MAC 1147 Final Exam Review

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1 MAC 1147 Final Exam Review nstructions: The final exam will consist of 15 questions plu::; a bonus problem. Some questions will have multiple parts and others will not. Some questions will be multiple choice and some will be free response. For the free response questions, be sure to show as much work as possible in order to demonstrate that you know what you are doing. The multiple choice questions will be graded partly on whether or not you circle the correct answer and partly on your work. So be sure to show your work for the multiple choice questions as well. The point value for y,ach question is listed after each question. There will be no bonus on this exam. A scientific calculator may be used but no graphing calculators or calculators on any device (cell phone, ipod, etc.) which can be used for any other purpose. The exam will be similar to this review, although the numbers and functions may be different so the steps and details (and hence the answers) may work out different. But the ideas and concepts will be the same. Additionally, you will be allowed to use the Trigonometric dentities and Unit Circle which have posted on my website, along with the formulas on the last page of this review. The formulas provided on the review will also be provided on the test but the Trigonometric dentities and Unit Circle will not. So if you wish to use either or both, you must bring them with you. The Trigonometric dentities and the Unit Circle may not be shared nor can they be used if they have any writing on them.

2 (1) Determine what transformations have been done to tbe function g(x) = Vi to get the function f(x) below. ndicate how you know. Then graph f(x). (8 points) f(x) = ij-x - 3 J -c.. d.ow~ ) CD rdi.tlo'" ~t 1-O.«.1~

3 (2) Determine what transformations have been done to the function g(x) = xl to get the function f(x) below. ndicate how you know. Then graph f(x). (8 points) 1 ~Q) ~ ;l. f(x) = ~ J X~ (4t1 \..31 rtfftui ~ ~ ~~S 10 \ \ 5-10 o

4 (3) (i) Find the vertical asymptotes, if any, of the rational function. (8 points - 4 points for the answer and 4 points for the steps) f(x) = x + 1 x -.,];2 (a) x = x=o, x= 1 (c).x=o, x=-l (d) x=o, x =-l, x=l (Ji) Give the equation of the horizontal asymptote, if any, of the function. (8 points - 4 points for the answer and 4 points for the steps). 3x 3 + 5x 2-7x - 4 f(x)= x2+4x+4 (a) y = 3 (b) y = -2 (c) y = 0 G None of (he above 1)t3r<.L n"..wwor:'" ~ o~~ ~ flo hlvi~ (")7""f4<ric V,m ~11\t>..ht'=:J.

5 (4) Graph thc function, (Hint: You will need the determine the domain, whether each point not in the domain of the function is a hole in the graph OT a vettical asymptote, intercepts, symmetry, horizontal or oblique aymptotes and locatwns where the function is positive or negative,) (8 points - 4 points for t he answer and 4 points for t he steps) (a) ; \; 0 10 ' - - 0,0 0 " 1 t''--. (c) (d) " 0 l 10 - ) if 0 - l ; v ; " ("" ~\,)l (.,. -1,)' :; 0 ~ t v - --\ tnj\t-"? ~ W",'W QS1 ",.. -, X-;.l, ~ \t:::' x-,~t', ~ J:;-O 0= ~~'" ("'-~ X.-:-0) ly\u \t -z-4 )<,-::'3, ~",\t~2 x:,.-'2. fw\u\t::. 2 ()on) Ujf"tl t'\~ -= ~ ~('t.t ~-..ahr ~ ~ ),.)~ ~-~.J ~~, ~-roj,t: yr:>o

6 (5) Solve the inequalit. (8., steps) y pomts each - 4 points for the answer and 4 points for the (i) x 5 + 4x 3 < 4.1:4 Q o,2) U (2, (0) (b) (-00, 0) U (2,00) (c) (0,2) (d) (2,00) XS- -~x~f~~~ LO K~( ~1. -~X +'~) 0 'i-~( X.-l}(~-1)LO x~:::o $~ro X~D ~\t;1 };.-1-=O!1 +2 k-=-l. hu \t-=-f L(? 0 ~~ ~ r 0.,l1l W ~~ C~-1) ~O ~(O,~) U{l/ txj ) (ii) x3(x-5)(x-2) (x+l)2(x +2) < 0 (a) (-2,0] U (-2, -1) U (-1,0) U (2,5) (c) (-2,-1)U(-l,O]U[2,5] (d) (-oo,-2)u(0,2)u(5,00) x." ~ x.,-'f-=-o x-t~o.21.! ~~ ~ ')1. Ck~l-) ~ J& \f ::00 X.\.j=-'O - -l ~ -:...~l. ~. ;;.:<JO ~v\t ~ ~~ Mv\t'";;1 ~=-),""vlt::\ ).-:; z.. 'Nlt:; luou J ~ ~~;h' f ~ (-2-,-)V t"-'j o)v(2$)

7 (6) For the given functions f and g, find the composite function (f 0 g)(x). (8 points - 4 points for the answer and 4 points for the steps) f(x) = x 2-3x + 2; g(x) = x 2-6 (a) X4-3x 2-16 (b) X4-3x (c) X4-6x x 2-12x - 2 ex'-15x (Jc,.)Y-J(jC>lJ ~ 1. - ~1oY -"3(/-0)+2 x~ - f2x\-3 b - 3 l t (8 t 1 X'4 -(~~ +-(10

8 (7) Find the inverse of j. State the domain and range of f and f-l. (8 points - 4 points for the answer and 4 points for the steps) f(x) = ~ x-4 (a) f(x) = ~ X~4 (c) [(x) ~ -,,,:, (d) f(x) = - x~4 "\,x ~ X-~ (~4\F tfj1r4 &jj ~ ~y 4~ X- \.f

9 (8) (i) Solve the equation. (8 points - 4 points for the answer and 4 points for the steps) 2 2x _ 3. 2x = 0 (a).1: = 2 (b).1: = = 2,3 (cl) No solution ( e ) None of the above ~1; ~J~ \.1. ~ ~ 2"t~ _ \1. 2.'t. f32'::0 z'x -::..~ u/\.{ 2 - ::.y '( 1\ ::-{,?, '1 \.A~~ ~~ lo,~~ L U _ \~~ -\-32 :::..0 X>3 (~_~\ (~-'i):;v ~~2. (ii) Solve the equation. (8 points - 4 points for the answer and 4 points for the steps) log3 (.1: + 10) = 2 - log3 (.1: + 2) (a) x~ -l1 Q X ~-1 (e) x~ -l1,-1 (cl) No solution 11l')~ (H.}) =0 ;Ho,} Cx J) \-"'i~ (V"') ~(f 3 ( t;l) [tij~ [X tf~ 1- ~J (ltt-))>j. \OjJ~1b)(U-~\ ~ J.,,~+-["l \ +-'1:;'0 (x.vll} (X H):::' 0 x+j-=o X-k'::: 0 J:1L r-- \ 'X.:: -H ~:~ 1 (X H\)) (~t J) ~ 3 k2u~~1=j

10 (9) St ate the amplit ude, period, and phase shift of t he function. Graph t he fu nction. Be sure to label t he intervals on the x- and y-axes and show at least two cycles. (8 points) y = - 2 cos ( 7r X + ~) ~rt~\t~~ oaoavt- ~-~\> ~ ( V 10.. "\, / ~ J \ J., r\ [\ L ~ \ V,, -:t..j 2 l\ ) l \.. \1\ 1f. JZ. '(1 3[1 " \ J k"l - \ V \ V ~ ~\i\ Vha.~ 5~,h-"-~. ~ 171r=l[;:, "f, t~t-l zt 1 - "') )f'.. '

11 (10 ) Establish the identity (8 pomts). 1 + sin.x 1 cos 2 x -1 =- esc x \+sk"x. { -~~\.'( -\ \ - (!-5\"~ \ - t\r\ X ~A-+S~X \-bl"-.)c -

12 (11) Find the exact value of sin( a - (3) under the given conditions. (8 points - 4 points for the answer and 4 points for the steps) tsj.,..,j E 10 31f. 10 ~ sec ex = 6 ' 2 < ex < 21f; S111 [3 = 26 ' "2 < [3 < 1f 61 (a) ~~ (b) -~~ ~ (d) -130 Q None of the l:1above.:::,.. i' S\f'rJ.. ': - To CoSrJ..~ 1D \Q \ ~L :;(b t 10 1 t-b 1 :; 7.L 2 2'{ o.'v V~~ ::. (00 a'l. :'--{g"1 ~ -=--~ ~i" (~-i):o slw..,l Ca; f - Co~.,("''''r tjo)(-~) -( ~oj({t) (Db -{,"L ::: ~"}<O bl':.~~ b;."t\{ ~_ l3 - JiL - ~o tho?,o l~l Lc,O t';d - c,s"

13 (12) Find the exact value of each expression. (8 points each - 4 points for the answer and 4 points for the steps).. _ " (1) cos (cot- 1 i~ - 8m 1 1~) (i) 2 (~b) 608 ( ~ ~ ~. 'tt e- ~ d..:- cot.- N ~f t~~ ~, l\{tti,tj.. ~t~ {~(,r l30~ :;' (1. ~S1>O=-t1. t~51> cj... -:::: ~v...,f E:,-.../'-- (ii) cos [2 sin -1 ( - 1~,) 1 _ ( -~\ f,, -d \ - - CPSltoi (~) COSL~"\ \:tl} C)\~lto () 608 (d) 87 ( ) C e ~ ~uff ~ ~~i'~- ' n. ~~r \~~ ~ S'L+-t > (1'2 j.~s" t'j,'j... ~:a~ bl-=-l..~ b-::::-g ) ~ f,.t-' ~ t ~ """""...!! (( Nfl one 0 above t 1e >"'.t.l' J J '- ~ [. r:;-\ C)(...'=- <;,\ ~ S1. t- \,").. ~\~ 1 1,,5' ~b ~ ::: \(0 ~ (a) 1 (b) ~~ \~ \? j, Of' - 11 J ~~ ()C (e) None of the above (),s LJ.L}-= cm2.. rj..- c,\,..l q.. :0 (!JjY- ~)' \U,-\ 1..1) ~ -- l~" \\~ -=-~ ((,'\ \ 'L ~ l'1 ~ ~ \1

14 (13) Solve the equation on the interval 0 :; B < 27f. (8 points each - 4 points for the answer and 4 points for the steps) (i) 1C 21C 1C 51C ( ) ( ) a 3' 3" c 6'6 S~ t'\&~~{ (ii) Solve the equation on the interval 0 :; B < 27f. (6 points - 3 points for the answer and 3 points for the st.eps) sin (4B) = 1 (a) ~ (b) ~ (\ (c) ~ 51C 91C 13". ~ 8' 8 ' 8' 8 (d) 31f 7". 1l7r 151f 8' 8' 8 '8 f..\.- ~ n-=--o ~ v - ~ \{s- -== ~\n-' (\) v\-:;.. : e ~ 1'r t - g + & - ~ 1r lc- Jt 1!L- Dr f'1. _.ll+ 11 -= lr+!!t: -= ~ 'T' n-;>-2 ~ v - '" "" ~ \ r lll ~ "t+ 12 rr ::::..rur- V\ -:;;. ~, '9:: g t"t. f $ ~

15 (iii) Solve the equation on the interval 0 ::; x < 27r. (6 points - 3 points for the answer and 3 points for the steps) sin e + sin (2e) = 0 (b) 0 7r 7f 57f (c) ~ 37f 57f 77f, '3' 3 2'2'6'6 7r 37r 7r 57r ( d) of the above Co'> &:. - "i \ f7= ~ ~~ ~~S&::~ los e~ - \. (iv) Solve the equation. Give a general formula for all the solutions. (6 points 3 points for the answer and 3 points for the steps) (a) e = 2; + 2n7r, e= 4; + 2n7r (c) e= ~ + 2mr, e= i + 2nn sec (4e) = -2 Sic l'1.\1\~-2 c.o~ (~G)~ - ~ ~G- ~ Cbs' t- ~')

16 (14) Find the sum of the series. (8 points each - 4 points for the answer and 4 points for the steps) (i) L20 3 (k + 3k + 2) =- ttli,fod + <O~O H{Q:: \(41~u k=l "00 L 2 -:?.l1,())~l{o ~~ (ii ) (a) 400 (e) None of the above q l-r --

17 (15) Expand the expression using the Binomial Theorem. (8 points - 4 points for the answer and 4 points for the steps),,-.-v --., '",. (2x + 4)6 \f\t-\ -3J l \,,? l.~ \ 1. \ 64X x x , 240x , 360x , 288x ".,'} ~ \ ~ 3 1 (b) 2x x x , 240x , 360x , 288x + 4 t'~~ / '1 6 't, (c) 2x x x x x , 288x + 4 r\~~-...) \ ) to 10.s- (cl) 64x x x x x , 288x \-:.-c, -, C:; S- 20 rs- lo \ ~K-+ti) ::: \. (1)C)\. ~o + ~{1:~.)S_ ~ +(~{'l.x)~. ~1.+ ~O{"b)' ~3 +-1) (b.y V'1 +- ~ {Zk)' ttrt"{lk.)~y" -:: \ -(to~)c(,) t+ (.,62~5). ~ + ~(ll,~ ~).,~ +Jt,(~~}~~ tis"{'b:')ozslo «'{1.\) tol~ t-ol o ~o~b

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