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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:
(1) Find the limit of the function as x approaches infinity
35
53lim xxx
(2) Find the limit of the function as x approaches negative infinity
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:
(1) Find the limit of the function as x approaches infinity
35
53lim xxx
(2) Find the limit of the function as x approaches negative infinity
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.
Any additional critical points identified here will be inflection points.
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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.
Any additional critical points identified here will be inflection points.
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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.
Any additional critical points identified here will be inflection points.
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
dxxgdxxfdxxgxf )()()()(
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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 )()()()(
cxGxF )()(
example
dxxx3
dxxdxx3
cx44
1
2124
2
1
4
1ccxx
1
cx22
1
2
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properties of indefinite integrals
dxxgdxxfdxxgxf )()()()(
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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)()()()( 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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)()()()( aFbFxFdxxf b
a
b
a
fundamental theorem of calculus
)()(' xfxF where
If is continuous on the interval from a to b)(xf
limits of integration
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
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