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Homework 1

# Homework 1 - Version 106 Homework 01 Radin(58415 This...

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Version 106 – Homework 01 – Radin – (58415) 1 This print-out should have 14 questions. Multiple-choice 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) = 2 . 1. f (0) = - 2 2. f (0) = - 5 3. f (0) = - 1 correct 4. f (0) = - 4 5. f (0) = - 3 Explanation: The most general anti-derivative 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 = 3 . From this it follows that f ( t ) = 6 t 2 - 4 t + 3 , and the most general anti-derivative of the latter is f ( t ) = 2 t 3 - 2 t 2 + 3 t + D , where D is an arbitrary constant. But if f (1) = 2, then f (1) = 2 - 2 + 3 + D = 2 , i.e., D = - 1 . Consequently, f ( t ) = 2 t 3 - 2 t 2 + 3 t - 1 . At x = 0, therefore, f (0) = - 1 . 002 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 4 , ( B ) F 2 ( x ) = sin 2 x 2 , ( C ) F 3 ( x ) = - cos 2 x 2 . Which are anti-derivatives of f ( x ) = sin x cos x ? 1. F 2 only 2. F 1 and F 3 only 3. F 3 only 4. all of them 5. none of them 6. F 1 and F 2 only 7. F 1 only 8. F 2 and F 3 only correct 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.

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Version 106 – Homework 01 – Radin – (58415) 2 ( B ) Anti-derivative. ( C ) Anti-derivative. 003 10.0 points Find f ( x ) on ( - π 2 , π 2 ) when f ( x ) = 2 cos x + 4 sec 2 x and f ( π 4 ) = 2. 1. f ( x ) = 5 - 4 tan x + 2 sin x 2. f ( x ) = 7 - 4 tan x - 2 cos x 3. f ( x ) = 4 tan x + 2 sin x - 3 correct 4. f ( x ) = 4 tan x - 2 sin x - 1 5. f ( x ) = 4 tan x - 2 cos x - 3 Explanation: The most general anti-derivative of f ( x ) = 2 cos x + 4 sec 2 x is f ( x ) = 2 sin x + 4 tan x + C with C an arbitrary constant. But if f parenleftBig π 4 parenrightBig = 2, then f parenleftBig π 4 parenrightBig = 1 + 4 + C = 2 , so C = - 3 . Consequently, f ( x ) = 4 tan x + 2 sin x - 3 . 004 10.0 points Find the unique anti-derivative F of f ( x ) = e 5 x - 4 e 2 x + 3 e 3 x e 2 x for which F (0) = 0. 1. F ( x ) = 1 3 e 3 x - 4 x - 3 5 e 5 x + 4 15 correct 2. F ( x ) = 1 5 e 5 x + 4 x + 1 3 e 3 x - 2 5 3. F ( x ) = 1 3 e 3 x - 4 x - 1 3 e 3 x 4. F ( x ) = 1 5 e 5 x - 4 x - 3 5 e 5 x - 2 5 5. F ( x ) = 1 3 e 3 x + 4 x + 3 5 e 5 x - 14 15 6. F ( x ) = 1 3 e 3 x + 4 x + 3 5 e 3 x - 4 15 Explanation: After division, e 5 x - 4 e 2 x + 3 e 3 x e 2 x = e 3 x - 4 + 3 e 5 x . Since d dx e αx = αe αx , it thus follows that F ( x ) = 1 3 e 3 x - 4 x - 3 5 e 5 x + C where the constant C is determined by the condition F (0) = 0. For then F (0) = 1 3 - 3 5 + C = 0 . Consequently, F ( x ) = 1 3 e 3 x - 4 x - 3 5 e 5 x + 4 15 .
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