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Exam 2 Practice 2 - i I f f(x =(U4)x4 213)x3 5/2)x2(x 6 x 5...

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i. If f(x) =(U4)x4- (213)x3 + (5/2)x2 - 6x + 5 then f '(x) is: a) x3 -2x2 +5x -6 b) (l/4)x3- (2/3)x2 + (5/2)x- 6 c) 1,6x3-9x2 + 4x - 6 d) 3x3-2x + 7 e) (L/4\xa-(2/3)x3+(5/2\x2 - 6x + 5 Z. If f(x) = xtet" then f '(x) is: a) 2xes' b) zxes' + x2e5* c) sxes" d) 2xes' + 5x2es* e) 10x2es* 3. If f(x) = ln (3x2 + 7 + eu) then f '(x) is: a) tn( 6x + 2e2") b) (6x + 2eu\ ln(3x2 + 7 + eb) c) 6x + Zeb d) (6x + 2e2)/(3x2 +7 + eu\ e) 1/(3x2 + 7 + eh) }' 4. If f(x) = et' , then f '(x) = l - x l - x 1 - x ' 1 , - 2 x L - Z x 5. If y = f(x) = x ln(2x), then y'= f '(x) is: 6. If y=f(x) = {tf * (Zx)'l = [1 + (Zx)']u', then y'=f '(x) is: 7. Theequationof thetangentlinetothegraphof f(x)=2x2 -x+10 at x=0 is: a ) y = - x + 1 , 0 b ) y = x - L 0 c ) y = 1 0 x d ) y = 1 0 x - 1 e ) y = 1 0 x - L 0 8. Suppose f'(x)=0 when x=2 andinaddition f"(x)>0 for x<4 and f'(x)<0 for x>4. Given this information, the graph of y = f(x): d) has a local minimum at x = 4
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0. Suppose f(x), f '(x) and f '(x) are all continuous with the following additional information: f '(3;=6, f'(-z) =0, f '(-1)=0 and f"(-z)=0. Furthermoresupposethesignsof f '(x) and f '(x) are given in the charts below: f '(x) > 0 0 f '(x) > o 0 f ' ( x ) < 0 I -2 f " ( x ) > O 0 f " ( x ) < O 0 f " ( " ) . 0 t l -2 -1 Given all this information the function y = f(x) has: a) Alocalminimum at x=-2 andinflectionpoints at x=-2 and x=-1 b) Alocalmaximumat x=3 andinflectionpoints at x=-2 and x=-1 c) Alocalminimum at x=-2, alocalmaximumat x=3 andnoinflectionpoints d) A local maximum at x = 3 and only one inflection point which is at x = -2 e) A local maximum at x = -2 and only one inflection point which is at x = -1 1. Suppose f(x) = ;3 + (3/2)xt - 6x + 36 is defined only on the dosed interval -4 3 xS 2. Then for f(x) defined on this interval: a) The global maximum of f(x) occurs at x = -4 and the global minimum occurs at x = -2 b) The global maximum of f(x) occurs at x = -4 and the global minimum occurs at x = 2 c) The global maximum of f(x) occurs at x = -2 and the global minimum occurs at x = -4
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