HW15 - Jolley Garrett – Homework 15 – Due Dec 4 2007...

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: Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 1 This print-out should have 18 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 if lim x →∞ tan- 1 µ 1 + 2 x 2 5 + 4 x ¶ exists, and if it does, find its value. 1. limit = π 6 2. limit = 0 3. limit = π 3 4. limit does not exist 5. limit = π 4 6. limit = π 2 correct Explanation: Since 1 + 2 x 2 5 + 4 x-→ ∞ as x → ∞ , we see that lim x →∞ tan- 1 µ 1 + 2 x 2 5 + 4 x ¶ exists, and that the limit = tan- 1 ∞ = π 2 . keywords: limit, limit at infinity, arctan, 002 (part 1 of 1) 10 points Find the derivative of f ( x ) = 2sin- 1 ( e 3 x ) . 1. f ( x ) = 6 e 3 x 1 + e 6 x 2. f ( x ) = 2 e 3 x √ 1- e 6 x 3. f ( x ) = 2 1 + e 6 x 4. f ( x ) = 2 e 3 x 1 + e 6 x 5. f ( x ) = 6 √ 1- e 6 x 6. f ( x ) = 6 1 + e 6 x 7. f ( x ) = 6 e 3 x √ 1- e 6 x correct 8. f ( x ) = 2 √ 1- e 6 x Explanation: Since d dx sin- 1 x = 1 √ 1- x 2 , d dx e ax = ae ax , the Chain Rule ensures that f ( x ) = 6 e 3 x √ 1- e 6 x . keywords: 003 (part 1 of 1) 10 points Determine f ( x ) when f ( x ) = tan- 1 ‡ x √ 5- x 2 · . ( Hint : first simplify f .) 1. f ( x ) = 1 √ 5- x 2 correct 2. f ( x ) = √ 5 √ 5- x 2 Jolley, Garrett – Homework 15 – Due: Dec 4 2007, 3:00 am – Inst: R Heitmann 2 3. f ( x ) = √ 5 √ 5 + x 2 4. f ( x ) = x x 2 + 5 5. f ( x ) = x √ x 2- 5 Explanation: If tan θ = x √ 5- x 2 , then by Pythagoras’ theorem applied to the right triangle p 5- x 2 x θ √ 5 we see that sin θ = x √ 5 . Thus f ( x ) = sin- 1 ‡ x √ 5 · . Consequently, f ( x ) = 1 √ 5- x 2 . Alternatively, we can differentiate f using the Chain Rule and the fact that d dx tan- 1 x = 1 1 + x 2 . . keywords: 004 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = 5tan- 1 ( e 2 x ) + 3 e- 2 x . 1. f ( x ) = 1 √ 1- e 4 x ( 2 e 2 x + 3 e- 2 x ) 2. f ( x ) = 1 1 + e 4 x ( 2 e 2 x + 3 e- 2 x ) 3. f ( x ) = 2 1 + e 4 x ( 2 e 2 x + 3 e- 2 x ) 4. f ( x ) = 2 √ 1- e 4 x ( 2 e 2 x + 3 e- 2 x ) 5. f ( x ) = 2 1 + e 4 x ( 2 e 2 x- 3 e- 2 x ) cor- rect 6. f ( x ) = 1 1 + e 4 x ( 2 e 2 x- 3 e- 2 x ) Explanation: By the Chain Rule, f ( x ) = 2 µ 5 e 2 x 1 + e 4 x- 3 e- 2 x ¶ since d dx tan- 1 x = 1 1 + x 2 ( e 2 x ) 2 = e 4 x . The expression for f can now be simplified by bringing the right hand side to a common denominator, for then f ( x ) = 2 • 5 e 2 x- 3 e- 2 x (1 + e 4 x ) 1 + e 4 x ‚ = 2 µ 5 e 2 x- 3 e- 2 x- 3 e 2 x 1 + e 4 x ¶ . Consequently, f ( x ) = 2 1 + e 4 x ( 2 e 2 x- 3 e- 2 x ) . keywords: 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = ˆ tan- 1 x 2- ln r 2 + x 2- x !...
View Full Document

This note was uploaded on 10/29/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.

Page1 / 11

HW15 - Jolley Garrett – Homework 15 – Due Dec 4 2007...

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