Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given equation. (Round your answer to four decimal places.)

Question

x⁵ − x − 7 = 0,    x₁ = 1

Answer 1

x_n+1 = x_n - f(x_n)/f'(x_n)

f(x) = x⁵-x-7 f'(x) = 5x⁴-1

Usually, you start with guess x₀ and n is the iteration number. I will do the first iteration:

x₂ = 1 - (-7)/4 = 2.75

This is not a good example of the power of Newton-Raphson. It does need a good guess if the function increases sharply. This iteration takes you farther from the root.