fsolve: avoid useless numerical evaluation of the function

Don't compute f(x) if new x is outside of the interval. We don't need that
value anyway, and the function might be difficult to compute numerically or
even ill defined outside the interval.

As a result fsolve is able to find root(s) of some weird functions.
For example

fsolve((1/(sqrt(2*Pi)))*integral(t, 0, x, exp(-1/2*t^2)) == 0.5, x, 0, 100)

actually works now.