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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 $2 2Lﬁi} +29% swim V: Maw HIV
'gzganutasawas VZEZL; we a as it, mam am we as s see m gas. =5 2. H < L +123 coo = 2% V: iZL< Zagreb) L wtllin a H i as w a; o saw5m
Lad” ‘2“) ﬂ I V “m” he‘l' recs“; moi
. m I WW Tags? g
Answer: V(L) : i L E2“
2. The graph of piecewise~deﬁned function f is 3
shown for x 2 —4. Assume the ﬁrst piece is Y 5
semicircular and the second linear. 5, (a 5)
4 1
a) Complete the formula for f . 3 (4’3)
1 auto) Wham?“ if—‘iéxé‘t ‘
“WWW ————e s 1H 12 ﬁx): Egg"X44 if K>L§
» 28+ Ms
M was Y; a (my (as? y
\j 3%“ {WWW _ x1 ‘5! 351 fans“ _ w KWX
WWW; M‘ﬁc a
Eo‘i‘i‘bm remlwclmivsw r. Emi l yaw—3 hmi ” :52;
y; mm ymg a: E1 x
v m» .L a a i {3.3
II w
3
m 1)) Evaluate each limit: lim f (x): ﬁg limf(x)
x44" — 3: Sketch the graph of one possible function f(x) satisfying all of the given conditions. Label
holes with both coordinates and asymptotes with their equations. x ; if; 0 domain off is (—oo,2)U(2‘,4)U(4,+oo) :  332: male; f<x>=2 *: 4 tweew and awe{~00 :  1};ng =1 “ 0 Graph is decreasing over the domain of f 0 Graph is concave down over (H003) U (2, 4‘) and concave up over (4,+oo) 4. Graphs of f and g are shown. Evaluate each limit. Showing work is not required. Graphoff ‘ _ ‘Graphofg a) 31332 f<x)= DNES b) 132m: —%~ w c) 133me I d) Elgg(x)= 2;
7 e) lim[4g(x)+5]= t) lim+ f(x) = Mia—m? x“ «93? g) Iim53/29+g(x) : 3
Jew>0 Y {3 x91 hﬁj W; 3—» 3i {E ‘ m b We _ 1 MW «
tisﬁ‘m 7;; “so w? i l . i E . '
 7 _ ‘\ mzwm Miami? miewgooo 3 ., "ﬁeuk‘qm Jon juggmﬂ 2. EH 5. Find each limit by any method. Partial credit will be awarded only if the answer is supported
by Workbut showing is not required. Circle each ﬁnal answer. ValM a) ENE—3 ‘W *3) Keo KW} Mlﬁﬁ‘mg‘lﬁﬂ “a? (See section 2.3 example 6) . _ a, r
g. >4 aha «s; 3%
:32; y m W a a a L Kemp K
‘2: \‘w' m e 3}
K “a; wawme an+3w (see inverse trig summary on p.63) .431
. t We 4% MUC
c) lin ace—x; gem C} Wmﬁﬁr M 91% “km”; a Xﬁm a iasuiee “’W‘aw‘» K “item ’ 6. Consider f(x)=—f:g—=mﬂ$ ) 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 «wt1A we a $53M? \f :5 .
f) Sketch the graph. Represent
asymptotes using dashed lines, and
label each asymptote with its
equation. Represent the hole with a
small open circle, and label it with
both coordinates. _ is continuous. 7. a) Identify the x—axis interval(s) over which the function f (x) = 2
' ' 4 —— x I m gsgmgaxa 7 must smegma. “that b) Identify the x—values Where f (x) = 1n(sin2 x) is discontinuous. Answer: mall’s?! ‘3 any .53??gi “gig? may ‘.a§Wviw~ §lnxe~rag 13;, {amalﬁamﬁéis s a mm k+cosx, ifx_<_0 8. Find the k—Value that makes f (x): 5Sinx _f >0 continuous at x='0. k: A; 1 t x M ' I
_. im Kaaasa a: lﬂw‘ve‘asﬁ a: gal '
l» Mm “xxam” mar—5" 27> KatiJ amt aim? at: hm mg: a when a: 4;. Mama w”? 0+ X i ) gnaw
2' dials—a Ktiwso :Kaei g‘ mam K34) ¥{9)m g?” . f
3 ' WWW ii”) “M £0?) z: {(0) ‘35,) .Séff‘l’f’lg'gﬁimﬂ ﬁling dﬁ‘plhn’l*icym a? cominwéy >€m¥ C)
9. Consider f (x) = 1 In technology this is authored as 1/(1 + e A (1 / x)) . l+e%‘ I
a) Evaluate each limit.
mg} f(x)=© lim f(x)= V; '
I x—a ‘ x—>+oc
a 12 : l l
I a as “gum a . 7; _ was m.“ 3
1+ €4€gﬁeﬂm$ 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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 Spring '11
 Teague
 Calculus

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