2311_Test1PracticeSOLUTIONS_F11

2311_Test1PracticeSOLUTIONS_F11 - gow‘tteem Fall, 2011...

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Unformatted text preview: gow‘tteem Fall, 2011 MAC2311 Test 1 Practice Problems 1. A closed box must have width W = 12 inches, and its surface area must be S = 600 square inches. Write a formula that gives volume V as a single—variable function of length L. (333 gmmetry facts 01]'p,5) A Sezo~+aui+2ws VfiLWE- $2 2Lfii} +29% swim V: Maw HIV 'gzganutasawas VZEZL; we a as it, mam am we as s see m gas. =5 2. H < L +123 coo -=- 2% V: iZL< Zagreb) L wtllin a H i as w a; o saw-5m Lad” ‘2“) fl I V “m” he‘l' recs“; moi . m I WW Tags? g Answer: V(L) : i L E2“ 2. The graph of piecewise~defined function f is 3 shown for x 2 —4. Assume the first piece is Y 5 semi-circular and the second linear. 5, (a 5) 4 1 a) Complete the formula for f . 3 (4’3) 1 auto) Wham?“ if—‘iéxé‘t ‘ “WWW ————e s 1H 12 fix): Egg-"X44 if K>L§ » 28+ Ms M was Y; a (my (as? y \j 3%“ {WWW _ x1 ‘5! 351 fans“ _ w KWX WWW; M‘fic a Eo‘i‘i‘bm remlwclmivsw r. Emi- l yaw—3 hmi ” :52; y; mm ymg a: E1 x v m» .L a a i {3.3 II w 3 m 1)) Evaluate each limit: lim f (x): fig limf(x) x44" — 3: Sketch the graph of one possible function f(x) satisfying all of the given conditions. Label holes with both coordinates and asymptotes with their equations. x ; if; 0 domain off is (—oo,2)U(2‘,4)U(4,+oo) : - 332: male; f<x>=2 *: 4 twee-w and awe-{~00 : - 1};ng =1 “ 0 Graph is decreasing over the domain of f 0 Graph is concave down over (H003) U (2, 4‘) and concave up over (4,+oo) 4. Graphs of f and g are shown. Evaluate each limit. Showing work is not required. Graphoff ‘ _ ‘Graphofg a) 31332 f<x)= DNES b) 132m: —%~ w c) 133me I d) Elgg(x)= 2; 7 e) lim[4g(x)+5]= t) lim+ f(x) = Mia—m? x“ «93? g) Iim53/29+g(x) : 3 Jew->0 Y {3 x91 hfij W; 3—» 3i {E ‘ m b We _ 1 MW «- tisfi‘m 7;; “so w? i l . i E -. ' - 7 _ ‘\ mzwm Miami? miewgooo 3 ., "fieuk‘qm Jon juggmfl 2. EH 5. Find each limit by any method. Partial credit will be awarded only if the answer is supported by Workbut showing is not required. Circle each final answer. Val-M a) ENE—3 ‘W *3) Keo KW} Mlfifi‘mg‘lfifl “a? (See section 2.3 example 6) . _ a, r g. >4 aha «s; 3% :32; y m W a a a L Kemp K ‘2: \‘w' m e 3} K “a; wawme an+3w (see inverse trig summary on p.63) .431 . t We 4% MUC c) lin ace—x; gem C} Wmfifir M 91% “km”; a Xfim a iasuiee “’W‘aw‘» K “item ’ 6. Consider f(x)=—f:g—=mfl$ ) her}? CloMaax Kg; Elilx$a3g p x +x—6 ‘ (Wag—2) five a) Evaluate each limit: lim f (x): (egg. lim f(x)= é/g x4»? iim f(x)= @535“ xv~+73 b) Evaluate each 11m1t. xlirzn f (x) ‘5» OD f (x) Ob f (x) D l 0) Evaluate each limit: lim f ()5): lim f(x)= X214. xvi? ‘ {g a? x+w xa+w NW? ' m 7 fl d) The graph of f has a hole at the point( “*3 , git}? ). E; m a l i {a «pm W W” E e) The graph of f has vertical asymptote x = 2?» KW gm 5 X KW E «wt-1A we a $53M? \f :5 . f) Sketch the graph. Represent asymptotes using dashed lines, and label each asymptote with its equation. Represent the hole with a small open circle, and label it with both coordinates. _ is continuous. 7. a) Identify the x—axis interval(s) over which the function f (x) = 2 ' ' 4 —— x I m gsgmgaxa 7 must smegma. “that b) Identify the x—values Where f (x) = 1n(sin2 x) is discontinuous. Answer: mall’s?! ‘3 any .53??gi “gig? may ‘.a§W-viw~ §lnxe~rag 13;, {amalfiamfiéis s a mm k+cosx, ifx_<_0 8. Find the k—Value that makes f (x): 5Sinx _f >0 continuous at x='0. k: A; 1 t x M ' I _. im Ka-aasa a: lflw‘ve‘asfi a: gal ' l» Mm “xx-am” mar—5" 27> Kati-J amt aim? at: hm mg: a when a: 4;. Mama w”? 0+ X i ) gnaw 2' dials—a Ktiwso :Kaei g‘ mam K34) ¥{9)m g?” . f 3 ' WWW ii”) “M £0?) z: {(0) ‘35,) .Séff‘l’f’lg'gfiimfl filing dfi‘plhn’l*icym a? cominwéy >€m¥ C) 9. Consider f (x) = 1 In technology this is authored as 1/(1 + e A (1 / x)) . l+e%‘ I a) Evaluate each limit. mg} f(x)=© lim f(x)= V; ' I x—a ‘ x—>+oc a 12 : l l I a as “gum a . 7; _ was m.“ 3 1+ €4€gfieflm$ 3_ \ b) If the graph has any holes, idenify them by giving both coordinates. (.0! Q ing (o g l‘} I c) If the graph has any asymptotes, identify them by their equations. \[ 1::- V2. ' Invefse Trigenumetric Functions 1 y=si11‘-x"c>x=siny where ~13x51 and W§Sysgw y=cos“1x<:>x=cosy Where —1Sx$1 and ()3ny y=tan"1x<:>x'=tany Where -—oo<x<+<>o and ~§<y<§ y = 56071 x<2>x=secywhexexswlorx21andUSysyr,y¢ @Omejfi‘y Form Mag Rectangle with Length l. and Width Rectangular Box with Length L. Width W,l and Height H Area A = LW Surface Area A = ZLW +2LH + 2WH Perimeter P = 2L + 2W Volume V = LWH L. Triangle With Base b and Height 11 Sghere of Radius r 1 = 2 Area A = —-bh Surface Area A 4;” 2 4 3 _ Volume V m «53—17:? b P that mean Theorem C linder of Radius r and Hei ht h For a right triangle with legs a and b 2 2 Surface Area A = 22:?2 +2m'h and hypotenuse c, a +5 = (:2. w Volume V = 7:142}; W Right Circular Cone of‘Radius r and Height h Surface Area A = 71%.er +1112 . . h ' h I Volume V m a???" \‘zh 5 Area A m 7W2 Circumference C = 27:1“ ...
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This note was uploaded on 12/12/2011 for the course MAC 2311 taught by Professor Teague during the Spring '11 term at Santa Fe College.

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2311_Test1PracticeSOLUTIONS_F11 - gow‘tteem Fall, 2011...

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