Answer
See graph and explanations.
Work Step by Step
a. See graph; we can find $x=\frac{1}{3}$ when the curve crosses the x-axis, which is the reciprocal of $3$.
b. The Newton’s formula gives $x_{n+1}=x_{n}-\frac{y(x_n)}{y’(x_n)}$. Since $y’=-\frac{1}{x^2}$, we have $x_{n+1}=x_{n}+\frac{1/x_n-3)}{1/ x_n^2)}=x_n+x_n-3x_n^2=x_n(2-3x_n)$.
