# 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 |