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Exam 1A Solutions

Exam 1A Solutions - Mt100 Calculus I October 3 2007...

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Mt100 - Calculus I Exam 1 October 3, 2007 Variation: A Solutions 1. (10) a. y = " 3 2 x + 5 2 b. y = 2 3 x 2 . (10) a. 3 " 4 b. Solving y = f ( x ) = x + 1 x " 1 for x gives f " 1 ( x ) = x + 1 x " 1 , so the domain of the inverse is "# ,1 ( ) $ 1, # ( ) = R " 1 { } and the range is the same, the domain of f . 3 . (10) a. " 2 + 1 2 = " 3 2 b. 2 + x 2 = " 3 x # x = " 1, " 2 4 . (10) a. 48 + 1 " 8 + 1 48 " 8 = 1 10 b. Greater: 10 20 30 40 50 60 2 3 4 5 6 7
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5 . (5) Since lim x " # + f ( x ) = lim x " # + 7cos x = 7cos # = $ 7 and lim x " # $ f ( x ) = lim x " # $ x 2 $ 2 # x + # 2 $ 7 = $ 7 , the two sided limit of f ( x ) at x = " exists, so the discontinuity is removable. 6 . (15) a. lim x " 5 x 3 # 125 x # 5 = lim x " 5 x # 5 ( ) x 2 + 5 x + 25 $ % & ( ) x # 5
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