1271Sol_final2

1271Sol_final2 - Page 2 1. Let f(x) = (31:2 + 551: - 6)3....

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Unformatted text preview: Page 2 1. Let f(x) = (31:2 + 551: - 6)3. Then f’(1) is equal to (A) 12 (B) 3(3 + 5 — 6)2 w 6+5—6f @(3 + 5 — 6)2(6 + 5) 144 . - L gm: 3 (3“*5""“é) (ex/1+?) 2. The tangent line to the curve 3; = 3:3 —- 2932 + 233 + l at the point (2,5) has equation (A) y—5=(3x2—4a:+2)(m—2) (B y=5m/2 @~—5=(12~8+2)($—2) )z—2=(l2—8+2)(y—-5) (my-5=—N$-m Page 3 _ 972-4 3' iflm3—8 (A) 0 (B) 1/2 x 'z‘ (x '2. 2;: gm ) + ) l (EH/3 NZ (x—2)( xl+2x + a) is equal to 4. Let f be defined by lat—2], 1fx<3 f(z)= (x—2)2, 1f35x$4 x—4, 1f:z:>4. Then f is continuous m VI ( g ) - (A) except atmz2; X ‘>3‘ B e et t =3; ‘ ()xcpaz (K {(x)‘ @except atx24; V “a 3* (D) except atw=3andrc=4; (E)exceptatz=2,m:3anda:=4. ah” f(x) K-> l" —. b ‘H ‘rH/ Page 4 . 5. Let f(x) = 2:53 —— 3:1:2 — 122:. Then f'(a:) = 6(z ~— + l) and f”(x) = 6(22: — 1). Then the absolute maximum of f (2:) on the interval [—2, 2] occurs (A) at :1: = —2 ~‘ t a: = ~1 \ \ I ’3’ i 4 i f ( at :1; = 0 v M kmmwmmmnwmwh.“WWW”...w MW,Wellnmfn w at {B = 2 \ l w , (E) nowhere ’ 6. The equation 7372313 — 5x312 —— 4y = 7 defines y implicitly as a function of z. Find dy/dm. 1495313 + 5y2 4 — 21332312 — 10.133; /‘(x 3 +7 94X" 1 ‘ ’ @ 5y2 -— 14my3 (7 J Cf .1m2y2 -— 102:3; —— 4 . I .. r “l. D lax t V __ (7 I T (c) * 5y2+14a=y3 <7 ’ d3 (7 O 21m2y2 — ley —— 4 if“ a” ‘a- v (D) (72722/3 - 5my2)/4 ' :1 (E)0 ; ZIXLJZ -/0x_J~7 (A) l Page 5 . a: 7. Suppose that f (as) is a function with first derivative f’(:z) 2 increasing on (A) (—00, 1) and [3,00) (B) [0,1) and [3,00) (C) (—oo,0] and (1,3] (D) (—oo,0] and (1,00) ——oo,0] and [3,00) 8. Let f(.'1:) : (a: + 2)e$. Then, there is at least one number 0 bet (B) 364 ~ (3/2)e (o) 364 + (3/2)e (D) 664 — 36 (E) 664 2 — 3w . (a: _ 1)2~ Then 15 + — — + ..+-_., -— 4 -—— . , 0 4 3 using the Mean Value Theorem, we can conclude that ween 1 and 4 such that f’(c) is equal to 10. Let flap) : fix/7152 +8dt. Then f’(2) = (A) 0 (B) 2 Page 7 11. The substitution ac = U2 turns f: tan fidm into 'dx: lea/6w ‘ e 2"; = CHM—5J3; (C)f\>/§§%utanudu X / j ) ‘2 (D) ff tanu du (E) ff 2u tanu du 12. Find the volume of the solid obtained by rotating about the m—axis the region under the curve y = from O to l. (B) W (C) 3777 (D) 27r (E) % Page 8 Hand—graded part 13.(16 points) a) If .732 + 3/2 = 2, find £1. 1x +zJ<7":0 => (7': —.X_ O‘L b) Find an equation of the tangent to the circle :52 + y2 = 2 at the point (1, —1). 'f[’0)75‘0'7~§‘f§/0 >\ f :‘5 cont/haw»; 0 _ {fa m MWWJW “M” 74m, @ gfu’ far}: 0 4/1! m r 9/106 n73, 1) f((s<): 5’WX >5 Page 10 15.(17 points) A particle moves along a line so that its velocity at time t is v(t) : t2 —t— 6. Find the distance traveled during the time period 1 _<_ t g 4. (7 TOM Guam away f Mew/055 Z ’ 75 ‘é'é : 0 (:5 {’12 : ¢I '+Zl{ 3. 1 4!: 9m _2 2.... w J + ++ , V f 4 f + ~a' 3 L’ A/ "s ‘7 fwm; d/f : f(é+1‘-{‘)M ifwfimm I 1 I 3 z .3 3 3 ’1 (¢£+§,é‘) mixing); a 3 I 3 z 3 : [\wflfiq -g-.L+J +5-5»..m / '2 Z 3 3 {Fly 1 3 - M e i 2 +17 — ‘ 5 qt I 3 ‘ We] Page 12 17.(18 points) Consider the function $2 132-4. f($) = We have —8m 8(4 + 3m2) f’l-T) = W and fill-T) = W a) Find the domain of f i b Determine the x-intercept and y—intercept of y = f ) c) Determine the horizontal and vertical asymptotes of y = f d) Determine the critical points, intervals of increase or decrease of f e) Determine the concavity intervals of f and points of inflection. ' f) j. Sketch the curve 9 = f A) x a 3: '2 ‘qu z / “'9 : j (*2! 00 l/avfrcuj Myra/fl»ch X; :2] s I?» b5 + 7pc: ) 2 00 \\E}M¥—%+¥ gaw'x {ilk/l?” \U ad; Vfi'z‘ A) X:0 +- 4» — "‘ ...
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1271Sol_final2 - Page 2 1. Let f(x) = (31:2 + 551: - 6)3....

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