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Unformatted text preview: Cheung, Anthony Homework 14 Due: Dec 5 2006, 3:00 am Inst: David Benzvi 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Determine the degree 2 Taylor polynomial T 2 ( x ) centered at x = 1 for the function f when f ( x ) = p 3 + x 2 . 1. T 2 ( x ) = 2 1 2 ( x + 1) + 3 16 ( x + 1) 2 2. T 2 ( x ) = 2 1 2 ( x 1) + 3 16 ( x 1) 2 3. T 2 ( x ) = 2 + 1 2 ( x 1) + 3 16 ( x 1) 2 correct 4. T 2 ( x ) = 2 1 2 ( x + 1) + 3 8 ( x + 1) 2 5. T 2 ( x ) = 2 + 1 2 ( x 1) + 3 8 ( x 1) 2 6. T 2 ( x ) = 2 + 1 2 ( x + 1) + 3 8 ( x + 1) 2 Explanation: For a function f the degree 2 Taylor poly nomial centered at x = 1 is given by T 2 ( x ) = f (1) + f (1)( x 1) + 1 2 f 00 (1)( x 1) 2 . Now when f ( x ) = p 3 + x 2 , f ( x ) = x 3 + x 2 , while f 00 ( x ) = 3 + x 2 x 2 3 + x 2 3 + x 2 = 3 (3 + x 2 ) 3 / 2 . But then f (1) = 2 , f (1) = 1 2 , f 00 (1) = 3 8 . Consequently, T 2 ( x ) = 2 + 1 2 ( x 1) + 3 16 ( x 1) 2 . keywords: 002 (part 1 of 1) 10 points Suppose p 4 ( x ) = 6 4( x 3) + 5( x 3) 2 3( x 3) 3 + 7( x 3) 4 is the degree 4 Taylor polynomial centered at x = 3 for a certain function f . Use p 4 to estimate the value of f (3 . 1). 1. f (3 . 1) 5 . 7477 2. f (3 . 1) 5 . 9477 3. f (3 . 1) 5 . 8477 4. f (3 . 1) 5 . 5477 5. f (3 . 1) 5 . 6477 correct Explanation: Since p 4 ( x ) is an approximation for f ( x ) we see that f (3 . 1) 6 4 10 + 5 10 2 3 10 3 + 7 10 4 . Consequently, f (3 . 1) 5 . 6477 . keywords: Cheung, Anthony Homework 14 Due: Dec 5 2006, 3:00 am Inst: David Benzvi 2 003 (part 1 of 1) 10 points Suppose T 4 ( x ) = 2 7( x 2) + 2( x 2) 2 6( x 2) 3 + 2( x 2) 4 is the degree 4 Taylor polynomial centered at x = 2 for some function f . What is the value of f (3) (2)? 1. f (3) (2) = 2 2. f (3) (2) = 6 3. f (3) (2) = 2 4. f (3) (2) = 36 5. f (3) (2) = 6 6. f (3) (2) = 36 correct Explanation: Since T 4 ( x ) = f (2) + f (2)( x 2) + f 00 (2) 2! ( x 2) 2 + f (3) (2) 3! ( x 2) 3 + f (4) (2) 4! ( x 2) 4 , we see that f (3) (2) = 3! 6 = 36 . keywords: 004 (part 1 of 5) 10 points When f is the function f ( x ) = 4ln(3 x + 2) , (i) find the third derivative f 000 of f . 1. f 000 ( x ) = 108 (3 x + 2) 3 2. f 000 ( x ) = 27 3. f 000 ( x ) = 108 (3 x + 2) 2 4. f 000 ( x ) = 216 (3 x + 2) 3 5. f 000 ( x ) = 216 (3 x + 2) 3 correct Explanation: Applying the Chain Rule successively we see that f ( x ) = 12 3 x + 2 , f 00 ( x ) = 36 (3 x + 2) 2 , and f 000 ( x ) = 216 (3 x + 2) 3 . 005 (part 2 of 5) 10 points (ii) Find the n thderivative f ( n ) of f ....
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 Spring '08
 RAdin
 Calculus

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