131f11x1ANS - Math 131 Exam 1 Solutions 10/6/11 1. (a) [8%]...

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Math 131 Exam 1 Solutions 10/6/11 1. (a) [8%] The average rate of change is: P (200) - P (100) 200 - 100 = 2000 / 200 - 2000 / 100 100 = 10 - 20 100 = - 1 10 lb / in 2 in 3 (b) [8%] The instantaneous rate of change is the limit: lim h 0 P (100 + h ) - P (100) h = lim h 0 2000 100 + h - 2000 100 h = 2000 lim h 0 1 100 + h - 1 100 h = 2000 lim h 0 100 - (100 + h ) h 100 (100 + h ) = 2000 lim h 0 - h h 100 (100 + h ) = 2000 lim h 0 - 1 100 (100 + h ) = 2000 - 1 100(100 + 0) = - 1 5 lb / in 2 in 3 2. [16%] f 0 ( x ) = lim h 0 f ( x + h ) - f ( x ) h = lim h 0 [( x + h ) 2 - ( x + h ) + 5] - [ x 2 - x + 5] h = lim h 0 [( x 2 + 2 xh + h 2 ) - x - h + 5] - [ x 2 - x + 5] h = lim h 0 x 2 + 2 xh + h 2 - x - h + 5 - x 2 + x - 5 h = lim h 0 2 xh + h 2 - h h = lim h 0 h (2 x + h - 1) h = lim h 0 (2 x + h - 1) = 2 x + 0 - 1 [by Direct Substitution] = 2 x - 1 1
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3. [16%] There are many possibilities. Here is one: M 4 M 3 M 2 M 1 1 2 3 4 x M 4 M 3 M 2 M 1 1 2 3 4 y 2
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4. (a) [5%]
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This note was uploaded on 11/22/2011 for the course MATH 131 taught by Professor Hall-seelig during the Fall '08 term at UMass (Amherst).

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131f11x1ANS - Math 131 Exam 1 Solutions 10/6/11 1. (a) [8%]...

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