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 $f(x_0)\ge f(x)$ for all $x$ with . Hence for $h\in (0,\delta )$ we notice that it holds

$$\frac{f(x_0+h)-f(x_0)}{h}\le 0.$$

Since the limit of this ratio as $h\to 0^{+}$ exists and is equal to $f^{\prime }(x_0)$ we conclude that $f^{\prime }(x_0)\le 0$. 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 $h\to 0^{+}$ exists and is equal to $f^{\prime }(x_0)$ so we also have $f^{\prime }(x_0)\ge 0$.

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.

Title	proof of Fermat’s Theorem (stationary points)
Canonical name	ProofOfFermatsTheoremstationaryPoints
Date of creation	2013-03-22 13:45:09
Last modified on	2013-03-22 13:45:09
Owner	paolini (1187)
Last modified by	paolini (1187)
Numerical id	5
Author	paolini (1187)
Entry type	Proof
Classification	msc 26A06