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