Show solutionHide solution+ Question 1 Use Newton’s Method to approximate a root of the function f(x)=x3−x−1f(x) = x^3 - x - 1 starting with x0=1.5x_0 = 1.5, and compute the first two iterations: x1x_1 and x2x_2. Question 1 - Solution Newton’s iteration is xn+1=xn−f(xn)f′(xn),f(x)=x3−x−1,f′(x)=3x2−1.x_{n+1}=x_n-\frac{f(x_n)}{f\prime(x_n)},\qquad f(x)=x^3-x-1,\quad f\prime(x)=3x^2-1. Keeping full precision internally gives x1=1.5000000000−0.87500000005.7500000000≈1.3478x2=1.3478260870−0.10068217314.4499054820≈1.3252\begin{aligned}x_{1}&=1.5000000000-\frac{0.8750000000}{5.7500000000}\approx\boxed{1.3478}\\[6pt]x_{2}&=1.3478260870-\frac{0.1006821731}{4.4499054820}\approx\boxed{1.3252}\end{aligned}