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How do you demonstrate that a local minimum of a function has its derivative equal to zero with the definition of the derivative.

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You can't, because it's not true. $x=0$ is a local minimum of $y=|x|$, but it is not a stationary point of the function. –  Arturo Magidin Jan 17 '11 at 23:44
    
we are not talking about x = 0 here –  DogDog Jan 18 '11 at 0:45
    
It is not important whether or not the minimum occurs where $x=0$, but it is important that the a function need not be differentiable (or even continuous) at a point where it has a local minimum. If you want to modify Arturo's example so that $x$ is not zero, consider $y=|x-1|$, which has a local minimum at $x=1$ but is not differentiable there. –  Jonas Meyer Jan 18 '11 at 1:15
    
As 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. –  Arturo Magidin Jan 18 '11 at 2:29
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up vote 3 down vote accepted

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 $0$.

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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.

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