Use all three methods in this Section to find solutions to within 107 for the following problems.
Question:
a. x2 − 4x + 4 − ln x = 0 for 1 ≤ x ≤ 2 and for 2 ≤ x ≤ 4
b. x + 1 − 2 sin πx = 0 for 0 ≤ x ≤ 1/2 and for 1/2 ≤ x ≤ 1
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a We have b We have Newtons Secant False Position ...View the full answer
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