To use the N-R for function evaluation $f(x)$ requires an extra step, which is to construct a function F such that $F(f(x)) = 0$. Is there a systematic way to do this? As an example, take $f(x) = 1/\sqrt{x}$. Construct a function $F(q)$ such that $F(1/\sqrt{x}) = 0$. The straightforward approach would give something like $$F_1(q) = q - 1/\sqrt{x}.$$ Computing the N-R update from $F_1$ is rather useless $$u_1(q) = q - \frac{q - {1/\sqrt{x}}}{1} = 1/\sqrt{x}.$$ The trick seems to be to take this trivially obtained but useless $F_1$ and change it in a way that turns the dependency on $x$ into elementary functions, without removing the zero of $F_1$ we are interested in. The following two steps seem appropriate: remove the square root $$F_2(q) = (q - 1/\sqrt{x})(q +1/\sqrt{x}) = q^2 - 1/x$$ and remove the division operation $$F_3(q) = xq^2 - 1$$ Which of the two is best? There is no way to decide until computing the N-R update steps. It turns out that both lead to the same function $$u_2(q) = u_3(q) = \frac{1}{2}(q + \frac{1}{xq}).$$ Now $u_2(q)$ still contains a division, which is not optimal. There are other ways. In [1] the function $F_4(q) = 1/q^2 - x$ is used, which leads to $$u_4(q) = \frac{1}{2}(3q - xq^3).$$ The heuristic to distill from this is that if there is a division involved, it might be best to express $F(q)$ in terms of $1/q$. The derivative will have a decreasing negative power. The negative power is then cancelled by $F'(q)$ appearing in a denominator in the update formula. That trick should then also work for evaluating $f(x) = 1/x$. Indeed, with $F_5(q) = 1/q - x$ we get $u_5(q) = q(2 - xq)$. %[1] http://en.wikipedia.org/wiki/Fast_inverse_square_root