HW12-solutions - white (taw933) HW12 ben-zvi (55600) 1 This...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: white (taw933) HW12 ben-zvi (55600) 1 This print-out should have 18 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 + x + 3 x . 1. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x 3 parenrightbigg + C 2. g ( x ) = x parenleftbigg 4 5 x 2 + 1 3 x + 3 parenrightbigg + C 3. g ( x ) = x ( 4 x 2 + x + 3 ) + C 4. g ( x ) = 2 x ( 4 x 2 + x + 3 ) + C 5. g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 3 parenrightbigg + C cor- rect 6. g ( x ) = 2 x ( 4 x 2 + x 3 ) + C Explanation: After division g ( x ) = 4 x 3 / 2 + 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 + 2 3 x 3 / 2 + 6 x 1 / 2 = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 3 parenrightbigg is an antiderivative of g . Consequently, g ( x ) = 2 x parenleftbigg 4 5 x 2 + 1 3 x + 3 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Determine f ( t ) when f ( t ) = 6(3 t 1) and f (1) = 6 , f (1) = 5 . 1. f ( t ) = 9 t 3 6 t 2 + 3 t 1 2. f ( t ) = 3 t 3 3 t 2 + 3 t + 2 correct 3. f ( t ) = 3 t 3 + 6 t 2 3 t 1 4. f ( t ) = 9 t 3 + 6 t 2 3 t 7 5. f ( t ) = 9 t 3 3 t 2 + 3 t 4 6. f ( t ) = 3 t 3 + 3 t 2 3 t + 2 Explanation: The most general anti-derivative of f has the form f ( t ) = 9 t 2 6 t + C where C is an arbitrary constant. But if f (1) = 6, then f (1) = 9 6 + C = 6 , i.e., C = 3 . From this it follows that f ( t ) = 9 t 2 6 t + 3 . The most general anti-derivative of f is thus f ( t ) = 3 t 3 3 t 2 + 3 t + D , where D is an arbitrary constant. But if f (1) = 5, then f (1) = 3 3 + 3 + D = 5 , i.e., D = 2 . Consequently, f ( t ) = 3 t 3 3 t 2 + 3 t + 2 . white (taw933) HW12 ben-zvi (55600) 2 003 10.0 points Consider the following functions: ( A ) F 1 ( x ) = cos 2 x 2 , ( B ) F 2 ( x ) = cos 2 x 4 , ( C ) F 3 ( x ) = sin 2 x 2 . Which are anti-derivatives of f ( x ) = sin x cos x ? 1. F 1 only 2. F 3 only correct 3. F 2 only 4. none of them 5. F 1 and F 3 only 6. F 1 and F 2 only 7. all of them 8. F 2 and F 3 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 anti-derivative. ( B ) Not anti-derivative. ( C ) Anti-derivative. 004 10.0 points Find the most general anti-derivative of the function f ( x ) = 7 cos x 3 sin x ....
View Full Document

This note was uploaded on 12/05/2011 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.

Page1 / 10

HW12-solutions - white (taw933) HW12 ben-zvi (55600) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online