# Name 2 2 compute approximate derivatives of f x at x

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error since there is no previous approximation).

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NAME 2 2. Compute approximate derivatives of f ( x ) at x = 1 by the forward difference, backward difference, and centered difference methods with a step size of h = 0 . 1. Compare to the true derivative at x = 1 and compute the true absolute error in the approximation. Explain this in terms of the O () of the theoretical error in the three cases. Solution We have f (1) = - 6 f ( . 9) = - 4 . 872 f (1 . 1) = - 7 . 068 f (1) = - 11 The three approximations are ˆ f F (1) = ( - 7 . 068) - ( - 6) 0 . 1 = - 10 . 68 , | E t | = 0 . 32 ˆ f B (1) = ( - 6) - ( - 4 . 872) 0 . 1 = - 11 . 28 , | E t | = 0 . 28 ˆ f C (1) = ( - 7 . 068) - ( - 4 . 872) 0 . 2 = - 10 . 98 , | E t | = 0 . 02 The centered difference quotient, which has error O ( h 2 ), has a much smaller true error than the forward and backward difference quotients, which have O ( h ) error.
NAME 3 3. Verify by the signs of the function values that a root of f ( x ) lies between 0.2 and 0.7. Perform three iterations of the bisection algorithm to obtain an approximate answer. What is the estimated absolute error in each case? If the true root is 0.5, what is the true absolute error? Solution Here are the iterates. Note that the first two are of opposite signs as required: f (0 . 2) = 3 . 696 f (0 . 7) = - 2 . 484 f (0 . 45) = 0 . 62475 , E t = 0 . 05 , E a = 0 . 25 f (0 . 575) = - 0 . 9366 , E t = 0 . 075 , E a = 0 . 125 f (0 . 5125) = - 0 . 15625 , E t = 0 . 0125 , E a = 0 . 0625

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NAME 4 4. Using the same starting interval of 0.2 to 0.7, perform two iterations of the false position algorithm (use 6 significant figures). What are the estimated and true absolute errors at the second iteration?
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