- Fermat's theorem (stationary points)
**Fermat's theorem**is atheorem inreal analysis , named afterPierre de Fermat . It gives a method to find local maxima and minima ofdifferentiable function s by showing that every local extremum of the function is astationary point (the functionderivative is zero in that point). So, by using Fermat's theorem, the problem of finding a function extremum is reduced to solving anequation .It is important to note that Fermat's theorem gives only a

necessary condition for extreme function values. That is, some stationary points are not extreme values, they areinflection point s. To check if a stationary point is an extreme value and to further distinguish between a function maximum and a function minimum it is necessary to analyse the second derivative (if it exists).**Fermat's theorem**Let $fcolon\; (a,b)\; ightarrow\; mathbb\{R\}$ be a function and suppose that $displaystyle\; x\_0\; in\; (a,b)$ is a local extremum of $displaystyle\; f$. If $displaystyle\; f$ is differentiable at $displaystyle\; x\_0$ then $displaystyle\; f\text{'}(x\_0)\; =\; 0$.

**Application to optimization**As a corollary, global extrema of a function "f" on a domain "A" occur only at boundaries, non-differentiable points, and stationary points.If $x\_0$ is a global extremum of "f", then one of the following is true:

***boundary:**$x\_0$ is in the boundary of "A"

***non-differentiable:**"f" is not differentiable at $x\_0$

***stationary point:**$x\_0$ is a stationary point of "f"**Intuition**The intuition is based on the behavior of

polynomial functions. Assume that function "f" has a maximum at "x"_{0}, the reasoning being similar for a function minimum. If $displaystyle\; x\_0\; in\; (a,b)$ is a local maximum then there is a (possibly small) neighborhood of $displaystyle\; x\_0$ such as the function is increasing before and decreasing after $displaystyle\; x\_0$. As the derivative is positive for an increasing function and negative for a decreasing function, $displaystyle\; f\text{'}$ is positive before and negative after $displaystyle\; x\_0$. $displaystyle\; f\text{'}$ doesn't skip values (by Darboux's theorem), so it has to be zero at some point between the positive and negative values. The only point in the neighbourhood where it is possible to have $displaystyle\; f\text{'}(x)\; =\; 0$ is $displaystyle\; x\_0$.Note that the theorem (and its proof below) is more general than the intuition in that it doesn't require the function to be differentiable over a neighbourhood around $displaystyle\; x\_0$. As stated in the theorem, it is sufficient for the function to be differentiable only in the extreme point.

**Proof**Suppose that $displaystyle\; x\_0$ is a local maximum (a similar proof applies if $displaystyle\; x\_0$ is a local minimum). Then there $exists\; ,\; delta\; >\; 0$ such that $(x\_0\; -\; delta,x\_0\; +\; delta)\; subset\; (a,b)$ and such that we have $f(x\_0)\; ge\; f(x),\; forall\; x$ with $displaystyle\; |x\; -\; x\_0|\; <\; delta$. Hence for any $h\; in\; (0,delta)$ we notice that it holds

:$frac\{f(x\_0+h)\; -\; f(x\_0)\}\{h\}\; le\; 0.$

Since the limit of this ratio as $displaystyle\; h$ gets close to 0 from above exists and is equal to $displaystyle\; f\text{'}(x\_0)$ we conclude that $f\text{'}(x\_0)\; le\; 0$. On the other hand for $h\; in\; (-delta,0)$ we notice that

:$frac\{f(x\_0+h)\; -\; f(x\_0)\}\{h\}\; ge\; 0$

but again the limit as $displaystyle\; h$ gets close to 0 from below exists and is equal to $displaystyle\; f\text{'}(x\_0)$ so we also have $f\text{'}(x\_0)\; ge\; 0$.

Hence we conclude that $displaystyle\; f\text{'}(x\_0)\; =\; 0$.

**ee also***

Derivative

*Extreme value

*Stationary point

*Inflection point **External links***

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