(a) Use Newton's method with $\mathrm{x}_{1}=1$ to find the root of the
equation $\mathrm{x}^{3}-\mathrm{x}=1$ correct to six decimal places.
b) Solve the equation in part (a) using $x_{1}=0.6$ as the initial
approximation.
(c) Solve the equation in part (a) using $x_{1}=0.57 .$ (You defi-
nitely need a programmable calculator for this part.)
(d) Graph $f(x)=x^{3}-x-1$ and its tangent lines at $x_{1}=1$
$0.6,$ and 0.57 to explain why Newton's method is so sen-
sitive to the value of the initial approximation.v