find the real root of equation x³-5x+1 using bisection method correct to two decimal places
Answers
Answer:
f(x) = x³ -5x + 1
f’(x) = 3x² -5
f’’ = 6x
From any credible starting x, we will iterate x -f(x)/f’(x). But we need a credible x, and we have one or three real roots.
Looking at the equation, for |x| > 3, the equation is dominated by the cubic term, the derivative has zeroes at ±sqrt(5/3), and from the 2nd derivative, a minimum at ≈ +1.291, and a max at -1.291. Plugging in a few values and looking for sign changes, there must be roots between -3 and -2, 0 and +1, and between +2 and +3, so there are three real roots. Using those values and slope, we guess starting values of -2.6, +0.3, and +2.2. Set up a spreadsheet in Excel to perform iterations, and the three starting values converge to -2.33006, 0.20164, +2.128419.
Note that my starting values were not great estimates, but they were fairly close to the roots, and well away from the zeroes of the derivative, which are very problematic in Newton’s method.