# HW 14 - Portillo Tom – Homework 14 – Due Dec 5 2006...

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Unformatted text preview: Portillo, Tom – Homework 14 – Due: Dec 5 2006, 3:00 am – Inst: David Benzvi 1 This print-out should have 17 questions. Multiple-choice 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 8 ( x + 1) 2 2. T 2 ( x ) = 2 + 1 2 ( x- 1) + 3 16 ( x- 1) 2 correct 3. T 2 ( x ) = 2- 1 2 ( x- 1) + 3 16 ( x- 1) 2 4. T 2 ( x ) = 2 + 1 2 ( x + 1) + 3 8 ( x + 1) 2 5. T 2 ( x ) = 2- 1 2 ( x + 1) + 3 16 ( 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 ) = 5- 3( x- 1) + 5( x- 1) 2- 3( x- 1) 3 + 7( x- 1) 4 is the degree 4 Taylor polynomial centered at x = 1 for a certain function f . Use p 4 to estimate the value of f (1 . 1). 1. f (1 . 1) ≈ 4 . 7477 correct 2. f (1 . 1) ≈ 5 . 0477 3. f (1 . 1) ≈ 4 . 8477 4. f (1 . 1) ≈ 4 . 9477 5. f (1 . 1) ≈ 4 . 6477 Explanation: Since p 4 ( x ) is an approximation for f ( x ) we see that f (1 . 1) ≈ 5- 3 10 + 5 10 2- 3 10 3 + 7 10 4 . Consequently, f (1 . 1) ≈ 4 . 7477 . keywords: Portillo, Tom – Homework 14 – Due: Dec 5 2006, 3:00 am – Inst: David Benzvi 2 003 (part 1 of 1) 10 points Suppose T 4 ( x ) = 6- 8( x- 5) + 4( x- 5) 2- 5( x- 5) 3 + 4( x- 5) 4 is the degree 4 Taylor polynomial centered at x = 5 for some function f . What is the value of f (3) (5)? 1. f (3) (5) = 30 2. f (3) (5) = 5 3 3. f (3) (5) =- 30 correct 4. f (3) (5) = 5 5. f (3) (5) =- 5 3 6. f (3) (5) =- 5 Explanation: Since T 4 ( x ) = f (5) + f (5)( x- 5) + f 00 (5) 2! ( x- 5) 2 + f (3) (5) 3! ( x- 5) 3 + f (4) (5) 4! ( x- 5) 4 , we see that f (3) (5) =- 3! × 5 =- 30 . keywords: 004 (part 1 of 5) 10 points When f is the function f ( x ) = 4ln(4 x + 3) , (i) find the third derivative f 000 of f . 1. f 000 ( x ) = 512 27 2. f 000 ( x ) = 256 (4 x + 3) 3 3. f 000 ( x ) = 256 (4 x + 3) 2 4. f 000 ( x ) =- 512 (4 x + 3) 3 5. f 000 ( x ) = 512 (4 x + 3) 3 correct Explanation: Applying the Chain Rule successively we see that f ( x ) = 16 4 x + 3 , f 00 ( x ) =- 64 (4 x + 3) 2 , and f 000 ( x ) = 512 (4 x + 3) 3 ....
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HW 14 - Portillo Tom – Homework 14 – Due Dec 5 2006...

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