mac 2233 final exam review no graphing calculators or

MAC 2233 Final Exam Review

Instructions The final exam will consist of 10 questions and be worth 150 points The point value for each part of each question is listed in the question and the total point value of each question is listed below Only a basic calculator or scientific calculator may be used although a calculator is not necessary for the exam NO GRAPHING CALCULATORS OR CALCULATORS ON A DEVICE (SUCH AS iPOD CELL PHONE ETC) WHICH CAN BE USED FOR ANY PURPOSE OTHER THAN AS A CALCULATOR WILL BE ALLOWED The concepts and types of questions on the final exam will be similar to the previous tests previous reviews and this review although the numbers and functions may be different on each question on the exam The questions on the exam will be taken from the previous four exams Each of the 10 questions can be found in the following places

1 Exam 1 Problem 2 or Section 15 (18 points)

2 Exam 1 Problem 6 or Section 24 (16 points)

3 Exam 2 Problem 1 or Section 25 (8 points)

4 Exam 2 Problem 2 or Section 26 (12 points)

5 Exam 2 Problem 5 or Section 31 amp 32 (24 pOints)

6 Exam 2 Problem 6 or Section 33 (8 points)

7 Exam 3 Problem 4 or Section 43 amp 45 (12 points)

8 Exam 4 Problem 1 or Section 51 (16 points)

9 Exam 4 Problem 2 or Section 52 (16 points)

10 Exam 4 Problem 4 or Section 54 (20 points)

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1 Find each limit If a limit does not exist explain why it does not exist (6 points each)

(a)

- ~ - 10 ~ I-J1 -5 Ll

(c) xIIm--shy

x-gt-2 (x + 2)2

)( shylim - ~ -J -00 X~-2 - tnY pos ~

lif) X- - 1~Alj -00( - 2+ ~t2)L ~

vx + 3 - 2lim shyx-tl x-I shy

X~ - Y ~ ~+b -gt ~~ - IIM shy iV~ - - shyX~ ()- I) Ch+ t))~ti -t)) -- (

~

-- i M

x- Jkt3 +-~ ~+-~ fit-dshy

(b) x2 - 6

I1m x-gt - 4 X - I

fjH - J Jl-~ _ ~_Q-() 0 0l -

INkJuMI~

kil r 4X ~(JiH +l)

- -L - j+L - [l

2 Find each derivative (8 points each)

(a) h(t) = (t3 - 4) (4t 3 - 2t2 + 5)

(b) f(x) = ~=~

t (l() (~lc -5) (3) - (3 ( -SVI) (~X s)2

- 12x- l~ - (2( -W)

~ y-s-)

- ~- IS--0 +-010

(h- S-Vmiddot

3 Find slope of the tangent line to the graph of f(x) = Jx2 - 6x + 9 at the point (52) (8 points)

-f (X)-= (--Gx+4)

-r (X= ~ (~-Co x-r~( (2K -iJ)

-f (~) ~ (S-- s +1r i (2S-- ~)

- ~ (~~-30+~)-~

(10 -Co)

- ~ (yr t y

l --~fi

- y - ~ d-

i --

~

--- ill

4 (a) Find the second derivative of the function h(3) = 33 (33 - 43 - 3) (6 points)

~t~)~Cos5-I(os~- (os

[ (s )= 305~ -lfgs~- Co]

(b) Find the third derivative of the function f (x) = 2x5 + 4x3 shy

2x (6 points)

~(~~ Jfb +~3 -2x

f (X) ~ lOx4 +12 xz - 2

til (X) 4Dx ~ +Jlx

(a) Find the critical numbers (6 points)

-f(X)~ lt~-(~)-2X-)j)~) =- Dlx~-(qX4-32l _ 2x 4 -gyenyUZk

(4~)l- ~kll l(k-l

_ K-4+32~ _ W(k 3T9) X3T~ ~ (XJ()(X2J~X~) X34(ll - J4xgtf3 bull )(3

-f (K)~~tJ middotJ0=fO (x=oJ

(b) Find the open intervals on which the function is increasing or decreasing (6 points)

m pos y-

O

r ri(k x --3 +(-3)~ g =pos

II VIV 1( - I + t (-I) -f J lIT fiG~ x~ I + I (1) yen pOs

flLreASi1 (~(XJI - d) U(0) (0)

JtcCUl s ~ (- i ( 0)

(c) Find the local maximum and local minimum values of the function (6 points)

middot((x) ~~ t1gt~ -b ~Vt of x~ -J So tellt) ~s~

lVtr~i1 -t ckc((~~j oJ-- X--() 80 ~ jigt 0 1mJ ~ at x-=--J

~ (-ir= a(-l~ -g _ _ ~(-8) -~ - - rfa--lt6 _ --y ~ _~ J[-d-)Z - ~(-t) lt6 ~

ilurJ VWN( =(-Jr -3)J

)lt 00 i 0 ~CJ 1l~tfM 50 + ~ I c I~ rgt4Nt or [~VI

n~ MM~v~] (d) Find the extreme maximum and extreme minimum of the

function on the interval [-4 -1] (6 points)

txtrl~ ~ (-)-~)

~+rtJN ~ (- t - )

6 Find the inflection points and the open intervals on which the function f(x) = x4 - 6x3 + 12x2 + 7x - 6 is concave upward or concave downward (8 points)

f II (X-) =- 12xz - 3(0)( +-~~

I~ [X- 3X tJ)

~ 12 (X -02)C )(-1)

t~(X-~) (X-I D

1 ict 1lt0 - I (0)= (po5)(~)(JJ-= rs Jr f Pid x -f II ()~ CfOS (j)(peS) =-J ][ Piel x= 3 middotfl C3 - (FJ(pos)Ceos) ~p05

CiNCQvt ~ (-pound)0 J ) u((l ~)

~CAve ~ ()

7 Differentiate the function (6 points each)

(a) f(x) = in (V4x + 6)

f(Xk I~ ((l~Ho) I~)

I - --~ y -- i 4tH ~ ttt~

8 Find the indefinite integral (8 points each)

(a)

J(4x3 - 5x + l)dx

~J X~~ - s JXrk +JItNx

i(4) -s(tl H +C

[ X~- ~X~rltCJ

(b)

9 Find the indefinite integral (8 points each)

(a)

(b)

_ l

-~ - ~

JA ~ 3 Lol

Ju oI~ Ju-s)~ _ -l

JL +c -y

10 Evaluate each definite integral (10 points)

(a)

3LgttoW -Dd0lt

i(~)- k I_~

X~-llt I [l -21-( I~-Il

~[~- J1-[H1

(b)

13

(y + 3)2dy

UCJ+S ~ iu ~

~r 3 Ct - ~-t 3 ~

d-=-O U=-o --1~ ~

r uk =- ~ 1 -=- -Co 3 O~ 3

dlb ~+ - ~-3

~t - q ~3]