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I understand how to calculate it, but I am just curious, why actually it works? Do we have a proof that it always works?

hey
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    It's the other way around: if a differentiable function at $;x_0;$ has an extreme value there then $;f'(x_0)=0;$ . This is Fermat's (go figure!) Theorem. – DonAntonio Jan 28 '14 at 19:23

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Take $f\colon x\in\mathbb{R}\mapsto x^3$. Then $f^\prime(0)=0$, yet $0$ is not an extremum (only an inflection point).

As DonAtonio said, the converse is true — if $x$ is an extremum of a differentiable function $f$, then $f^\prime(0)=0$. A way to see it is that the curve of $f$ goes from "going up" to "going down" (or vice-versa), so the slope (derivative) must be zero (horizontal) at the extremum. Or, to prove it, consider the definition of the derivative as the limit of $$ \frac{f(x+h)-f(x)}{h} $$ when $h\to 0$. If $f(x)$ is a maximum, then for $h < 0$ this is $\geq 0$ (so the limit when $h\nearrow 0$ is), while for $h > 0$ this is $\leq 0$ (so the limit when $h\searrow 0$ is). Hence the limit is $0$.

(similarly if $f(x)$ is a minimum)

Clement C.
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Take an interval and draw in there a continuous function. Draw a line that is tangent to an extreme of that function. What is the value of your line's slope? Match it with the definition of derivative. What does that suggest to you?

Daniel Muñoz Parsapoormoghadam
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Basically, a maximum or minimum is defined as when the function goes from going up (positive slope) to going down (negative slope) or vice versa. When the slope is 0 it's exactly between going up and going down, or a point where the curve flattens out but doesn't change direction (as in C), so the places that have slope 0 and change direction (aren't inflection points) are maximums and minimums.

qwr
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  • This explanation, besides being way non-formal, is seriously wrong, and the drawing itself gives a hint: at the point $;C;$ the function' derivative (assuming $;f;$ is differentiable and etc.) is zero, and that point is not an extreme point but in fact, if we're going by the drawing, an inflection point, thus rendering the last words above wrong. – DonAntonio Jan 29 '14 at 04:08
  • Modified my statement. My goal was to be understandable, so I avoided a traditional proof. – qwr Jan 29 '14 at 04:54
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Consider the case where a differentiable function $f(x)$ has a minimum at the point $x_{0}$. By definition, there exists a $\epsilon-$neighborhood $$ x = \left\{ x\in\mathbb{R} \colon \left| x - x_{0} \right| < \epsilon \right\} $$ with $\epsilon>0$ where $f(x) > f(x_{0})$.

If we constrain $\delta < \epsilon$, then $$ \frac{f(x+\delta) - f(x)} {\delta} > 0. $$ So within the $\epsilon-$neighborhood, a $\delta-$sequence produces a limit which approaches 0 from above.

A similar argument holds for approaching the minimum from the left.

dantopa
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