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Unformatted text preview: Version 049 – Homework 01 – Radin – (58415) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Welcome to Quest Learning and As sessment. Best of luck this semester. 001 10.0 points Find the value of f (0) when f ′′ ( t ) = 4(3 t + 1) and f ′ (1) = 5 , f (1) = 4 . 1. f (0) = 3 2. f (0) = 5 correct 3. f (0) = 1 4. f (0) = 4 5. f (0) = 2 Explanation: The most general antiderivative of f ′′ has the form f ′ ( t ) = 6 t 2 + 4 t + C where C is an arbitrary constant. But if f ′ (1) = 5, then f ′ (1) = 6 + 4 + C = 5 , i.e., C = 5 . From this it follows that f ′ ( t ) = 6 t 2 + 4 t 5 , and the most general antiderivative of the latter is f ( t ) = 2 t 3 + 2 t 2 5 t + D , where D is an arbitrary constant. But if f (1) = 4, then f (1) = 2 + 2 5 + D = 4 , i.e., D = 5 . Consequently, f ( t ) = 2 t 3 + 2 t 2 5 t + 5 . At x = 0, therefore, f (0) = 5 . 002 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 2 , ( B ) F 2 ( x ) = sin 2 x , ( C ) F 3 ( x ) = cos 2 x 4 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. none of them 2. F 2 only 3. F 3 only correct 4. F 1 and F 3 only 5. all of them 6. F 1 only 7. F 1 and F 2 only 8. F 2 and F 3 only 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 antiderivative. ( B ) Not antiderivative. Version 049 – Homework 01 – Radin – (58415) 2 ( C ) Antiderivative. 003 10.0 points Find f ( x ) on ( π 2 , π 2 ) when f ′ ( x ) = 2 √ 2 sin x + sec 2 x and f ( π 4 ) = 5. 1. f ( x ) = tan x + 2 √ 2 sin x + 6 2. f ( x ) = tan x + 2 √ 2 cos x + 2 3. f ( x ) = 8 tan x 2 √ 2 cos x 4. f ( x ) = 4 tan x + 2 √ 2 sin x 5. f ( x ) = tan x 2 √ 2 cos x + 6 correct Explanation: The most general antiderivative of f ′ ( x ) = 2 √ 2 sin x + sec 2 x is f ( x ) = 2 √ 2 cos x + tan x + C with C an arbitrary constant. But if f parenleftBig π 4 parenrightBig = 5, then f parenleftBig π 4 parenrightBig = 2 + 1 + C = 5 , so C = 6 . Consequently, f ( x ) = tan x 2 √ 2 cos x + 6 . 004 10.0 points Find the unique antiderivative F of f ( x ) = e 5 x + 2 e 2 x + 5 e − 3 x e 2 x for which F (0) = 0. 1. F ( x ) = 1 3 e 3 x 2 x + e − 5 x 4 3 2. F ( x ) = 1 5 e 5 x 2 x + 1 3 e − 3 x 4 5 3. F ( x ) = 1 5 e 5 x + 2 x e − 5 x 4 5 4. F ( x ) = 1 3 e 3 x + 2 x e − 5 x + 2 3 correct 5. F ( x ) = 1 3 e 3 x 2 x + e − 3 x 2 3 6. F ( x ) = 1 3 e 3 x + 2 x 1 3 e − 3 x Explanation: After division, e 5 x + 2 e 2 x + 5 e − 3 x e 2 x = e 3 x + 2 + 5 e − 5 x ....
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This note was uploaded on 03/18/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.
 Spring '08
 RAdin

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