Homework 01

# Homework 01 - Su, Yung Homework 1 Due: Sep 4 2007, 3:00 am...

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Su, Yung – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Shinko Harper 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. Make sure you get started as early as possi- ble! 001 (part 1 oF 1) 10 points ±ind the most general Function f such that f 00 ( x ) = 16 cos 4 x. 1. f ( x ) = - cos 4 x + Cx + D correct 2. f ( x ) = sin x + Cx + D 3. f ( x ) = cos x + Cx + D 4. f ( x ) = - 4 cos 4 x + Cx 2 + D 5. f ( x ) = - 4 sin x + Cx 2 + D 6. f ( x ) = 4 sin 4 x + Cx + D Explanation: When f 00 ( x ) = 16 cos 4 x then f 0 ( x ) = 4 sin 4 x + C with C an arbitrary contant. Consequently, the most general Function f is f ( x ) = - cos 4 x + Cx + D with D also an arbitrary constant. keywords: antiderivative, trigonometric Func- tions 002 (part 1 oF 1) 10 points ±ind f ( x ) on ( - π 2 , π 2 ) when f 0 ( x ) = 6 + 5 tan 2 x and f (0) = 3. 1. f ( x ) = 8 - x - 5 sec x 2. f ( x ) = 3 + x + 5 tan x correct 3. f ( x ) = - 2 + 6 x + 5 sec x 4. f ( x ) = - 2 + 6 x + 5 sec 2 x 5. f ( x ) = 3 - x - 5 tan x 6. f ( x ) = 3 + x + 5 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 0 ( x ) as f 0 ( x ) = 1 + 5 sec 2 x, For then the most general anti-derivative oF f 0 is f ( x ) = x + 5 tan x + C, with C an arbitrary constant. But iF f (0) = 3, then f (0) = C = 3 . Consequently, f ( x ) = 3 + x + 5 tan x . keywords: antiderivatives, trigonometric Functions, particular values 003 (part 1 oF 1) 10 points Determine f ( t ) when f 00 ( t ) = 2(3 t + 4) and f 0 (1) = 6 , f (1) = 3 . 1. f ( t ) = t 3 - 4 t 2 + 5 t + 1 2. f ( t ) = t 3 + 4 t 2 - 5 t + 3 correct

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Su, Yung – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Shinko Harper 2 3. f ( t ) = t 3 - 8 t 2 + 5 t + 5 4. f ( t ) = 3 t 3 + 4 t 2 - 5 t + 1 5. f ( t ) = 3 t 3 + 8 t 2 - 5 t - 3 6. f ( t ) = 3 t 3 - 8 t 2 + 5 t + 3 Explanation: The most general anti-derivative of f 00 has the form f 0 ( t ) = 3 t 2 + 8 t + C where C is an arbitrary constant. But if f 0 (1) = 6, then f 0 (1) = 3 + 8 + C = 6 , i.e., C = - 5 . From this it follows that f 0 ( t ) = 3 t 2 + 8 t - 5 . The most general anti-derivative of f is thus f ( t ) = t 3 + 4 t 2 - 5 t + D , where D is an arbitrary constant. But if f (1) = 3, then f (1) = 1 + 4 - 5 + D = 3 , i.e., D = 3 . Consequently, f ( t ) = t 3 + 4 t 2 - 5 t + 3 . keywords: 004 (part 1 of 1) 10 points Find the unique anti-derivative F of f ( x ) = e 5 x - 2 e 2 x + 2 e - 3 x e 2 x for which F (0) = 0. 1. F ( x ) = 1 3 e 3 x + 2 x + 2 5 e - 5 x - 11 15 2. F ( x ) = 1 3 e 3 x - 2 x - 1 3 e - 3 x 3. F ( x ) = 1 5 e 5 x - 2 x - 2 5 e - 5 x - 1 5 4. F ( x ) = 1 5 e 5 x + 2 x + 1 3 e - 3 x - 1 5 5. F ( x ) = 1 3 e 3 x + 2 x + 2 5 e - 3 x - 1 15 6. F ( x ) = 1 3 e 3 x - 2 x - 2 5 e - 5 x + 1 15 correct Explanation: After division, e 5 x - 2 e 2 x + 2 e - 3 x e 2 x = e 3 x - 2 + 2 e - 5 x .
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## This note was uploaded on 04/15/2009 for the course M 59685 taught by Professor Harper during the Spring '09 term at University of Texas at Austin.

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Homework 01 - Su, Yung Homework 1 Due: Sep 4 2007, 3:00 am...

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