# HWK1 - Valenica Daniel Homework 1 Due Sep 4 2007 3:00 am...

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Valenica, Daniel – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Cheng 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. 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 ) = 4 sin 4 x + Cx + D 3. f ( x ) = cos x + Cx + D 4. f ( x ) = - 4 sin x + Cx 2 + D 5. f ( x ) = sin x + Cx + D 6. f ( x ) = - 4 cos 4 x + Cx 2 + 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 + tan 2 x and f (0) = 4. 1. f ( x ) = 4 - 5 x - tan x 2. f ( x ) = 3 + 6 x + sec 2 x 3. f ( x ) = 5 - 5 x - sec x 4. f ( x ) = 3 + 6 x + sec x 5. f ( x ) = 4 + 5 x + tan 2 x 6. f ( x ) = 4 + 5 x + tan x correct 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 ) = 5 + sec 2 x, For then the most general anti-derivative oF f 0 is f ( x ) = 5 x + tan x + C, with C an arbitrary constant. But iF f (0) = 4, then f (0) = C = 4 . Consequently, f ( x ) = 4 + 5 x + tan x . keywords: antiderivatives, trigonometric Functions, particular values 003 (part 1 oF 1) 10 points Determine f ( t ) when f 00 ( t ) = 6( t + 1) and f 0 (1) = 1 , f (1) = 4 . 1. f ( t ) = 3 t 3 + 3 t 2 - 8 t + 6 2. f ( t ) = t 3 - 3 t 2 + 8 t - 2

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Valenica, Daniel – Homework 1 – Due: Sep 4 2007, 3:00 am – Inst: Cheng 2 3. f ( t ) = 3 t 3 + 6 t 2 - 8 t + 3 4. f ( t ) = 3 t 3 - 6 t 2 + 8 t - 1 5. f ( t ) = t 3 - 6 t 2 + 8 t + 1 6. f ( t ) = t 3 + 3 t 2 - 8 t + 8 correct Explanation: The most general anti-derivative of f 00 has the form f 0 ( t ) = 3 t 2 + 6 t + C where C is an arbitrary constant. But if f 0 (1) = 1, then f 0 (1) = 3 + 6 + C = 1 , i.e., C = - 8 . From this it follows that f 0 ( t ) = 3 t 2 + 6 t - 8 . The most general anti-derivative of f is thus f ( t ) = t 3 + 3 t 2 - 8 t + D , where D is an arbitrary constant. But if f (1) = 4, then f (1) = 1 + 3 - 8 + D = 4 , i.e., D = 8 . Consequently, f ( t ) = t 3 + 3 t 2 - 8 t + 8 . keywords: 004 (part 1 of 1) 10 points Find the unique anti-derivative F of f ( x ) = 2 e 5 x - 2 e 2 x + 5 e - 3 x e 2 x for which F (0) = 0. 1. F ( x ) = 2 5 e 5 x - 2 x - e - 5 x - 3 5 2. F ( x ) = 2 3 e 3 x - 2 x - e - 5 x + 1 3 correct 3. F ( x ) = 2 3 e 3 x + 2 x + e - 5 x - 5 3 4. F ( x ) = 2 5 e 5 x + 2 x + 2 3 e - 3 x - 3 5 5. F ( x ) = 2 3 e 3 x + 2 x + e - 3 x - 1 3 6. F ( x ) = 2 3 e 3 x - 2 x - 2 3 e - 3 x Explanation: After division, 2 e 5 x - 2 e 2 x + 5 e - 3 x e 2 x = 2 e 3 x - 2 + 5 e - 5 x . Since
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## This homework help was uploaded on 04/16/2008 for the course CALC 303L taught by Professor Cheng during the Fall '07 term at University of Texas.

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HWK1 - Valenica Daniel Homework 1 Due Sep 4 2007 3:00 am...

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