Fermats_Theore := Let f be differentiable on (a, b). If f has a local maximum or minimum at
c ∈ (a, b),
then f′(c) = 0.
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proof. Suppose c is a local maximum: Since it is a local maximum, the values on the left should be less, and the values on the right should be less. Then stretch this constant by $\frac{1}{x-c}$ due to foresight, the notational equivalence to this is.$$ \frac{f(x)-f(c)}{x-c} \ge 0 \forall x \in (c-\epsilon,c) \text{and} \frac{f(x)-f(c)}{x-c} \le 0 \forall x \in (c,c+\epsilon) $$ It follows that $$ \lim_{x \to c_{_}} \frac{f(x)-f(c)}{x-c} \ge 0 and \lim_{x \to c_{+}} \frac{f(x)-f(c)}{x-c} \le 0 $$
and so $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x-c} = 0$
Note: |x| is not differentiable at 0 since :todo: Why is |x| not differentiable here?