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).