# proof of Fermat’s Theorem (stationary points)

Suppose that $x_{0}$ is a local maximum (a similar proof applies if $x_{0}$ 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 $f(x_{0})\geq f(x)$ for all $x$ with $|x-x_{0}|<\delta$. Hence for $h\in(0,\delta)$ we notice that it holds

$\frac{f(x_{0}+h)-f(x_{0})}{h}\leq 0.$

Since the limit of this ratio as $h\to 0^{+}$ exists and is equal to $f^{{\prime}}(x_{0})$ we conclude that $f^{{\prime}}(x_{0})\leq 0$. On the other hand for $h\in(-\delta,0)$ we notice that

$\frac{f(x_{0}+h)-f(x_{0})}{h}\geq 0$

but again the limit as $h\to 0^{+}$ exists and is equal to $f^{{\prime}}(x_{0})$ so we also have $f^{{\prime}}(x_{0})\geq 0$.

Hence we conclude that $f^{{\prime}}(x_{0})=0$.

To prove the second part of the statement (when $x_{0}$ is equal to $a$ or $b$), just notice that in such points we have only one of the two estimates written above.

Major Section:
Reference
Type of Math Object:
Proof

## Mathematics Subject Classification

26A06 One-variable calculus

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## Info

Owner: paolini
Added: 2003-07-15 - 12:58
Author(s): paolini

## Versions

(v5) by paolini 2013-03-22
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