Homework 1 - Version 065 Homework 01 Gompf (58370) This...

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Version 065 – Homework 01 – Gompf – (58370) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. Welcome to Quest Learning and As- sessment. Best of luck this semester. 001 10.0 points ±ind the value of f (0) when f ′′ ( t ) = 2(3 t - 1) and f (1) = 5 , f (1) = 2 . 1. f (0) = - 2 correct 2. f (0) = 2 3. f (0) = 0 4. f (0) = - 1 5. f (0) = 1 Explanation: The most general anti-derivative of f ′′ has the form f ( t ) = 3 t 2 - 2 t + C where C is an arbitrary constant. But if f (1) = 5, then f (1) = 3 - 2 + C = 5 , i.e., C = 4 . ±rom this it follows that f ( t ) = 3 t 2 - 2 t + 4 , and the most general anti-derivative of the latter is f ( t ) = t 3 - t 2 + 4 t + D , where D is an arbitrary constant. But if f (1) = 2, then f (1) = 1 - 1 + 4 + D = 2 , i.e., D = - 2 . Consequently, f ( t ) = t 3 - t 2 + 4 t - 2 . At x = 0, therefore, f (0) = - 2 . 002 10.0 points Consider the following functions: ( A ) F 1 ( x ) = sin 2 x , ( B ) F 2 ( x ) = cos 2 x 2 , ( C ) F 3 ( x ) = cos 2 x 4 . Which are anti-derivatives of f ( x ) = sin x cos x ? 1. F 1 only 2. F 2 and F 3 only 3. none of them correct 4. F 2 only 5. F 1 and F 2 only 6. F 3 only 7. F 1 and F 3 only 8. all of them Explanation: By trig identities, cos 2 x = 2 cos 2 x - 1 = 1 - 2 sin 2 x , while sin 2 x = 2 sin x cos x . But d dx sin x = cos x, d dx cos x = - sin x . Consequently, by the Chain Rule, ( A ) Not anti-derivative. ( B ) Not anti-derivative.
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Version 065 – Homework 01 – Gompf – (58370) 2 ( C ) Not anti-derivative. 003 10.0 points Find f ( x ) on ( - π 2 , π 2 ) when f ( x ) = 2 sin x + 2 sec 2 x and f ( π 4 ) = 5. 1. f ( x ) = 2 tan x - 2 cos x + 4 correct 2. f ( x ) = 2 tan x + 2 sin x + 4 3. f ( x ) = 8 - 2 tan x - 2 cos x 4. f ( x ) = 6 - 2 tan x + 2 sin x 5. f ( x ) = 2 tan x + 2 cos x + 2 Explanation: The most general anti-derivative of f ( x ) = 2 sin x + 2 sec 2 x is f ( x ) = - 2 cos x + 2 tan x + C with C an arbitrary constant. But if f p π 4 P = 5, then f p π 4 P = - 1 + 2 + C = 5 , so C = 4 . Consequently, f ( x ) = 2 tan x - 2 cos x + 4 . 004 10.0 points Find the unique anti-derivative F of f ( x ) = 3 e 3 x + 2 e 2 x + 5 e x e 2 x for which F (0) = 0. 1. F ( x ) = 3 e x + 2 x - 5 3 e 3 x - 4 3 correct 2. F ( x ) = e 3 x - 2 x + 3 e x - 2 3 3. F ( x ) = 3 e x - 2 x + 5 3 e 3 x + 14 3 4. F ( x ) = 3 e x - 2 x + 5 3 e x + 4 3 5. F ( x ) = 3 e x + 2 x - 3 e x 6. F ( x ) = e 3 x + 2 x - 5 3 e 3 x - 2 3 Explanation: After division, 3 e 3 x + 2 e 2 x + 5 e x e 2 x = 3 e x + 2 + 5 e 3 x .
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This note was uploaded on 12/04/2009 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.

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Homework 1 - Version 065 Homework 01 Gompf (58370) This...

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