Math 118 Fall 2005 Final Answers

# Math 118 Fall 2005 Final Answers - Fall, 2005 Final Exam...

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Fall, 2005 Final Exam Solutions Page 1 1a When x is close to 2 the numerator will be positive. When x is less than 2 the denominator is negative and as x approaches 2 it will approach zero. Hence the limit will be negative infinity. 1b () ( ) 2 42 22 2 2 2 2 2 2 4 lim lim 2 2 lim 2 lim 2 2 2 64 xx x x x x x x x x x →→ −+ + + = +− ++ + + = =+ + + = 1c 11 lim ln ln lim ln(0) x x →+∞ →+∞  =   = =−∞ 1d 3 55 5 lim 420 4 t t e →+∞ == 2a 3 3 (3) 9 3 f 2b lim ( ) lim 4 1 9 fx x −− = 2c In order for f to be continuous at x = 2 it is sufficient to show that 2 lim ( ) 9 x + = and that (2) 9 f = . Exploring the limit, 3 8 lim ( ) lim 33 x . Since the limit from the right is not 9, the limit does not exist and the function is not continuous at x = 2 . 2d Since the function is not continuous at 2 it cannot be differentiable at 2 since every function that is differentiable at a point is also continuous there. 3a The value (200) 0.002 f = provides an estimate for the marginal of f at 200. Our estimate for the level of copper when the population is 210 thousand is: (210) (200) (200) 10 0.95 (0.002)10 0.97 milligrams per liter fff +⋅ =

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Fall, 2005 Final Exam Solutions Page 2 3b The estimated marginal of 0.002 indicates that an increase in population of one thousand would result in an increase of 0.002 mg per liter.
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## This note was uploaded on 09/26/2009 for the course MATH 118x taught by Professor Vorel during the Fall '07 term at USC.

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Math 118 Fall 2005 Final Answers - Fall, 2005 Final Exam...

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