1) Replace x with (-x)

(2) Find the limit as x goes to positive infinity instead


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Horizontal asymptotes:


35

53lim xxx


35 53lim xxx

35)(5)(3lim xxx

35 53lim xx

xno horizontal asymptote

example

35

53 xxy


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Find the points where the function has a restriction on the domain

Vertical asymptotes:

),( x

no vertical asymptote

example

35

53 xxy


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Points of maximum or minimum, and the functions value at these points.

example

35

53 xxy


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recipe for findingcritical points

(1) Find the derivative of the function Set the derivative equal to 0 and solve for x. This will help us find points where the tangent

line is flat.

Plug in values between critical points to determine if the identified critical points are local

minima, local maxima, or inflection points.

(2) Find the second derivative of the function Set the second derivative equal to 0 to find any points where a change in concavity occurs.



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Find the derivative of the function

Set the derivative equal to 0 and solve for x. This will help us find

points where the tangent line is flat.

Plug in values between critical points to determine if the identified

critical points are local minima, local maxima, or inflection points.24 1515' xxy

)1(15 22 xx

0)1)(1(15

2 xxx

015?

2 x

0)1(

?

x

0)1(?

x

0x

1x

1x

criticalpoints

minimum, maximum, or inflection point?

example

35

53 xxy


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example

0x1x 1x

)1,( )0,1( )1,0( ),1(

221

21 2

24 1515' xxy

24 )2(15)2(15 24 )2

1(15)

2

1(15 24 )

2

1(15)

2

1(15 24 )2(15)2(15

4151615 )4

1(15)

16

1(15 )

4

1(15)

16

1(15 4151615

1801645

1645

180

maximum inflectionpoint minimum

Range

Test Point

Derivative ofthe function at

the test point

Increasing (+)or Decreasing (-)

35

53 xxy


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example

0x1x 1x

maximum inflectionpoint minimum

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

35 )1(5)1(3 y

2y

24 1515' xxy

35

53 xxy


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example

c. Inflection points and the value of the function at these points.

35

53 xxy


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recipe for findingcritical points

(1) Find the derivative of the function Set the derivative equal to 0 and solve for x. This will help us find points where the tangent

line is flat.

Plug in values between critical points to determine if the identified critical points are local

minima, local maxima, or inflection points.

(2) Find the second derivative of the function Set the second derivative equal to 0 to find any points where a change in concavity occurs.



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example

Find the second derivative of the function

Set the second derivative equal to 0 to find any points where a change

in concavity occurs.


xxy 3060'' 3

)12(30 2 xx

0

030?

x

0)12(?

2 x

0x

21x

inflectionpoints12 2 x2

12 x

212 x

35

53 xxy

24

1515' xxy


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-5

-4

-3

-2

-1

0

1

2

3

4

5

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

0

0

example

maximum

inflection point

minimum

0x1x 1x


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-5

-4

-3

-2

-1

0

1

2

3

4

5

-1.5 -1 -0.5 0 0.5 1 1.5 2

-5

-4

-3

-2

-1

0

1

2

3

4

5

-2 -1.5 -1 -0.5 0 0.5 1 1.5 2

example

maximum

inflection point

minimum

inflection point

inflection point


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the setting: antiderivatives

Given: )(xf

Find: )(xF such that )()(' xfxF dxxf )(


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example

Find: such that)(xF

The derivative of what function will give us x3?

Given: )(xf3)(' xxF

3

x dxx

3


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Given: )(xf

Find: )(xF such that )()(' xfxF

the setting: antiderivatives

cxFdxxf )()(

integral signintegrand

take integralwith respect to x

antiderivativeconstant ofintegration

dxxf )(


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example

Find: such that3)(' xxF )(xF

3

x dxx

3

Given: )(xf

cxFdxxf )()(

cxdxx

43

4

14

4

1)(, xxF

31444

1)(' xxxF

The derivative of what function will give us x3?


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antiderivative power rule

dxxnxn

xn 1n 1n

1n


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example

dxx3

c

x

4

4

cx 44

1

cn

xdxx

nn

11

cn

xdxx

nn

11

3, n

c

x

dxx

1313

34

4

1)( xxF

31444

1)(' xxxF


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example

dxx

c

x

2

2

cx 22

1

cn

xdxx

nn

11

cn

xdxx

nn

11

1, n

c

x

dxx

1111

12

2

1)( xxF

31222

1)(' xxxF


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dxxgdxxfdxxgxf )()()()(

properties of indefinite integrals


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example

dxxx3

dxxdxx3



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example

dxx3

c

x

4

4

cx 44

1

cn

xdxx

nn

11

cn

xdxx

nn

11

3, n

c

x

dxx

1313

34

4

1)( xxF

31444

1)(' xxxF


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example

dxx

c

x

2

2

cx 22

1

cn

xdxx

nn

11

cn

xdxx

nn

11

1, n

c

x

dxx

1111

12

2

1)( xxF

31222

1)(' xxxF


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cxGxF )()(

example

dxxx3

dxxdxx3

cx44

1

2124

2

1

4

1ccxx

1

cx22

1

2


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properties of indefinite integrals


dxxfkdxxfk )()(

cxGxF )()(


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example

Find: such that 2)(' xF)(xF2

dx2

Given: )(xf

The derivative of what function will give us 2?

dxx02

dxxfkdxxfk )()( 2,)( 0

kxxf

dxxdxx 00 22 c

x

12

1

cx 2


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helpful hints

When you are asked for the antiderivative of a function, f(x),

then you are looking for the function whose derivative is

equal to f(x)

You can always (and should!) check your work to see if youranswer is correct by taking the derivative to make sure that it

is equal to f(x)

Use the properties of derivatives to help break the problem

down into more approachable parts

Dont forget the constant of integration (the +c)


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example

Find: such that 2)(' xF)(xF2

dx2

Given: )(xf

The derivative of what function will give us 2?

dxx02

dxxfkdxxfk )()( 2,)( 0

kxxf

dxxdxx 00 22 c

x

12

1

cx 2


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antiderivative rules

cn

xdxx

nn

11

dxex

cex

dx1 cxdx dxx0

cdxxgdxxfdxxgxf )()()()(

cdxxfkdxxfk )()(

1, n


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the setting: indefinite integral

cxFdxxf )()(


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)()()()( aFbFxFdxxf b

a

b

a

the setting: definite integral

)()(' xfxF where

If is continuous on the interval from a to b)(xf

limits of integration


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a

b

a

the setting: definite integral

)()(' xfxF where




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a

b

a

fundamental theorem of calculus

)()(' xfxF where



recipe for performing


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1.Integrate the function

2.Evaluate the antiderivative at the upper limit

3.Evaluate the antiderivative at the lower limit

4.Subtract the value found in step 3 from the value found in

step 2.

recipe for performingdefinite integration


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example

dxx2

02 c

x

22

2

2

0 cx 2

2

0c 22 c 20

4


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example

dxx2

1

3

4

4x 2

1 4

24

4

14

4

14

4

33


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example

422

0

dxx

2

4

Area hb 2

1

422

1

4


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example

440)2(02 220

2

20

2

xdxx

2

4

Area hb 2

1

422

1

4

l


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example

dxx

2

1

3

4

3

3


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properties of definite integrals

l


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dxx

1

1

3

example

041

41

1

1

4

4x

Module 2 Examples

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