Question

If f′(x) ≥ 0 for all x, then f(a) ≤ f(b) whenever a≤b.

Answer 1

Use the mean value theorem to prove this rigorously. If a < b, then because f is differentiable on the interval [a,b], there exists a number c such that:\[a<c<b\]and \[\frac{f(b)-f(a)}{b-a}=f^{\prime}(c)\]Multiplying both sides of the statement by b-a yields:\[f(b)-f(a)=f^{\prime}(c)(b-a)\]Note that b-a is positive, and also f'(c) is positive because we are assuming that f'(x) is greater than zero for all x. Since a positive times a positive is still positive, we see that:\[f(b)-f(a)\ge0\]But this is the same as saying:\[f(b)\ge f(a).\]