Final Review - Chapter 4(Applications of Differentiation)

• Find critical numbers of a function.

• Find the absolute maximum and absolute minimum of a function on a closed interval.

Example 1: Find the absolute maximum and minimum of f(x) = xex/2 on [�3, 1]

Example 2: Find the absolute maximim and minimum of f(x) = x3 � 3

px on the interval [�1, 8]

1

t.IT#Ikl : chart

:p

.gg#2flf'=1.eN2+x.e42.tzZEI e

"

y T [ positive .

= e# ( 1 + tzx) = 0 definitely

close to - ( so Max

ljnrserthan TSo min

.

X= -2.

( Recall e×/2 -1-0 ) * Note fl→ confirms this conclusion 9,

since f '< o for ×< -2

and f '> 0 on × > -2 .

Chectf .

. -2 does lie in [-3/1] answer : maximum off on E3,D : ehminimum of f on [-3 , if : - %

1/3

Findcrit#'sforfC==zt×- × chart :

xf ' # to - text"3=± ( i - EB)

f=,|§t||¥|¥f 1=0 when 1 - x# 3=0 or

ftp.tz.3f fc 8) = 8g -2--85--23×43=1.

SoX=±l -

= 13+1=25flis undefined when to .

cvit # s : -1,0 ,I

. answer : maximum of fCx) on El ,8] is 43 .

chef : -1,0 ,1 all fall in interval

minimum of FCD on E 1,8 ] Is -43.

• Determine where a function is increasing decreasing.

• Determine where a function is concave up and concave down.

Example 3: Given G(x) = 5x2/3 � 2x5/3

(a) Find the intervals of increase/ decrease.

(b) Find the local maximum and minimum values.

(c) Find the intervals of concavity and the inflection points.

2

; a '= . kaE÷);G'' indeedcrit#I : f '=o when x=l ;

f ' undefined when # o ;G is increasing on ( 0 , 1) and

.. . . { + + + o . . . -

← sign of decreasing on C- a ,o)u ( 1,6 ) .EE fl- l o Yz I 2

- - - += - =, ±\ chewer : G101=0 is a local minimum

- . . { + + + o . - . -

← sign of G( 1) = 5-2=3 is a local maximum

EE fl- l o Yz 1 2

Gll , -1012¥ we see :

Using 9×413 Answer-

:tr

denominator- 1

G "=o when ×=- Yz

always #r ! G is concave up on C- x. E) and

concave downonftz ,X ).

G" undefined when X=o .

+ + to-.

. { - - -

n sign of(E)GKD Is an inflection point .

sit G"

t -

Yz it 0 1

T Thu

here,

inflection inflectionpt pt .

Example 4: Given g(x) =x

x2 � 9, g0(x) =

�x2 � 9

(x2 � 9)2and g00(x) =

2x3 + 54x

(x2 � 9)3find the following.

(a) Determine the intervals of increase/ decrease.

(b) Find the relative maxima/ minima and indicate whether it is a maxima or minima. Ifthere are no local exrema explain why.

(c) Determine the intervals of concavity.

(d) Find the inflection points. If there are no inflection points, explain why.

(e) State the vertical asymptotes for g(x)

(f) State the horizontal asymptotes for g(x)

(g) Sketch the graph of g(x).

3

- (

i.g)

← always -

doesn't factor✓ -

2×(1+17)calwayst .

tralwayst

g' Is never 0 .

g1(×)< o for all × 's in its domain . . .

g' undefined at

×=Iz answer : GCA is decreasing onto ,3)Ut3 , 3) v ( 3,6 )

and increasing never .

glx) has no local extrema because it is always decreasing .

g"=0 when X=0

. - - { + +

+0. . . { + + +

' - sign ofg

"

¥147g

"Is undefined when

- z0 3

×=±3 answer : g is ccaveup on C- 3,0 )U( 3 ,~ )

c cave down on C- x ,- 3) UCO , 3)

.

( 0,0 ) is the only inflection point .

X= 3,

X= -3

y=o

^r

*

• Solve max/ min optimization problems.

Example 5: A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide,by cutting a square from each of the four corners and bending up the sides. Find the largest volume thatsuch a box can have.

Example 6: Suppose a box with a square base and open top must have a volume of 32 m3. Find thedimensions of the box that minimize the amount of material used.

4

.

maximize Volume !

Edta!i -

±¥÷÷F¥¥→*.

+ + +0 - - - -

¥÷#i¥ itV=(3-2x)2x= @-12×+4×2)x ° Yz

1 3kImax at #

'12=9×-12×2.4×3vltt :( 3-2 .tt?tz=22.tz=2#V

1

( × )=9 .

24×112×2=3(3-8×+4×2)

=3 (3-2×74-4)=0X= } or tz 32

÷18

EIIEEI' Eleusis'¥¥¥±i:¥

¥' s 'Cx)=2× - 128×-2=0

dimensions : ZX = 1¥02 or ×3= 121--64=43

×jhiYuIt: |sox¥mt÷.

✓' 0 4

so S has a minimum @ X=4-

Example 7: A rectangular storage container with an open top is to have a volume of 10 m3. Thelength of the base is twice the width. Material for the base costs $10 per square meter. Material for thesides costs $ 6 per square meter. Find the costs of materials for the cheapest such container.

Suggested Probs: Ch. 4 Rvw. Pg 359 # 1, 5, 7, 9, 15, 17, 19, 27 & Pg 337 # 15, 19

✓ = 10 m3 base $ !/m2 V=l0=lwh=2w2h

=sides $6/m2

# h minimize costSo h=5w2

It.¥w c=|o.lw+6°(2lh)+6l2wh)]l=2wl=2w

=l0(2w)wtl2(2w)(5w2)+|zw.sw2h=5w2

C(w)=Z0w2tl2ow'

+60W '=2ow2t180wt

C 'lw)=40w - 18062=0 minCost : C(F%)

40W -- 18%2 or w3elETo=£ =zo(9z)%tl8%%w=F% fristhelocatim

of a minimum

- - - - 0 ttt

#013✓% 9