AnsHW14 - Cheung, Anthony Homework 14 Due: Dec 5 2006, 3:00...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Cheung, Anthony 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 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 th-derivative f ( n ) of f ....
View Full Document

Page1 / 9

AnsHW14 - Cheung, Anthony Homework 14 Due: Dec 5 2006, 3:00...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online