111FinalSpring2008-Solutions

# We also know that f t 1 t when t 0 because the

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We also know that f ( t ) < 1 t when t > 0. Because the function 1 t = t - 1 / 2 is decreasing , we know that its largest value on the interval [4 , 4 . 5] occurs at the left endpoint, and thus this largest value is 1 4 = 1 2 . Because 4 s 4 . 5, this means f ( s ) < 1 s 1 2 , so E (4 . 5) = f ( s ) 8 < 1 / 2 8 = 1 16 . 10

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MATH 111, Spring 2009 Final Exam Solutions Therefore f (4 . 5) - L (4 . 5) = E (4 . 5) < 1 16 , so f (4 . 5) < 1 16 + L (4 . 5) = 9 1 16 . Thus 9 f (4 . 5) < 9 1 16 , so f (4 . 5) lies in the interval 9 , 9 1 16 ) . (c) P 2 ( x ) = f (4) + f (4) ( x - 4 ) + f (4) 2 ( x - 4 ) 2 = 7 + 4( x - 4) + 1 3 ( x - 4) 2 . (d) Let x > 4 be fixed. Then by Taylor’s Formula , we know the error E 2 ( x ) = f ( x ) - P 2 ( x ) in our order 2 Taylor approximation centered at 4 of the f ( x ) is E 2 ( x ) = f ( s ) 3! ( x - 4) 3 = f ( s ) 6 ( x - 4) 3 for some number s with 4 s x . Since s > 0, we are given that f ( s ) < 0. Since x > 4, we also know that ( x - 4) 3 > 0. Therefore E 2 ( x ) < 0. Since E 2 ( x ) = f ( x ) - P 2 ( x ), we conclude that f ( x ) < P 2 ( x ), so P 2 ( x ) is an over-approximation of f ( x ). 11
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• Fall '06
• MARTIN,C.
• lim, dx

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