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Unformatted text preview: kim (tk5895) – HW02 – Henry – (54974) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find all functions g such that g ′ ( x ) = x 2 + 4 x + 2 √ x . 1. g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 4 3 x − 2 parenrightbigg + C 2. g ( x ) = √ x parenleftbigg 1 5 x 2 + 4 3 x + 2 parenrightbigg + C 3. g ( x ) = √ x ( x 2 + 4 x + 2 ) + C 4. g ( x ) = 2 √ x ( x 2 + 4 x − 2 ) + C 5. g ( x ) = 2 √ x ( x 2 + 4 x + 2 ) + C 6. g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 4 3 x + 2 parenrightbigg + C cor rect Explanation: After division g ′ ( x ) = x 3 / 2 + 4 x 1 / 2 + 2 x − 1 / 2 , so we can now find an antiderivative of each term separately. But d dx parenleftbigg ax r r parenrightbigg = ax r − 1 for all a and all r negationslash = 0. Thus 2 5 x 5 / 2 + 8 3 x 3 / 2 + 4 x 1 / 2 = 2 √ x parenleftbigg 1 5 x 2 + 4 3 x + 2 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 4 3 x + 2 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Consider the following functions: ( A ) F 1 ( x ) = sin 2 x , ( B ) F 2 ( x ) = cos 2 x 4 , ( C ) F 3 ( x ) = cos 2 x 2 . Which are antiderivatives of f ( x ) = sin x cos x ? 1. none of them correct 2. F 3 only 3. F 1 and F 2 only 4. F 2 and F 3 only 5. F 1 and F 3 only 6. all of them 7. F 1 only 8. F 2 only 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 antiderivative. ( B ) Not antiderivative. ( C ) Not antiderivative. 003 10.0 points kim (tk5895) – HW02 – Henry – (54974) 2 Find f ( t ) when f ′ ( t ) = 5 3 cos 1 3 t − 6 sin 2 3 t and f ( π 2 ) = 3. 1. f ( t ) = 11 sin 1 3 t − 9 cos 2 3 t + 2 2. f ( t ) = 11 cos 1 3 t − 9 sin 2 3 t + 2 3. f ( t ) = 5 sin 1 3 t + 9 cos 2 3 t − 4 correct 4. f ( t ) = 5 cos 1 3 t + 9 sin 2 3 t − 4 5. f ( t ) = 5 sin 1 3 t + 3 cos 2 3 t − 1 6. f ( t ) = 5 cos 1 3 t + 3 sin 2 3 t − 1 Explanation: The function f must have the form f ( t ) = 5 sin 1 3 t + 9 cos 2 3 t + C where the constant C is determined by the condition f parenleftBig π 2 parenrightBig = 5 sin π 6 + 9 cos π 3 + C = 3 . But by known trig values sin π 6 = cos π 3 = 1 2 , so 7 + C = 3. Consequently, f ( t ) = 5 sin 1 3 t + 9 cos 2 3 t − 4 . 004 10.0 points Find f ( x ) on ( − π 2 , π 2 ) when f ′ ( x ) = 7 + 5 tan 2 x and f (0) = 2. 1. f ( x ) = 2 + 2 x + 5 tan x correct 2. f ( x ) = − 3 + 7 x + 5 sec 2 x 3. f ( x ) = 2 − 2 x − 5 tan x 4. f ( x ) = 2 + 2 x + 5 tan 2 x 5. f ( x ) = 7 − 2 x − 5 sec x 6. f ( x ) = − 3 + 7 x + 5 sec x Explanation: The properties d dx (tan x ) = sec 2 x, tan 2 x = sec 2 x − 1 , suggest that we rewrite...
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This note was uploaded on 04/09/2011 for the course M 408 L taught by Professor Cepparo during the Fall '08 term at University of Texas.
 Fall '08
 Cepparo
 Calculus

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