If you can bound $f'(x)$, you only need to numerically evaluate finitely many points to show that the function is positive everywhere. (Given $-A < f'(x) < A$, if $f(x_0) = C$, take $x_1 = x_0 + C/A$.)
This doesn't work at $f(1) = 0$, but showing the negativity of $f'(x)$ around that point (between the last x-value you check and x=1) would suffice. Unfortunately, $f'(1) = 0$, so you'd need to show the positivity of $f''(x)$ near $1$. $f''(x)$ is positive at $x=1$, so the original strategy will work for it. It just unfortunately means you'd have to bound $ f'''(x)$.
