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HW02-solutions

# HW02-solutions - griffin(ang684 HW02 Gilbert(56650 This...

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grifn (ang684) – HW02 – Gilbert – (56650) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. 001 10.0 points ²ind all ±unctions g such that g ( x ) = 4 x 2 + 5 x + 4 x . 1. g ( x ) = 2 x p 4 5 x 2 + 5 3 x 4 P + 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 p 4 5 x 2 + 5 3 x + 4 P + C 5. g ( x ) = 2 x p 4 5 x 2 + 5 3 x + 4 P + C cor- rect 6. g ( x ) = 2 x ( 4 x 2 + 5 x + 4 ) + C Explanation: A±ter division g ( x ) = 4 x 3 / 2 + 5 x 1 / 2 + 4 x 1 / 2 , so we can now Fnd an antiderivative o± each term separately. But d dx p ax r r P = ax r 1 ±or all a and all r n = 0. Thus 8 5 x 5 / 2 + 10 3 x 3 / 2 + 8 x 1 / 2 = 2 x p 4 5 x 2 + 5 3 x + 4 P is an antiderivative o± g . Consequently, g ( x ) = 2 x p 4 5 x 2 + 5 3 x + 4 P + C with C an arbitrary constant. 002 10.0 points ²ind the most general ±unction f such that f ′′ ( x ) = 16cos 4 x . 1. f ( x ) = 4 sin4 x + Cx + D 2. f ( x ) = cos x + + D 3. f ( x ) = sin x + + D 4. f ( x ) = cos 4 x + + D correct 5. f ( x ) = 4 sin x + 2 + D 6. f ( x ) = 4 cos 4 x + 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 ±unction f is f ( x ) = cos 4 x + + D with D also an arbitrary constant. 003 10.0 points ²ind the value o± f ( π ) when f ( t ) = 5 3 cos 1 3 t 6 sin 2 3 t and f ( π 2 ) = 1. 1. f ( π ) = 21 2 + 7 2 3 2. f ( π ) = 21 2 5 2 3 3. f ( π ) = 21 2 + 5 2 3 correct

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grifn (ang684) – HW02 – Gilbert – (56650) 2 4. f ( π ) = 19 2 7 2 3 5. f ( π ) = 19 2 + 7 2 3 6. f ( π ) = 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 p π 2 P = 5 sin π 6 + 9 cos π 3 + C = 1 . But sin π 6 = cos π 3 = 1 2 , so 7 + C = 1 . Thus f ( t ) = 5 sin 1 3 t + 9 cos 2 3 t 6 . At t = π , thereFore, f ( π ) = 21 2 + 5 2 3 . 004 10.0 points ±ind f ( x ) on ( π 2 , π 2 ) when f ( x ) = 6 + tan 2 x and f (0) = 4. 1. f ( x ) = 4 5 x tan x 2. f ( x ) = 3 + 6 x + sec x 3. f ( x ) = 4 + 5 x + tan 2 x 4. f ( x ) = 3 + 6 x + sec 2 x 5. f ( x ) = 5 5 x sec 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 ( x ) as f ( x ) = 5 + sec 2 x, For then the most general anti-derivative oF f is f ( x ) = 5 x + tan x + C, with C an arbitrary constant. But iF f (0) = 4, then f (0) = C = 4 .
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HW02-solutions - griffin(ang684 HW02 Gilbert(56650 This...

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