How do you demonstrate that a local minimum of a function has its derivative equal to zero with the definition of the derivative.
Proving that a point is a Stationary point with the definition of the derivative
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0As Jonas points out, the issue here is non-differentiability, not the value or location of the local minimum. If $f(x)$ has a local minimum at $p$, and is defined on an open interval containing $p$, then we know that **either** $f'(p)=0$, or else that $f$ is not differentiable at $p$. But we cannot ignore the latter possibility. – 2011-01-18
2 Answers
You want to prove that if $f(x)$ has a local minimum at $a$, and $f$ is differentiable at $a$, then f'(a)=0.
Because $f(x)$ has a local minimum at $a$, that means that there exists a $\delta\gt 0$ such that for all $x$, if $|x-a|\leq \delta$, then $f(a)\leq f(x)$.
The derivative at $a$ is: $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$ Consider the limits from the left and from the right. If $x\gt a$ and is sufficiently close to $a$ (within $\delta$), then $f(x)-f(a)\geq 0$ and $x-a\gt 0$. So the fraction $\frac{f(x)-f(a)}{x-a}$ satisfies $\frac{f(x)-f(a)}{x-a}\geq 0$. In particular, the limit must be greater than or equal to $0$ (if it exists). Note that we can certainly restrict ourselves to being within $\delta$ of $a$, because the value of the limit only depends on what happens close to $a$.
Now consider what happens if $x\lt a$. If we are within $\delta$ of $a$, then $f(x)-f(a)\geq 0$ (because $f$ has a minimum at $a$), but $x-a\lt 0$. That means that the fraction $\frac{f(x)-f(a)}{x-a}$ satisfies $\frac{f(x)-f(a)}{x-a}\leq 0$. In particular, the limit, if it exists, must be less than or equal to $0$.
Since we are assuming that the derivative exists, both one-sided limits exist and are equal. The limit from the right must be greater than or equal to $0$, and the limit from the left must be less than or equal to $0$; the only possibility for them to be equal is for them both to be $0$; that is, for f'(a) to be equal to $0$.
Having a local minimum at $x$ implies that the right-hand difference quotients are nonnegative and the left-hand difference quotients are nonpositive (when taken sufficiently close to $x$). If the derivative exists, then it must equal the limit of both the left-hand and right-hand difference quotients, so it must be a limit of nonnegative numbers and a limit of nonpositive numbers.