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Unformatted text preview: griffin (ang684) HW02 Gilbert (56650) 1 This printout should have 15 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 ) = 4 x 2 + 5 x + 4 x . 1. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 5 3 x 4 parenrightbigg + C 2. g ( x ) = x ( 4 x 2 + 5 x + 4 ) + C 3. g ( x ) = 2 x ( 4 x 2 + 5 x 4 ) + C 4. g ( x ) = x parenleftbigg 4 5 x 2 + 5 3 x + 4 parenrightbigg + C 5. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 5 3 x + 4 parenrightbigg + C cor rect 6. g ( x ) = 2 x ( 4 x 2 + 5 x + 4 ) + C Explanation: After division g ( x ) = 4 x 3 / 2 + 5 x 1 / 2 + 4 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 8 5 x 5 / 2 + 10 3 x 3 / 2 + 8 x 1 / 2 = 2 x parenleftbigg 4 5 x 2 + 5 3 x + 4 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 4 5 x 2 + 5 3 x + 4 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Find the most general function f such that f ( x ) = 16cos 4 x . 1. f ( x ) = 4 sin4 x + Cx + D 2. f ( x ) = cos x + Cx + D 3. f ( x ) = sin x + Cx + D 4. f ( x ) = cos 4 x + Cx + D correct 5. f ( x ) = 4 sin x + Cx 2 + D 6. f ( x ) = 4 cos 4 x + Cx 2 + D Explanation: When f ( x ) = 16 cos 4 x then f ( x ) = 4 sin4 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. 003 10.0 points Find the value of f ( ) when f ( t ) = 5 3 cos 1 3 t 6 sin 2 3 t and f ( 2 ) = 1. 1. f ( ) = 21 2 + 7 2 3 2. f ( ) = 21 2 5 2 3 3. f ( ) = 21 2 + 5 2 3 correct griffin (ang684) HW02 Gilbert (56650) 2 4. f ( ) = 19 2 7 2 3 5. f ( ) = 19 2 + 7 2 3 6. f ( ) = 19 2 5 2 3 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 2 parenrightBig = 5 sin 6 + 9 cos 3 + C = 1 . But sin 6 = cos 3 = 1 2 , so 7 + C = 1 ....
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This note was uploaded on 09/09/2009 for the course M 408L taught by Professor Gilbert during the Spring '09 term at University of TexasTyler.
 Spring '09
 GILBERT

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