final_2009_sol

# B 2 points how fast was the ball thrown solution h t

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(b) [2 points] How fast was the ball thrown? Solution: h 0 ( t ) = - 7 2 t + 14, so h (0) = 14. Therefore, the ball was thrown upward at 14ft/sec. Page 4

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Math 171: Final Exam (c) [2 points] What is the acceleration due to gravity on the alien planet? Solution: h 00 ( t ) = - 7 / 2, so the acceleration due to gravity is 7 2 ft/sec 2 (in the downward direction, of course). 9. Let f ( x ) = 3 x + 1. (a) [6 points] Express f 0 ( x ) as a limit. Solution: By the definition of the derivative: f 0 ( x ) = lim h 0 p 3( x + h ) + 1 - 3 x + 1 h . (b) [6 points] Using the limit definition of the derivative, find f 0 ( x ). Solution: f 0 ( x ) = lim h 0 p 3( x + h ) + 1 - 3 x + 1 h = lim h 0 p 3( x + h ) + 1 - 3 x + 1 h p 3( x + h ) + 1 + 3 x + 1 p 3( x + h ) + 1 + 3 x + 1 = lim h 0 (3( x + h ) + 1) - (3 x + 1) h p 3( x + h ) + 1 + 3 x + 1 = lim h 0 3 x + 3 h + 1 - 3 x - 1 h p 3( x + h ) + 1 + 3 x + 1 = lim h 0 3 h h p 3( x + h ) + 1 + 3 x + 1 = lim h 0 3 p 3( x + h ) + 1 + 3 x + 1 = 3 p 3( x + 0) + 1 + 3 x + 1 = 3 2 3 x + 1 . 10. [12 points] Sketch a graph of y = (2 x - 1)( x - 1) ( x - 2)( x + 1) . Find and label the horizontal and vertical asymptotes and all intercepts. Solution: The numerator is zero when x = 1 / 2 , 1, while the denominator is zero when x = - 1 , 2. This tells us that the x -intercepts (zeros) are x = 1 / 2 , 1 and the vertical asymptotes are at x = - 1 , 2. Page 5
Math 171: Final Exam A rational function can only change sign at zeros and vertical asymptotes, so we can determine when the function is positive or negative by considering the inter- vals ( -∞ , - 1), ( - 1 , 1 / 2), (1 / 2 , 1), (1 , 2), and (2 , ). By checking points in those intervals we find that it is +, - , +, - , and +, respectively. To find the y -intercept we just evaluate the function at x = 0, at which point it’s - 1 / 2. For the horizontal asymptote, lim x →∞ (2 x - 1)( x - 1) ( x - 2)( x + 1) = lim x →∞ 2 x 2 + smaller order terms x 2 + smaller order terms = lim x →∞ 2 x 2 x 2 = lim x →∞ 2 = 2 .
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