# hw2 - padilla(tp5647 – HW02 – Gilbert –(56650 1 This...

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Unformatted text preview: padilla (tp5647) – HW02 – Gilbert – (56650) 1 This print-out should have 15 questions. Multiple-choice 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 + 3 √ x . 1. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C cor- rect 2. g ( x ) = 2 √ x ( 4 x 2 + 5 x + 3 ) + C 3. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x − 3 parenrightbigg + C 4. g ( x ) = √ x ( 4 x 2 + 5 x + 3 ) + C 5. g ( x ) = 2 √ x ( 4 x 2 + 5 x − 3 ) + C 6. g ( x ) = √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C Explanation: After division g ′ ( x ) = 4 x 3 / 2 + 5 x 1 / 2 + 3 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 + 6 x 1 / 2 = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + 5 3 x + 3 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Find the most general function f such that f ′′ ( x ) = 9 cos3 x . 1. f ( x ) = 3 sin3 x + Cx + D 2. f ( x ) = − 3 cos 3 x + Cx 2 + D 3. f ( x ) = − cos 3 x + Cx + D correct 4. f ( x ) = − 3 sin x + Cx 2 + D 5. f ( x ) = cos x + Cx + D 6. f ( x ) = sin x + Cx + D Explanation: When f ′′ ( x ) = 9 cos 3 x then f ′ ( x ) = 3 sin3 x + C with C an arbitrary contant. Consequently, the most general function f is f ( x ) = − cos 3 x + Cx + D with D also an arbitrary constant. 003 10.0 points Find the value of f ( π ) when f ′ ( t ) = 1 3 cos 1 3 t − 6 sin 2 3 t and f ( π 2 ) = 5. 1. f ( π ) = − 9 2 + 3 2 √ 3 2. f ( π ) = 7 2 − 1 2 √ 3 3. f ( π ) = − 7 2 + 3 2 √ 3 padilla (tp5647) – HW02 – Gilbert – (56650) 2 4. f ( π ) = 9 2 − 1 2 √ 3 5. f ( π ) = 7 2 − 3 2 √ 3 6. f ( π ) = − 9 2 + 1 2 √ 3 correct Explanation: The function f must have the form f ( t ) = sin 1 3 t + 9 cos 2 3 t + C where the constant C is determined by the condition f parenleftBig π 2 parenrightBig = sin π 6 + 9 cos π 3 + C = 5 . But sin π 6 = cos π 3 = 1 2 , so 5 + C = 5 ....
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## This note was uploaded on 05/03/2009 for the course M 408 L taught by Professor Gilbert during the Spring '09 term at University of Texas-Tyler.

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hw2 - padilla(tp5647 – HW02 – Gilbert –(56650 1 This...

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