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Unformatted text preview: pokharel (yp624) – HW12 – Radin – (56520) 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 Use Newton’s method to estimate the solu tion to x 3 − x − 7 = 0 starting with the initial guess x = 2 and applying one iteration. 1. estimate = 9 5 2. estimate = 21 11 3. estimate = 5 2 4. estimate = 23 11 correct 5. estimate = 3 2 6. estimate = 11 5 Explanation: If x n is one estimate of a solution to the equation f ( x ) = 0, then Newton’s method says that x n +1 = x n − f ( x n ) f ′ ( x n ) will usually be a better estimate. But when f ( x ) = x 3 − x − 7 . then f ′ ( x ) = 3 x 2 − 1 , so Newton’s method gives the iteration for mula x n +1 = x n − x 3 n − x n − 7 3 x 2 n − 1 . Consequently, with an initial guess of x = 2, x 1 = 2 − 2 3 − 1 · 2 − 7 3 · 2 2 − 1 = 23 11 . 002 10.0 points A differentiable function f has the proper ties (i) f ( x ) < 0 for x < x , (ii) f ( x ) > 0 for x > x , (iii) f (3) = 7 , f ′ (3) = 4. Use the tangent line at (3 , 7) to the graph of f to compute an approximate value for the value of x . 1. x ≈ 3 2 2. x ≈ 3 4 3. x ≈ 7 4 4. x ≈ 5 4 correct 5. x ≈ 1 Explanation: The tangent line at (3 , 7) to the graph of f is given by y − 7 = f ′ (3)( x − 3) , i . e ., y = 4 x − 5 . The basic idea underlying Newton’s Method is that the xintercept of this tangent line pro vides an approximate value for the xintercept of the graph of f . Since the graph of f crosses the xaxis at x = x , x ≈ 5 4 . 003 10.0 points Find all functions g such that g ′ ( x ) = 4 x 2 + 3 x + 1 √ x . pokharel (yp624) – HW12 – Radin – (56520) 2 1. g ( x ) = 2 √ x ( 4 x 2 + 3 x + 1 ) + C 2. g ( x ) = √ x ( 4 x 2 + 3 x + 1 ) + C 3. g ( x ) = √ x parenleftbigg 4 5 x 2 + x + 1 parenrightbigg + C 4. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + x + 1 parenrightbigg + C cor rect 5. g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + x − 1 parenrightbigg + C 6. g ( x ) = 2 √ x ( 4 x 2 + 3 x − 1 ) + C Explanation: After division g ′ ( x ) = 4 x 3 / 2 + 3 x 1 / 2 + 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 x 3 / 2 + 2 x 1 / 2 = 2 √ x parenleftbigg 4 5 x 2 + x + 1 parenrightbigg is an antiderivative of g ′ . Consequently, g ( x ) = 2 √ x parenleftbigg 4 5 x 2 + x + 1 parenrightbigg + C with C an arbitrary constant. 004 10.0 points Determine f ( t ) when f ′′ ( t ) = 6( t − 1) and f ′ (1) = 5 , f (1) = 4 . 1. f ( t ) = t 3 − 3 t 2 + 8 t − 2 correct 2. f ( t ) = t 3 + 6 t 2 − 8 t + 5 3. f ( t ) = t 3 + 3 t 2 − 8 t + 8 4. f ( t ) = 3 t 3 − 6 t 2 + 8 t − 1 5. f ( t ) = 3 t 3 − 3 t 2 + 8 t − 4 6. f ( t ) = 3 t 3 + 6 t 2 − 8 t + 3 Explanation: The most general antiderivative of f ′′ has the form f...
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This note was uploaded on 06/05/2010 for the course PHYS 92515 taught by Professor Tsoi during the Spring '10 term at University of Texas at Austin.
 Spring '10
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