MAC 2233 Exam #3

Name: !+Ntwv= K~

ID# SpOJ JD" HONOR CODE: On my honor, I have neither given nor received any aid on this examination.

Signature: _________________

Instructions: Do all scratch work on the test itself. Make sure your final answers are clearly labelled. Be sure to simplify all answers whenever possible. SHOW ALL WORK ON THIS EXAM IN ORDER TO RECEIVE FULL CREDIT!!!

No. Score 1 /20 2 /20 3 /16 4 /24 5 /20

Bonus

Total I /10

/ 100 I

1. A box has a square base and a surface area of 24m2. Find the dimensions that maximize the volume of the box. (20 points)

'2.

Y-= ;J.'1~'(

V-=-X'l v-4 (~~-~f-)

V-c X (;;1'1 -~~J

V - ~\{~ -;2.x 3

-­- 'i

V-= fox ­ t'l.~

V' -=- ~- ;X~

2. A local sports store sells 61 jerseys per week at $40 each. However, for each $4 increase in price, the sales decrease by 2 sales per week. It costs the store $2 each week for each jersey that sits on the shelf. W hat price should the store sell each jersey for in order to maximize their profit from selling jerseys? (20 points)

I- ( 101-)( \ lc" '-1-1(1';:\f" ~D+ ~ TJ) -;

x~ eol- 1 (f- lfD )OR l' = ~o +- Id.~ -.'2)<

1-=- l{,,~ - ;).)c

~2'( 4- '(.,2 -:::. f

1<='(-r ':: X(1~;2 -,?x) == 1(0'2)( _ :l~2

f= 1(o~-2'l- -:.2-~

f= IfoO-:1);2 f -=- IfgJ, - d..X

p' -=- "0 -~)< f ~ (&;l- ~(~,)

r-=- Ift,~ -<O~1(,0 -J~ x-=-O t'(~ ~ - fgal

~~¥ t\

)\~ 41

3(0)-;:)0 _ 0 - 0 ­ ~ _ -:...?Q. _ ± 00

0 - 0 - 0 - 0 ­

_ !.:2-p-;lO _ 0 _' ~ - ~ - G,~- tt-t - Ifo - 0 - 11\ ~1'\.Co

2.X+S' - .2('i)t~ ':;. ~ -:. ~ -=t ±r>o X(X+I) - 'fL4~1) "Ilr);W

3. Find the vertical and horizontal asymptotes of the function. (16 points)

- -

- J2 ­

f (x) = 2X2 - 3x - 20 x 3 3x2 4x

J.x"1.. - 3)( - ;).0 _ \i~ -(-0X~ - 3x'l. - Y)c: X~ tOo

'(~ - ?x.~- ~X:=:~ o

xCx.'l -~x - 4) ~o

X(X - ~) (~+\) ~ ()

X:;"0

2 'X..'2._ )'X - J.-O ;z.lo)" - 3 (b) - J-o _ :;" (0) ­

,,3 _3'('l_'{)t 0'3-3lo)"- ~lo) D-3lo)- \j(o)

rveri\~ o.Sr:..p-ln~ aJ- X~ tiM 1:X,L-3x- 'lO J.('i)"Z.- 3('i)- lO __ ~ (II.,) - r2-~O

'X....'\t ~~-3-t- '-\~ - I..t~-3{'i)1.- \f(\'{) - Col{-3(tt.) - t!o

\ilY\ ~-~X-';). O _ li~ (2.>q-~)~) \ iM

-X...."t x~ - 3x'2 - ~x. - )c.... 'fs )(~( ~+I) ::: X-''f ±- -

I tJ<> vtrtk.J c..5'1''''( to k oj )( ~ ~ }

\ 'I(V'\ 2x.'~ ,'K-;W .1l- l)"L ~3[-1) - ;2..0 _ ~(\) +3- :to _ c1.+-3 -~o- := ::..!!. ::: t-PQ X..:,- I!: x? - 3x'2-- '1'>c "=C-:: l)"- 3(-t)'2. - \.f(-1)-..... -f -~(I) +\{ - 1-> ~\.f 0

(b) f(x) = log3 (6x2 - 4x + 5)

t '()():: \ . (12X -4) · foX2 -L\~ +S" (~ '3

(c) f (x) = 4x2

- 2x+5

X~- 2.x+~

-f t(XJ~ ~ . (2X-d-) ·In ~

lfa

(d) f(x) = eJ2x - 4 ==- e (2)( -lf)

I (ZK \.f) 'fz \-f (X) :: e - . i(2~ _yyi (2)

/I 8(4x4-x• J (X) =

• Inflection points: undefined when x

l'~cr~SlJ : C-J,-l)

5. Sketch the graph of the function. (20 points)

J(x) = -4x3 + 4x2+ 20x + 12 = -4(x + 1)2(x - 3) x3 + 3x2 - 4 (x + 2)2(x - 1)

Hints:

• x-intercepts: (-1,0) and (3,0); y-intercept: (0, -3)

• Vertical asymptotes: x = -2,1

J'( ) - -8 x+I)(2x2-x+5) • X - (x+2 3 (x-IF

• Critical values: J' (x) = 0 when x = -1 and J'(x) = undefined when x = -2,1 "'(-1\ _ (1'I~)(t\~)(f05) =: 5

1 2-} - (Oo5) (('05) -poI ?O~ I I I

OIl neg -2 -1 ~ ~ neg • ) I(0) = C"lj) (r~) CfO~) =~ ~. lpos) (po.\)

3+27x2+29x+13) (x+2)4(x-l)3

Jff(x) = 0 has no solution and JI! (x) ­= -2, 1 " _ Cpos) (pes)

~-f (D)- (US\(t\lA).-__~~____+__~po_s~. ~ J neg

JV\lA..

-2

---- to\'\~-.;'(Lf : (L Od)

~~~ ~'. (-pa, -d-)V (- .21~kcraSo"j : (- tlD, -J)V (-(,I)V (I, (0)

IttJ ~ : (~lJO) \uc.J '/'tV.(\, '. V\0Wl

Ijli\ ) _ (f'\~)~\(rtf~) ::: ~~ =_00 Ii", .j'(~) ~ (~\(yos')(~':i) =~ :0 -DO

~-I)-~- flx - (r''')(''''j) J ~-J)I- (~)Ln~)

lim ~(~) = (n0)~~~ _ -. Ii'" t (~\ -= J~) (1)>5\ (~..,I :0 l"'" -= j.-Oo~4-2· (tm)(t\~) - n~ - - ()() k..,,\-- (f~)(P'~

5. (continued)

\ , 10 • t

I, I

I y l , 5

-10

hWV' - ~'lcl +\{'X~ +=U>lcH;l X.... t6:1

l; ~ o'('l - \{

_ ~~1 l{~z. 20)0; ~ lim ~ + ~ + X"S'" ): t...,t;oo

Y..'\ 3'l '" - + ~ - X~X1. 17'0)'10 /'()

-\.f f- ..i ZO ~ X + ~z + X')liM - '-i-- t =-~X-. tc:w "i- ----.141- X1.

~ ~ ()0

10

--

Bonus. Find the equation of the tangent line to the graph of f(x) =

ex2 - x In (x2 ) at the point (1,0).

2 7

+,'{X) = el -~'(2X - IJ ·1V\(l) ~ l -X, -1 '(2/)

~ ' (~) = e~':.. (21<-1) It> (X2 )+ (' l -. ( ~~

~\ (t\~ e' 1

-' (;;lh)-I) If\ (I~) t e' 1-

~I (-T)

:: e H

(;;'_1) lh [I) te H (J.)

:: eO (I) I~ ( I) +e 0

(J-) ~ ~ ~ \ 0 I

1~2Xlr'o

0'":. J (,) \--\:,

O~ ~-t· \:. - J, ~i

- ):-=-'0

I1~ J,( -:tj