Question 10 Consider the equation x3−3x+1=0.x^3-3x+1=0. (a) Explain why Newton’s method fails with the initial guess x1=1x_1=1. (b) Restart with x1=0x_1=0, and compute x2x_2, x3x_3, and x4x_4. Round the answers to four decimal places. Show solutionHide solution+Question 10 - Solution At x1=1x_1=1, f(1)=−1f(1)=-1 and f′(1)=0f\prime(1)=0, so the Newton step is undefined. Restart at x1=0x_1=0. Newton’s iteration is xn+1=xn−f(xn)f′(xn),f(x)=x3−3x+1,f′(x)=3x2−3.x_{n+1}=x_n-\frac{f(x_n)}{f\prime(x_n)},\qquad f(x)=x^3-3x+1,\quad f\prime(x)=3x^2-3. Keeping full precision internally gives x2=0.0000000000−1.0000000000−3.0000000000≈0.3333x3=0.3333333333−0.0370370370−2.6666666667≈0.3472x4=0.3472222222−0.0001955804−2.6383101852≈0.3473\begin{aligned}x_{2}&=0.0000000000-\frac{1.0000000000}{-3.0000000000}\approx\boxed{0.3333}\\[6pt]x_{3}&=0.3333333333-\frac{0.0370370370}{-2.6666666667}\approx\boxed{0.3472}\\[6pt]x_{4}&=0.3472222222-\frac{0.0001955804}{-2.6383101852}\approx\boxed{0.3473}\end{aligned}
