This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: mandel (tgm245) – HW12 – Radin – (56470) 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 ) = x 2 + 2 x + 4 √ x . 1. g ( x ) = 2 √ x ( x 2 + 2 x + 4 ) + C 2. g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 2 3 x + 4 parenrightbigg + C cor rect 3. g ( x ) = √ x ( x 2 + 2 x + 4 ) + C 4. g ( x ) = 2 √ x ( x 2 + 2 x − 4 ) + C 5. g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 2 3 x − 4 parenrightbigg + C 6. g ( x ) = √ x parenleftbigg 1 5 x 2 + 2 3 x + 4 parenrightbigg + C Explanation: After division g ′ ( x ) = x 3 / 2 + 2 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 2 5 x 5 / 2 + 4 3 x 3 / 2 + 8 x 1 / 2 = 2 √ x parenleftbigg 1 5 x 2 + 2 3 x + 4 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 1 5 x 2 + 2 3 x + 4 parenrightbigg + C with C an arbitrary constant. 002 10.0 points Find the value of f (0) when f ′′ ( t ) = 2(3 t + 4) and f ′ (1) = 3 , f (1) = 2 . 1. f (0) = 8 2. f (0) = 7 3. f (0) = 5 correct 4. f (0) = 9 5. f (0) = 6 Explanation: The most general antiderivative of f ′′ has the form f ′ ( t ) = 3 t 2 + 8 t + C where C is an arbitrary constant. But if f ′ (1) = 3, then f ′ (1) = 3 + 8 + C = 3 , i.e., C = − 8 . From this it follows that f ′ ( t ) = 3 t 2 + 8 t − 8 , and the most general antiderivative of the latter is f ( t ) = t 3 + 4 t 2 − 8 t + D , where D is an arbitrary constant. But if f (1) = 2, then f (1) = 1 + 4 − 8 + D = 2 , i.e., D = 5 . Consequently, f ( t ) = t 3 + 4 t 2 − 8 t + 5 . mandel (tgm245) – HW12 – Radin – (56470) 2 At x = 0, therefore, f (0) = 5 . 003 10.0 points Find the value of f (0) when f ′ ( t ) = 5 sin2 t , f parenleftBig π 2 parenrightBig = 3 . 1. f (0) = − 3 2. f (0) = 1 3. f (0) = − 2 correct 4. f (0) = 0 5. f (0) = − 1 Explanation: Since d dx cos mt = − m sin mt , for all m negationslash = 0, we see that f ( t ) = − 5 2 cos 2 t + C where the arbitrary constant C is determined by the condition f ( π/ 2) = 3. But cos 2 t vextendsingle vextendsingle vextendsingle t = π/ 2 = cos π = − 1 . Thus f parenleftBig π 2 parenrightBig = 5 2 + C = 3 , and so f ( t ) = − 5 2 cos 2 t + 1 2 . Consequently, f (0) = − 2 . 004 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 antiderivatives of f ( x ) = sin x cos x ? 1. F 2 only 2. F 1 only 3. all of them 4. F 2 and F 3 only 5. none of them 6. F 3 only correct 7. F 1 and F 2 only 8. F 1 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 ....
View
Full
Document
This note was uploaded on 06/14/2011 for the course MATH 305G taught by Professor Cathy during the Spring '11 term at University of Texas at Austin.
 Spring '11
 Cathy

Click to edit the document details