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# cacl12 - pokharel(yp624 HW12 Radin(56520 This print-out...

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pokharel (yp624) – HW12 – Radin – (56520) 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 Use Newton’s method to estimate the solu- tion to x 3 x 7 = 0 starting with the initial guess x 0 = 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 0 = 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 0 , (ii) f ( x ) > 0 for x > x 0 , (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 0 . 1. x 0 3 2 2. x 0 3 4 3. x 0 7 4 4. x 0 5 4 correct 5. x 0 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 x -intercept of this tangent line pro- vides an approximate value for the x -intercept of the graph of f . Since the graph of f crosses the x -axis at x = x 0 , x 0 5 4 . 003 10.0 points Find all functions g such that g ( x ) = 4 x 2 + 3 x + 1 x .

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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 anti-derivative of f ′′ has the form f ( t ) = 3 t 2 6 t + C where C is an arbitrary constant. But if f (1) = 5, then f (1) = 3 6 + C = 5 , i.e., C = 8 . From this it follows that f ( t ) = 3 t 2 6 t + 8 .
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