Question

Prove fermat's theorem for the case where f has a minimum at x0 (show f 0 (x0) = 0)

Answer 1

Suppose that some value, c, is a point of a local minimum point.

The theorem states that if a function f is differentiable at a point c of local extremum, then f'(c) = 0.

This implies that the function f is continuous over the given interval. So there must be some value h such that f(c + h) - f(c) >= 0, where h is some infinitesimally small quantity.
As h approaches 0 from the negative side, then: $\frac{f(c + h) - f(c)}{h} \leq 0 \text{, where h is approaching 0 from the negative side}$
As h approaches 0 from the positive side, then: $\frac{f(c + h) - f(c)}{h} \geq 0$

Thus, f'(c) = 0