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Unformatted text preview: Create assignment, 57321, Homework 6, Mar 22 at 4:56 pm 1 This printout should have 22 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. CalC4j00Ex2 51:01, calculus3, multiple choice, > 1 min, wordingvariable. 001 Find all functions g such that g ( x ) = 3 x 2 + 4 x + 1 √ x . 1. g ( x ) = 2 √ x µ 3 5 x 2 + 4 3 x + 1 ¶ + C cor rect 2. g ( x ) = √ x µ 3 5 x 2 + 4 3 x + 1 ¶ + C 3. g ( x ) = 2 √ x µ 3 5 x 2 + 4 3 x 1 ¶ + C 4. g ( x ) = 2 √ x ( 3 x 2 + 4 x + 1 ) + C 5. g ( x ) = √ x ( 3 x 2 + 4 x + 1 ) + C 6. g ( x ) = 2 √ x ( 3 x 2 + 4 x 1 ) + C Explanation: After division g ( x ) = 3 x 3 / 2 + 4 x 1 / 2 + x 1 / 2 , so we can now find an antiderivative of each term separately. But d dx µ ax r r ¶ = ax r 1 for all a and all r 6 = 0. Thus 6 5 x 5 / 2 + 8 3 x 3 / 2 + 2 x 1 / 2 = 2 √ x µ 3 5 x 2 + 4 3 x + 1 ¶ is an antiderivative of g . Consequently, g ( x ) = 2 √ x µ 3 5 x 2 + 4 3 x + 1 ¶ + C with C an arbitrary constant. keywords: antiderivative, power functions CalC4j00Ex4 51:01, calculus3, multiple choice, > 1 min, wordingvariable. 002 Stewart Section 4.10, Example, 4 page 302 Determine f ( t ) when f 00 ( t ) = 2(9 t + 1) and f (1) = 4 , f (1) = 6 . 1. f ( t ) = 3 t 3 + t 2 7 t + 9 correct 2. f ( t ) = 3 t 3 2 t 2 + 7 t 2 3. f ( t ) = 9 t 3 + t 2 7 t + 3 4. f ( t ) = 3 t 3 t 2 + 7 t 3 5. f ( t ) = 9 t 3 + 2 t 2 7 t + 2 6. f ( t ) = 9 t 3 2 t 2 + 7 t 8 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) = 4, then f (1) = 9 + 2 + C = 4 , i.e., C = 7 . From this it follows that f ( t ) = 9 t 2 + 2 t 7 . Create assignment, 57321, Homework 6, Mar 22 at 4:56 pm 2 The most general antiderivative of f is thus f ( t ) = 3 t 3 + t 2 7 t + D , where D is an arbitrary constant. But if f (1) = 6, then f (1) = 3 + 1 7 + D = 6 , i.e., D = 9 . Consequently, f ( t ) = 3 t 3 + t 2 7 t + 9 . keywords: antiderivatives, second order derivatives, particular values CalC4j14s 51:01, calculus3, multiple choice, > 1 min, normal. 003 Find the most general antiderivative of the function f ( x ) = 5 cos x 3 sin x. 1. F ( x ) = 5 sin x + 3 cos x + C correct 2. F ( x ) = 5 cos x + 3 sin x + C 3. F ( x ) = 5 sin x + 3 cos x + C 4. F ( x ) = 5 cos x + 3 sin x + C 5. none of these Explanation: Since d dx sin x = cos x, d dx cos x = sin x, we see immediately that the most general antiderivative of f ( x ) = 5 cos x 3 sin x is the function F ( x ) = 5 sin x + 3 cos x + C with C an arbitrary constant. keywords: antiderivative, trig function CalC4j29s 51:01, calculus3, multiple choice, > 1 min, wordingvariable....
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This note was uploaded on 03/20/2008 for the course M 408c taught by Professor Mcadam during the Spring '06 term at University of Texas.
 Spring '06
 McAdam

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