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Absolute extreme values are either maximum or
minimum points on a curve.
They are sometimes called global extremes.
They are also sometimes called absolute extrema.
(Extremais the plural of the Latin extremum.)
4.1 Extreme alues of
!unctions
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4.1 Extreme Values of
FunctionsDefinition Absolute Extreme Values
Let f be a function "ith domain #. Thenf(c) is the
a. absolute minimum value on # if and only
iff(x) $f(c) for allxin #.
b. absolute maximum value on # if and only
iff(x) %f(c) for allxin #.
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Extreme values can be in the interior or the end
points of a function.
&
y x=( )'D= Absolute
inimum
o Absolute
aximum
4.1 Extreme Values of
Functions
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&
y x=[ ]*'&D=
Absolute inimum
Absolute
aximum
4.1 Extreme alues of !unctions
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&y x=
( ]*'&D =
No Minimum
AbsoluteMaximum
4.1 Extreme Values of
Functions
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&y x=
( )*'&D =No Minimum
NoMaximum
4.1 Extreme Values of
Functions
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Extreme Value Theorem:
+f f is continuous over a closed interval' ,a'b- then f has a
maximum and minimum value over that interval.
aximum
minimum
at interior points
aximum
minimum
at endpoints
aximum at
interior point'
minimum atendpoint
4.1 Extreme alues of !unctions
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Local Extreme Values:
A local maximum is the maximum value "ithin some
open interval.
A local minimum is the minimum value "ithin some
open interval.
4.1 Extreme alues of !unctions
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Absolute minimum
(also local minimum)
Local maximum
Local minimum
Absolute maximum
(also local maximum)
Local minimum
Local extremes
are also called
relative extremes.
4.1 Extreme Values of
Functions
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Local maximum
Local minimum
Notice that local extremes in the interior of the function
occur where is zero or is undefined.f f
Absolute maximum
(also local maximum)
4.1 Extreme Values of
Functions
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Local Extreme Values:
+f a function f has a local maximum value or a
local minimum value at an interior point cof its
domain' and if exists at c' then
( ) *f c =
f
4.1 Extreme Values of
Functions
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Critical Point:
A point in the domain of a function f at which
or does not exist is a critical pointof f.
*f=f
Note:Maximum and minimum points in the interior of a function
always occur at critical points but critical points are not
always maximum or minimum values.
4.1 Extreme Values of
Functions
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EXAMPLE 3 FINDING ABSOLUTE EXTREMA
!ind the absolute maximum and minimum values of
on the interval .( ) & /0f x x= [ ]&'0
( ) & /0f x x=
( )1
0&0
f x x =
( )0
&
0
f x
x
=
"here are no values of x that will ma#e
the first derivative e$ual to zero.
"he first derivative is undefined at x%&
so (&&) is a critical point.
'ecause the function is defined over a
closed interval we also must chec# theendpoints.
4.1 Extreme Values of
Functions
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( )* *f
=
"o determine if this critical point is
actually a maximum or minimum we
try points on either side without
passin other critical points.
( ) & /0f x x=
( )1 1f = ( )1 1f =
ince &*+ this must be at least a local minimum and possibly a
lobal minimum.
[ ]&'0D=
At: *x=
At: &x= ( ) ( )&
0& & 1.234f =
At: 0x= ( ) ( )
&
00 0 &.*2**2f =
4.1 Extreme Values of
Functions
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( )* *f =
( ) & /0f x x= [ ]&'0D=
At: &x= ( ) ( )&0& & 1.234f =
At: 0x=
Absolute
minimum:
Absolute
maximum:
( )*'*
( )0'&.*2
( ) ( )&
00 0 &.*2**2f =
4.1 Extreme Values of
Functions
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4.1 Extreme Values of
Functionsyx&/0
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Findin Ma!i"#"$ and Mini"#"$ Anal%ticall%:
+ !ind the derivative of the function and determine
where the derivative is zero or undefined. "heseare the critical points.
, !ind the value of the function at each critical point.
- !ind values or slopes for points between thecritical points to determine if the critical points are
maximums or minimums.
!or closed intervals chec# the end points as
well.
4.1 Extreme Values of
Functions
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4.1 Extreme Values of
Functions
Find the absolute maximum and minimum of the functi
-&'1,'&4&)( &0 += onxxxxf
41*5)(6 & += xxxf
41*5* & += xxFind the critical numbers
&0* & += xx)1)(&0(* = xx 1
0
&== xx or
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4.1 Extreme Values of
Functions
!ind the absolute maximum and minimum of the function
-&'1,'&4&)( &0 += onxxxxf
7hec8 endpoints and critical numbers
The absolute maximum is & "henx 9&The absolute minimum is 910 "henx 91
( )
&&
11
&3
&5
0
&
101
xfx
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4.1 Extreme Values of
Functions
Find the absolute maximum and minimum of the functi
-0'*,'1
0)(
&
on++=
x
xxf &
&
)1(
)1)(0()&)(1()(6+
++=x
xxxxf
0&* & += xx
Find the critical numbers
)1)(0(* += xx 10 == xx or
&
&
)1(
0&)(6
+
+
= x
xxxf
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4.1 Extreme Values of
Functions
!ind the absolute maximum and minimum of the function
-0'*,'1
0)(
&
on++=
x
xxf ( )
00&1
0*
xfx
7hec8 endpoints and critical numbers
The absolute maximum is 0 "henx *' 0
The absolute minimum is & "henx 1
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4.1 Extreme alues of !unctions
!ind the absolute maximum and minimum of the function
[ ]&'*'sinsin)( & onxxxf =
xxxxf cossin&cos)(6 =
!ind the critical numbers
xxx cossin&cos* =
)sin&1(cos* xx =
*cos =x *sin&1 = x
&
0'
&
=x
5
'
5
=x
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4.1 Extreme Values of
Functions
!ind the absolute maximum and
minimum of the function
[ ]&'*'sinsin)( & onxxxf =
( )
*&
&&
0
4
1
5
*&
4
1
5
**
xfx
The absolute maximum is 1/4 "henx /5' /5The absolute minimum is :& "henx0/&
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Critical points are not always extremes!
0y x=
*f=(not an extreme)
4.1 Extreme Values of
Functions
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1/ 0y x=
is undefined.f
(not an extreme)
4.1 Extreme Values of
Functions
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+f f (x) is a differentiable function over ,a'b-'
then at some point bet"een aand b;
( ) ( )( )
f b f af c
b a
=
Mean Value Theorem for Derivatives
4.2 Mean Value Theorem
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+f f (x) is a differentiable function over ,a'b-'then at some point bet"een aand b;
( ) ( )( )
f b f af c
b a
=
Mean Value Theorem for Derivatives
Differentiableimplies that the function is also continuous.
4.2 Mean Value Theorem
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+f f (x) is a differentiable function over ,a'b-'then at some point bet"een aand b;
( ) ( )( )
f b f af c
b a
=
Mean Value Theorem for Derivatives
Differentiableimplies that the function is also continuous.
The ean alue Theorem only applies over aclose interval
4.& ean alue Theorem
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+f f (x) is a differentiable function over ,a'b-'then at some point bet"een aand b;
( ) ( )( )
f b f af c
b a
=
Mean Value Theorem for Derivatives
The ean alue Theorem says that at some point
in the close interval" the actual slope e#uals
the avera$e slope.
4.2 Mean Value Theorem
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y
x*
A
B
a b
lope of chord:
( ) ( )f b f a
b a
lope of tanent:
( )f c
( )y f x=
"anent parallel
to chord.
c
4.2 Mean Value Theorem
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+f f (x) is a differentiable function over ,a'b-'and iff(a)f(b) *' then there is at least one
point c bet"een aand b such thatf(c)=*;
%olle&s Theorem
4.2 Mean Value Theorem
(a'*) (b'*)
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4.2 Mean Value Theorem
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4.2 Mean Value Theorem(0,1)
(!,12)
atx .4=2' the slope
of the tangent line is
e>ual to the slope of
the chord.
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4.& ean alue Theorem
Definitions 'ncreasin$ (unctions" Decreasin$ (unctionsLet f be a function defined on an interval + and letx1andx&
be any t"o points in +.
1. f increases on + ifx1$x&f(x1) $f(x&).2. f decreases on + ifx1%x&f(x1) %f(x&).
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A function is increasing over an interval if thederivative is al"ays positive.
A function is decreasing over an interval if the
derivative is al"ays negative.
A couple of some"hat obvious definitions;
4.& ean alue TheoremCorollary 'ncreasin$ (unctions" Decreasin$ (unctions
Let f be continuous on ,a'b- and differentiable on (a'b).1. +ff? % * at each point of (a'b)' then f increases on ,a'b-.
&. +ff? $ * at each point of (a'b)' then f decreases on ,a'b-.
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4.& ean alue Theorem!ind "here the function
is increasing and decreasing and find the local
extrema.
xxxxf &4=)( &0 +=
xxxxf &4=)( &0 +=
&4120)(6 & += xxxf
)25(0* & += xx)25(* & += xx)&)(4(* = xx
& 4
* *f(x)
@9@
)'4()&'( inc)4'&(ec
x &' local maximum
x 4' local minimum
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4.2 Mean Value Theorem
(&'&*) local max
(4'15) local min
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y
x*
( )y f x=
( )y g x=
These t"o functions have the
same slope at any value ofx.
!unctions "ith the same
derivative differ by a constant.
C4.2 Mean Value Theorem
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!ind the function "hose derivative is and "hose
graph passes through
( )f x ( )sin x( )*'&
( ) ( )cos sind
x xdx
=
( ) ( )cos sind
x xdx
=so;
( ) ( )cosf x x C = +
( )& cos * C= +
4.2 Mean Value Theorem
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!ind the functionf(x) "hose derivative is sin(x) and
"hose graph passes through (*'&).
( ) ( )cos sind
x xdx
=
( ) ( )cos sind
x xdx =so:
( ) ( )cosf x x C = +
( )& cos * C= +
& 1 C= +0 C=
( ) ( )cos 0f x x= +Notice that we had to haveinitial values to determine
the value of C.
4.2 Mean Value Theorem
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"he process of findin the oriinal function from the
derivative is so important that it has a name:
Antid&ri'ati'&
A function is an antid&ri'ati'&of a function
if for all xin the domain off. "he process
of findin an antiderivative is antidi((&r&ntiation.
( )F x ( )f x
( ) ( )F x f x =
/ou will hear much more about antiderivatives in the future.
"his section is 0ust an introduction.
4.2 Mean Value Theorem
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ince acceleration is thederivative of velocity
velocity must be the
antiderivative of
acceleration.
1xample 2b: !ind the velocity and position e$uations
for a downward acceleration of 3.4 m5sec,and an
initial velocity of + m5sec downward.
( ) =.2a t =
( ) =.2 1v t t= +
( )1 =.2 * C= +1 C=
( ) =.2v t t C = +(6e let down be positive.)
4.2 Mean Value Theorem
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ince velocity is the derivative of position
position must be the antiderivative of velocity.
( ) =.2a t =
( ) =.2 1v t t= +
( )1 =.2 * C= +
1 C=
( ) =.2v t t C = +( ) &
=.2
&s t t t C= + +
"he power rule in reverse:
7ncrease the exponent by one and
multiply by the reciprocal of the
new exponent.
4.2 Mean Value Theorem
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( ) =.2a t =
( ) =.2 1v t t= +
( )1 =.2 * C= +
1 C=
( ) =.2v t t C = +
( ) &=.2
&
s t t t C= + +
( ) &4.=s t t t C= + +"he initial position is zero at time zero.
( )&* 4.= * * C= + +* C=
( ) &
4.=s t t t= +
4.2 Mean Value Theorem
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+n the past' one of the important uses of derivatives "as
as an aid in curve s8etching. e usually use a calculator
of computer to dra" complicated graphs' it is still
important to understand the relationships bet"een
derivatives and graphs.
4.0 7onnectingf? andf?? "ith the
Braph off
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!irst #erivative Test for Local Extrema at a critical point c
4.0 7onnectingf? andf?? "ith the
Braph off
1. +ff C changes sign from positive to
negative at c' thenf has a localmaximum at c.
local max
f?%* f?$*
&. +ff changes sign from negative to
positive at c' thenf has a local
minimum at c.
0. +ff changes does not change sign
at c' thenf has no local extrema.
local min
f?$* f?%*
no extreme
f?%* f?%*
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!irst derivative;
y is positive 7urve is rising.
y is negative 7urve is falling.
y is Dero ossible local maximum orminimum.
4.0 7onnectingf? andf?? "ith
the Braph off
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4.0 7onnectingf? andf?? "ith the
Braph offDefinition Concavity
The graph of a differentiablefunctionyf(x) is
a. concave up on an open interval
+ ify? is increasing on +. (y??%*)b. concave do"n on an open interval
+ ify? is decreasing on +. (y??$*)concave do"n
concave up
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4.0 7onnectingf? andf?? "ith the
Braph off
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4.0 7onnectingf? andf?? "ith the
Braph off
Definition )oint of 'nflection
A point "here the graph of a function has a tangent line and
"here the concavity changes is called a point of inflection.
inflection point
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( ) ( )&0 &0 4 1 &y x x x x= + = +
&0 5y x x=
*y=
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&0 5y x x=
*y=
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&0 5y x x=
*y=et
&* 0 5x x=
&* &x x=
( )* &x x=
*' &x=
8ossible extreme at .*' &x=
9r you could use the second derivative test:
maximum at *x= minimum at &x=
5 5y x= ( )* 5 * 5 5y = = neative
concave down
local maximum
( )& 5 & 5 5y = = positiveconcave up
local minimum
4.0 7onnectingf? andf?? "ith the
Braph off
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5 5y x=
e then loo8 for inflection points by setting the second
derivative e>ual to Dero.
* 5 5x=
5 5x=
1 x=
ossible inflection point at .1x=
y1
* +
( )* 5 * 5 5y = = negative
( )& 5 & 5 5y = = positive
inflection point at 1x=
4.0 7onnectingf? andf?? "ith the
Braph off
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Ma#e a summary table: x y y y
1 * = 1& risin concave down
* 4 * 5 local max
1 & 0 * fallin inflection point
& * * 5 local min
0 4 = 1& risin concave up
4.! #onnectin$ f% and f%%
&ith the 'rah of f
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A 7lassic roblem
Fou have 4* feet of fence to enclose a rectangular garden along
the side of a barn. hat is the maximum area that you can
encloseG
4.4 odeling and HptimiDation
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x x
4* &x
( )4* &A x x=
&4* &A x x=
4* 4A x=
* 4* 4x= 4 4*x=
1*x=
( )1* 4* & 1*A=
( )1* &*A=
&&** ftA=
4* &l x=
x= 1* ft=
&* ftl=
4.4 odeling and HptimiDation
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To find the maximum (or minimum) value of a function;
4.4 odeling and HptimiDation
1. Inderstand the roblem.&. #evelop a athematical odel.
0. Braph the !unction.
4. +dentify 7ritical oints and Endpoints.
.
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hat dimensions for a one liter cylindrical can "ill
use the least amount of materialG
e can minimiDe the material by minimiDing the area.
&& &A r r! = +
area of
ends
lateral
area
e need another
e>uation that relates
rand !;
&" r !=( )01 L 1*** cm=
&1*** r !=
&
1***!
r
=
&
&
1* *&
*&A r r
r = +
& &***&A rr
= +
&
&***4A r
r
=
4.4 odeling and HptimiDation
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&& &A r r! = +area of
ends
lateral
area
&" r !=
( )01 L 1*** cm=&1*** r !=
&
1***!
r=
&&
1* *&
*&A r r
r = +
& &***&A rr
= +
&
&***4A r
r=
&
&***
* 4 r r
=
&
&***4 r
r=
0&*** 4 r=
0** r
=
0 **r
=
.4& cmr
( )&
1***
.4&!
1*.20 cm!
4.4 Modelin$ and
timi*ation
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4.4 Modelin$ and
timi*ation!ind the radius and height of
the right9circular cylinder of
largest volume that can beinscribed in a right9circular
cone "ith radius 5 in. and
height 1* in.h
r
1* in
+ in
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4.4 odeling and HptimiDation
h
r
1* in
5 in
The formula for the volume of
the cylinder is !r" &=
To eliminate one variable' "e
need to find a relationship
bet"een rand !.
51*1* =r
!
r!0
1* =
5
h
1*9h
r1*
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4.4 Modelin$ and
timi*ation
h
r
10 in
+ in
!r" &=
0&&
0
1*0
1* rrrr" =
=
&&* rr
dr
d" =
)4(* rr =
4'* == rr
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4.4 odeling and HptimiDation
h
r
1* in
5 in
7hec8 critical points and endpoints.
Jr *' " *Jr 4 " 15*/0Jr 5 " *
The cylinder "ill have amaximum volume "hen
r 4 in. and ! 1*/0 in.
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#etermine the point on the
curveyx&that is closest to
the point (12' *).
4.4 odeling and HptimiDation
&&)12( yxd +=
4&)12( xxd +=
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#etermine the point on the
curveyx&that is closest to
the point (12' *).
4.4 odeling and HptimiDation
)05&4()0&405(
&
1 0&1
&4 +++=
xxxxx
dx
ds
*=dx
dsset 05&4*
0 += xx 12&* 0 += xx
&=x 4=y
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#etermine the point on the
curveyx&that is closest to
the point (12' *).
4.4 odeling and HptimiDation
12&* 0 += xx
&=x 4=y
)=4&)(&(* & ++= xxx
&
9 * @
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+f the end points could be the maximum or
minimum' you have to chec8.
otes;
+f the function that you "ant to optimiDe has morethan one variable' use substitution to re"rite the
function.
+f you are not sure that the extreme you?ve found is a
maximum or a minimum' you have to chec8.
4.4 odeling and HptimiDation
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!or any functionf (x)'the tangent is aclose approximation of the function for
some small distance from the tangent
point.
y
x* x a=
( ) ( )f x f a=e call the e>uation of the
tangent the lineariDation of
the function.
4. LineariDation and
e"ton?s ethod
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The lineariDation is the e>uation of the tangent line' and you
can use the old formulas if you li8e.
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!ind the lineariDation off(x) x4@ &xatx &
( ) ( ) ( ) ( )# x f a f a x a= +
4. LineariDation and
e"ton?s ethod
f(x) 4x0@ &
# (x) f(0) @f?(0)(x 9 0)
# (x) 23 @ 11*(x 9 0)
# (x) 11*x $ &40
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+mportant lineariDations forxnear Dero;
( )1%
x+ 1 %x+
sinx
cosx
tanx
x
1
x
( )1
&1
1 1 1
&
x x x+ = + +
( )
( )
10 4 4 0
4 4
1 1
1 1 1
0 0
x x
x x
+ = +
+ = +
( )f x ( )# x
This formula also leads to
non9linear approximations;
4. LineariDation and
e"ton?s ethod
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4. LineariDation and
e"ton?s ethodEstimate using local lineariDation.03
&
1
&
1)(6
)(
=
=
xxf
xxf ( ) ( ) ( ) ( )# x f a f a x a= +
)0503)(05(6)05()03( += ff#
)1(1&15)03( +=#
*200.5)03( =#
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4. LineariDation and
e"ton?s ethodEstimate sin 01K using local lineariDation.
xxf
xxf
cos)(6
sin)(
== ( ) ( ) ( ) ( )# x f a f a x a= +
+=12*
)0*(6)0*()01(
ff#
+=12*&
0&1)01( #
05*
012*)01(
+=#
eed to
be in radians
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Differentials:
hen "e first started to tal8 about derivatives' "e said
that becomes "hen the change in xand
change in y become very small.
y
x
dy
dx
dycan be considered a very small change iny.
dxcan be considered a very small change inx.
4. LineariDation and
e"ton?s ethod
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Let yf(x) be a differentiable function.
The differential dxis an independent
variable.
The differential dyis; dyf ?(x)dx
4. LineariDation and
e"ton?s ethod
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Example; 7onsider a circle of radius 1*. +f the radius increases by
*.1' approximately ho" much "ill the area changeG
&
A r=&dA r dr =
&dA dr rdx dx
=
very small change in A
very small change in r
( )& 1* *.1dA =
&dA
= (approximate change in area)
4. LineariDation and
e"ton?s ethod
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7ompare to actual change;
e" area;
Hld area;
( )&
1*.1 1*&.*1 =
( )&
1* 1**.** =
4. LineariDation and
e"ton?s ethod
*1.&=A
&=dA Absoluteerror&
1**
&==
A
dA
*1.&1**
*1.&==
A
Apercenterror
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4. LineariDation and
e"ton?s ethodTrue Estimate
Absolute 7hange
Melative 7hange
ercent 7hange
)()( afdxaff += dxafdf )(6=
)(af
f)(af
df
1**)(x
afdf1**
)(x
aff
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4. LineariDation and
e"ton?s ethod *ewton&s Metho
* x
y
yf(x)
%oot
sou$ht
x1
!irst
(x1&f(x1))
x&
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This is e"ton?s ethod of finding roots. +t is an
example of an algorithm (a specific set of
computational steps.)
e"ton?s ethod;( )
( )1
f xx x
f x+ =
This is a recursive algorithm because a set of steps are
repeated "ith the previous ans"er put in the next
repetition. Each repetition is called an iteration.
4. LineariDation and
e"ton?s ethod
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*ewton&s Metho
( )
&1
0&f x x=
!inding a root for;
e "ill use
e"ton?s ethod to
find the rootbet"een & and 0.
4. LineariDation and
e"ton?s ethod
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*ewton&s Metho( ) &
10
&f x x=
4. LineariDation and
e"ton?s ethod
xxf =)(6
Buessx1 &
)(6
)(
1
11&
xf
xfxx =
.&&
1&& =
=x
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*ewton&s Metho( ) &
10
&f x x=
4. LineariDation and
e"ton?s ethod
xxf =)(6
Buessx& &.
)(6
)(
&
&&0
xf
xfxx =
4.&.&
1&..&0 ==x
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!ind "here crosses .0y x x= 1y=
0
1 x x= 0
* 1x x=
( ) 0 1f x x x=
( ) &0 1f x x =
4. LineariDation and
e"ton?s ethod
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(x ( )(f x( ( )(f x( )
( )1
(
( (
(
f xx x
f x+
=
* 1 1 & 11 1.& =
1 1. .23 .3.23
1. 1.0432&51.3
=
& 1.0432&51 .1**52&& 4.44==* 1.0&&**4
( )0
1.0&&**4 1.0&&**4 1.**&*24 = 1
4. LineariDation and
e"ton?s ethod
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There are some limitations to e"ton?s ethod;
rong root found
Loo8ing for this root.
Nad guess.
!ailure to converge
4. LineariDation and
e"ton?s ethod
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!irst' a revie" problem;
7onsider a sphere of radius 1* cm.
+f the radius changes *.1 cm (a very small amount)
ho" much does the volume changeG
040
" r= &4d" r dr =
( )&
4 1*cm *.1cmd" = 04* cmd" =
The volume "ould change by approximately 4*cm0
4.5 Melated Mates
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o"' suppose that the radius is
changing at an instantaneous rate
of *.1 cm/sec.04
0" r= &4
d" dr r
dt dt =
( )& cm
4 1*cm *.1
sec
d"
dt
=
0cm4*
sec
d"
dt
=
The sphere is gro"ing at a rate of 4*cm0/sec .
ote; This is an exact ans"er' not an approximation li8e
"e got "ith the differential problems.
4.5 Melated Mates
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ater is draining from a cylindrical tan8
at 0 liters/second. Oo" fast is the surface
droppingG
L0
sec
d"
dt=
0cm0***
sec=
!indd!
dt
&" r !=
&d" d!rdt dt
=(ris a constant.)
0&cm0***
sec
d!r
dt
=
0
&
cm0***
secd!
dt r
=
(e need a formula to
relate "and !. )
4.5 Melated Mates
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+teps for %elate %ates )roblems:
1. #ra" a picture (s8etch).
&. rite do"n 8no"n information.
0. rite do"n "hat you are loo8ing for.
4. rite an e>uation to relate the variables.
. #ifferentiate both sides "ith respect to t.
5. Evaluate.
4.5 Melated Mates
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,ot Air -alloon )roblem:
Biven;
4
=
rad*.14
min
d
dt
=
Oo" fast is the balloon risingG
!indd!
dttan
**
!=
& 1sec**
d d!
dt dt
=
( )&
1sec *.14
4 **
d!
dt
=
!
**ft
4.5 Melated Mates
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,ot Air -alloon )roblem:
Biven;
4
=
rad*.14
min
d
dt
=
Oo" fast is the balloon risingG
!ind
d!
dt tan **
!
=
& 1sec**
d d!
dt dt
= ( )
&1
sec *.144 **
d!
dt
=
!
1**ft
( ) ( )&
& *.14 ** d!
dt =
1
1
&
4
sec &4
=
ft14*
min
d!
dt=
4.5 Melated Mates
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4x=
0y=
N
A
=
Truc.)roblem:
Truc8 A travels east at 4* mi/hr.
Truc8 N travels north at 0* mi/hr.
Oo" fast is the distance bet"een the
truc8s changing 5 minutes laterG
r t d =14* 4
1* = 10* 0
1* =
& & &0 4 )+ =
&= 15
)+ =
&&
)=
)=
4.5 Melated Mates
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4x=
0y=
0*dy
dt=
4*dx
dt
=
'
A
=
Tr#c) Pro*l&":
ow fast is the distance between the
truc#s chanin ; minutes later