Calculus Chapter 4

7/25/2019 Calculus Chapter 4

1/97

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


2/97

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 #.


3/97

Extreme values can be in the interior or the end

points of a function.

&

y x=( )'D= Absolute

inimum

o Absolute

aximum


Functions


4/97

&

y x=[ ]*'&D=

Absolute inimum

Absolute

aximum

4.1 Extreme alues of !unctions


5/97

&y x=

( ]*'&D =

No Minimum

AbsoluteMaximum


Functions


6/97

&y x=

( )*'&D =No Minimum

NoMaximum


Functions


7/97

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



8/97

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.



9/97

Absolute minimum

(also local minimum)

Local maximum

Local minimum

Absolute maximum

(also local maximum)

Local minimum

Local extremes

are also called

relative extremes.


Functions


10/97

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)


Functions


11/97

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


Functions


12/97

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.


Functions


13/97

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.


Functions


14/97

( )* *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 =


Functions


15/97

( )* *f =

( ) & /0f x x= [ ]&'0D=

At: &x= ( ) ( )&0& & 1.234f =

At: 0x=

Absolute

minimum:

Absolute

maximum:

( )*'*

( )0'&.*2

( ) ( )&

00 0 &.*2**2f =


Functions


16/97


Functionsyx&/0


17/97

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.


Functions


18/97


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


19/97


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


20/97


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


21/97


Functions


-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


22/97



[ ]&'*'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


23/97


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/&


24/97

Critical points are not always extremes!

0y x=

*f=(not an extreme)


Functions


25/97

1/ 0y x=

is undefined.f

(not an extreme)


Functions


26/97

+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


27/97

+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

=


Differentiableimplies that the function is also continuous.



28/97


( ) ( )( )

f b f af c

b a

=


Differentiableimplies that the function is also continuous.

The ean alue Theorem only applies over aclose interval

4.& ean alue Theorem


29/97


( ) ( )( )

f b f af c

b a

=


The ean alue Theorem says that at some point

in the close interval" the actual slope e#uals

the avera$e slope.



30/97

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



31/97

+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


(a'*) (b'*)


32/97



33/97

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.


34/97

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&).


35/97

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-.


36/97

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


37/97


(&'&*) local max

(4'15) local min


38/97

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


39/97

!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= +



40/97

!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.



41/97

"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.



42/97

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.)



43/97

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.



44/97

( ) =.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= +



45/97

+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


46/97

!irst #erivative Test for Local Extrema at a critical point c


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?%*


47/97

!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


48/97


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


49/97


Braph off


50/97


51/97


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


52/97

( ) ( )&0 &0 4 1 &y x x x x= + = +

&0 5y x x=

*y=


53/97

&0 5y x x=

*y=


54/97

&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


Braph off


55/97

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=


Braph off


56/97

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


57/97

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


58/97

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=



59/97

To find the maximum (or minimum) value of a function;


1. Inderstand the roblem.&. #evelop a athematical odel.

0. Braph the !unction.

4. +dentify 7ritical oints and Endpoints.

.


60/97

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

=



61/97

&& &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


62/97

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


63/97


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*


64/97

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


65/97


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.


66/97

#etermine the point on the

curveyx&that is closest to

the point (12' *).


&&)12( yxd +=

4&)12( xxd +=


67/97



the point (12' *).


)05&4()0&405(

&

1 0&1

&4 +++=

xxxxx

dx

ds

*=dx

dsset 05&4*

0 += xx 12&* 0 += xx

&=x 4=y


68/97



the point (12' *).


12&* 0 += xx

&=x 4=y

)=4&)(&(* & ++= xxx

&

9 * @


69/97

+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.



70/97

!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


71/97

The lineariDation is the e>uation of the tangent line' and you

can use the old formulas if you li8e.


72/97

!ind the lineariDation off(x) x4@ &xatx &

( ) ( ) ( ) ( )# x f a f a x a= +


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


73/97

+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;


e"ton?s ethod


74/97


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( =#


75/97


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


76/97

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.


e"ton?s ethod


77/97

Let yf(x) be a differentiable function.

The differential dxis an independent

variable.

The differential dyis; dyf ?(x)dx


e"ton?s ethod


78/97

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)


e"ton?s ethod


79/97

7ompare to actual change;

e" area;

Hld area;

( )&

1*.1 1*&.*1 =

( )&

1* 1**.** =


e"ton?s ethod

*1.&=A

&=dA Absoluteerror&

1**

&==

A

dA

*1.&1**

*1.&==

A

Apercenterror


80/97


e"ton?s ethodTrue Estimate

Absolute 7hange

Melative 7hange

ercent 7hange

)()( afdxaff += dxafdf )(6=

)(af

f)(af

df

1**)(x

afdf1**

)(x

aff


81/97


e"ton?s ethod *ewton&s Metho

* x

y

yf(x)

%oot

sou$ht

x1

!irst

(x1&f(x1))

x&


82/97

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.


e"ton?s ethod


83/97

*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.


e"ton?s ethod


84/97

*ewton&s Metho( ) &

10

&f x x=


e"ton?s ethod

xxf =)(6

Buessx1 &

)(6

)(

1

11&

xf

xfxx =

.&&

1&& =

=x


85/97

*ewton&s Metho( ) &

10

&f x x=


e"ton?s ethod

xxf =)(6

Buessx& &.

)(6

)(

&

&&0

xf

xfxx =

4.&.&

1&..&0 ==x


86/97

!ind "here crosses .0y x x= 1y=

0

1 x x= 0

* 1x x=

( ) 0 1f x x x=

( ) &0 1f x x =


e"ton?s ethod


87/97

(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


e"ton?s ethod


88/97

There are some limitations to e"ton?s ethod;

rong root found

Loo8ing for this root.

Nad guess.

!ailure to converge


e"ton?s ethod


89/97

!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


90/97

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


91/97

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


92/97

+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


93/97

,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


94/97

,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


95/97

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


96/97

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

Calculus Chapter 4

Documents