Use a fixed-point iteration method to find an approximation to $\sqrt{3}$ that is accurate to within $10^{-4}$?
For this method one is supposed to derive an $g(x)$. Typically $g(x)$ can be derived from $f(x)$ but the problem here is we do not not have an $f(x)$. The book says it got $g(x)=0.5(x+\frac{3}{x})$. I do not know how the book got this result. My best guess is.
$$x=\sqrt{3} \\ x^2=3 \\ x^2-3=0 \\ f(x)= x^2-3 \\ g(x) = \sqrt{3}$$
The given $g(x)$ is Newton's method for $f(x)=x^2-3$. You could get a different fixed-point method from the Newton method for $f(x)=x-3/x$ or $f(x)=x^{3/2}-3x^{-1/2}$.