f09fea - 92-131 Final Exam Answers Fall 2009 Brent Sec...

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Unformatted text preview: 92-131 Final Exam Answers Fall 2009 Brent Sec. 203 & 207 3) y′ = e−4 x (3 cos(3x) − 4 sin (3 x)) 1) f ′( x) =e x sec2 (e x ) t 4 + 3t 2 + 2t (t 2 + 1) 2 2) s′( x) = 3x 2 1− x 6 4) g ′(t ) = 5) y′ = −9 x − 4 ln x = − 9 ln x x4 6) f ′( x) = ln x 2x 7) f ′( x) = lim h →0 [( x + h) 3 + 1] − [ x 3 + 1] f ( x + h) − f ( x ) x 3 + 3hx 2 + 3h 2 x + h 3 + 1 − x 3 − 1 = lim = lim h →0 h →0 h h h = lim 3hx 2 + 3h 2 x + h 3 = lim (3x 2 + 3hx + h 2 ) = 3x 2 h →0 h→0 h 8) dy 2 x y 2e x y −1 , y = −2 x + 3 = 22 d x 1 − 2 x 2 ye x y −1 22 9) a) v ( t ) = 16t − t 3 b) t = 4 1 c) S ( t ) = 8t 2 − t 4 + 2 4 10) Let f ( x) = e a) b) c) d) e) − x2 / 2 . f increases for x < 0 , and decreases for x > 0 . f (0) = 1 is a relative maximum value. f is concave down for − 1 < x < 1 , there. For x < −1 U x > 1 , f is concave up. (±1, e −1/ 2 ) . Horizontal asymptote y = 0 . There are no vertical asymptotes. 11) r = h = 10 m 12) a) xn+1 = xn − 3 xn − 3 xn 2 3 xn − 3 b) x2 = 2 x13 2(9 / 5) 3 729 = = ≈ 1.735714 2 2 3 x1 − 3 3(9 / 5) − 3 420 ...
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This note was uploaded on 02/13/2012 for the course MATH 92.131 taught by Professor Staff during the Fall '09 term at UMass Lowell.

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