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Unformatted text preview: Version 010 – Homework 01 – Odell – (58340) 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 ) = 6( t + 1) and f ′ (1) = 6 , f (1) = 5 . 1. f (0) = 5 2. f (0) = 2 3. f (0) = 1 4. f (0) = 3 5. f (0) = 4 correct Explanation: The most general antiderivative of f ′′ has the form f ′ ( t ) = 3 t 2 + 6 t + C where C is an arbitrary constant. But if f ′ (1) = 6, then f ′ (1) = 3 + 6 + C = 6 , i.e., C = 3 . From this it follows that f ′ ( t ) = 3 t 2 + 6 t 3 , and the most general antiderivative of the latter is f ( t ) = t 3 + 3 t 2 3 t + D , where D is an arbitrary constant. But if f (1) = 5, then f (1) = 1 + 3 3 + D = 5 , i.e., D = 4 . Consequently, f ( t ) = t 3 + 3 t 2 3 t + 4 . At x = 0, therefore, f (0) = 4 . 002 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 2 , ( B ) F 2 ( x ) = sin 2 x 2 , ( C ) F 3 ( x ) = cos 2 x 4 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. none of them 2. F 3 only 3. F 1 and F 2 only 4. F 1 and F 3 only 5. F 1 only 6. all of them 7. F 2 and F 3 only correct 8. F 2 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. Version 010 – Homework 01 – Odell – (58340) 2 ( B ) Antiderivative. ( C ) Antiderivative. 003 10.0 points Find f ( x ) on ( π 2 , π 2 ) when f ′ ( x ) = 3 √ 2 sin x + 2 sec 2 x and f ( π 4 ) = 4. 1. f ( x ) = 3 2 tan x + 3 √ 2 sin x 2. f ( x ) = 2 tan x + 3 √ 2 sin x + 5 3. f ( x ) = 2 tan x 3 √ 2 cos x + 5 correct 4. f ( x ) = 9 2 tan x 3 √ 2 cos x 5. f ( x ) = 2 tan x + 3 √ 2 cos x 1 Explanation: The most general antiderivative of f ′ ( x ) = 3 √ 2 sin x + 2 sec 2 x is f ( x ) = 3 √ 2 cos x + 2 tan x + C with C an arbitrary constant. But if f parenleftBig π 4 parenrightBig = 4, then f parenleftBig π 4 parenrightBig = 3 + 2 + C = 4 , so C = 5 . Consequently, f ( x ) = 2 tan x 3 √ 2 cos x + 5 . 004 10.0 points Find the unique antiderivative F of f ( x ) = e 4 x + 3 e 2 x + 3 e − 2 x e 2 x for which F (0) = 0. 1. F ( x ) = 1 2 e 2 x 3 x + 3 4 e − 4 x 5 4 2. F ( x ) = 1 2 e 2 x +3 x 3 4 e − 4 x + 1 4 correct 3. F ( x ) = 1 4 e 4 x + 3 x 3 4 e − 4 x 1 2 4. F ( x ) = 1 4 e 4 x 3 x + 1 2 e − 2 x 1 2 5. F ( x ) = 1 2 e 2 x 3 x + 3 4 e − 2 x 1 4 6. F ( x ) = 1 2 e 2 x + 3 x 1 2 e − 2 x Explanation: After division, e 4 x + 3 e 2 x + 3 e − 2 x e 2 x = e 2 x + 3 + 3 e − 4 x ....
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This note was uploaded on 04/15/2011 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas.
 Spring '10
 ZHENG

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