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Unformatted text preview: Kim, Jin – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Diane Radin 1 This printout should have 16 questions. Multiplechoice 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 Find the most general function f such that f 00 ( x ) = 18 cos 3 x. 1. f ( x ) = 3 sin 3 x + Cx + D 2. f ( x ) = 2 cos x + Cx + D 3. f ( x ) = 3 cos 3 x + Cx 2 + D 4. f ( x ) = 2 cos 3 x + Cx + D correct 5. f ( x ) = 3 sin x + Cx 2 + D 6. f ( x ) = 2 sin x + Cx + D Explanation: When f 00 ( x ) = 18 cos 3 x then f ( x ) = 6 sin 3 x + C with C an arbitrary contant. Consequently, the most general function f is f ( x ) = 2 cos 3 x + Cx + D with D also an arbitrary constant. keywords: antiderivative, trigonometric func tions 002 (part 1 of 1) 10 points Find f ( x ) on ( π 2 , π 2 ) when f ( x ) = 8 + 7 tan 2 x and f (0) = 5. 1. f ( x ) = 12 x 7 sec x 2. f ( x ) = 5 + x + 7 tan x correct 3. f ( x ) = 2 + 8 x + 7 sec 2 x 4. f ( x ) = 2 + 8 x + 7 sec x 5. f ( x ) = 5 x 7 tan x 6. f ( x ) = 5 + x + 7 tan 2 x Explanation: The properties d dx (tan x ) = sec 2 x, tan 2 x = sec 2 x 1 , suggest that we rewrite f ( x ) as f ( x ) = 1 + 7 sec 2 x, for then the most general antiderivative of f is f ( x ) = x + 7 tan x + C, with C an arbitrary constant. But if f (0) = 5, then f (0) = C = 5 . Consequently, f ( x ) = 5 + x + 7 tan x . keywords: antiderivatives, trigonometric functions, particular values 003 (part 1 of 1) 10 points Determine f ( t ) when f 00 ( t ) = 2(9 t + 1) and f (1) = 6 , f (1) = 2 . 1. f ( t ) = 9 t 3 2 t 2 + 5 t 10 2. f ( t ) = 3 t 3 + t 2 5 t + 3 correct Kim, Jin – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Diane Radin 2 3. f ( t ) = 9 t 3 + 2 t 2 5 t 4 4. f ( t ) = 3 t 3 t 2 + 5 t 5 5. f ( t ) = 9 t 3 + t 2 5 t 3 6. f ( t ) = 3 t 3 2 t 2 + 5 t 4 Explanation: The most general antiderivative of f 00 has the form f ( t ) = 9 t 2 + 2 t + C where C is an arbitrary constant. But if f (1) = 6, then f (1) = 9 + 2 + C = 6 , i.e., C = 5 . From this it follows that f ( t ) = 9 t 2 + 2 t 5 . The most general antiderivative of f is thus f ( t ) = 3 t 3 + t 2 5 t + D , where D is an arbitrary constant. But if f (1) = 2, then f (1) = 3 + 1 5 + D = 2 , i.e., D = 3 . Consequently, f ( t ) = 3 t 3 + t 2 5 t + 3 . keywords: 004 (part 1 of 1) 10 points Find the unique antiderivative F of f ( x ) = e 3 x + 3 e 2 x + 2 e x e 2 x for which F (0) = 0. 1. F ( x ) = 1 3 e 3 x + 3 x 2 3 e 3 x 1 3 2. F ( x ) = e x 3 x + 2 3 e x + 1 3 3. F ( x ) = e x + 3 x 2 3 e 3 x 1 3 correct 4. F ( x ) = 1 3 e 3 x 3 x + e x 1 3 5. F ( x ) = e x + 3 x e x 6. F ( x ) = e x 3 x + 2 3 e 3 x + 5 3 Explanation: After division, e 3 x + 3 e 2 x + 2 e x e 2 x = e x + 3 + 2 e 3 x ....
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 Spring '09
 Turner
 Derivative, Fundamental Theorem Of Calculus, Velocity, Constant of integration, Diane Radin

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