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

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:
As h approaches 0 from the positive side, then:

Thus, f'(c) = 0

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The first digit is in the middle school 100s place