Math 118 Fall 2006 Final Answers

# Math 118 Fall 2006 Final Answers - Solutions to Fall 2006...

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Solutions to Fall, 2006 Math 118 Final Exam Page 1 1a Simply factor and cancel the common factor in the numerator and denominator: 2 11 1 1( 1 ) ( 1 ) lim lim lim 1 2 1 ) xx x x x →→ −− + = =+ = 1b Multiply top and bottom by the conjugate of 3 2 + x , then cancel the resulting common factor on top and bottom: ( ) ( ) () 77 7 7 23 lim lim 7 72 3 7 lim 3 1 lim 1 6 x x x x x x + −+ + +− = + = + = ++ = 1c Divide top and bottom by 2 x and then take the limit: 2 2 22 2 2 2 2 2 lim lim 31 200 30 2 3 x x x →+∞ →+∞ = + + = = 2a We first find the limits of ) ( x f as x approaches 1 from the right and left, and then solve for k so that those limits are equal. We then repeat that process at e in order to solve for c : 2 2 lim ( ) lim 1 and lim ( ) lim ln , so if is to be continuous at 1 then 1. If 1, then lim ( ) lim( 1 ln ) 2. lim ( ) lim , so 2, or 2 xe fx x k x k f k kf x x x c e c e c c e =− = = = = − − = + + = −=

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Solutions to Fall, 2006 Math 118 Final Exam Page 2 2b This function is differentiable at 1 but not at e. If we look at the derivative of x at 1 x = we have a slope of 1 . The derivative of 1l n x − − is 1 x , so the slope on the right is also -1. These slopes match, so the function is differentiable at 1 x = . The slope on the left at e is then 1 e , but on the right the derivative is 2 x so the slope on the right is 1 2 e e ≠− . Hence the function is not differentiable at e . 3a Differentiate with respect to x , treating y as a function of x , and then solve for the derivative of y with respect to x : () 22 1, 1 3 2' 2 ' 0 2 2 ' 2 2(1) 1 '1 12 xy xx y y x y xy yy x yy x y y x y == ++= + ++ = += −− = + = =− + 3b 1 1
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Math 118 Fall 2006 Final Answers - Solutions to Fall 2006...

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