# proof of Darboux’s theorem

Without loss of generality we migth and shall assume ${f}_{+}^{\prime}(a)>t>{f}_{-}^{\prime}(b)$. Let $g(x):=f(x)-tx$. Then ${g}^{\prime}(x)={f}^{\prime}(x)-t$, ${g}_{+}^{\prime}(a)>0>{g}_{-}^{\prime}(b)$, and we wish to find a zero of ${g}^{\prime}$.

Since $g$ is a continuous function^{} on $[a,b]$, it attains a maximum on $[a,b]$.
Since ${g}_{+}^{\prime}(a)>0$ and $$ Fermat’s theorem (http://planetmath.org/FermatsTheoremStationaryPoints) states that
neither $a$ nor $b$ can be points where $f$ has a local maximum^{}.
So a maximum is attained at some $c\in (a,b)$. But then ${g}^{\prime}(c)=0$ again by Fermat’s theorem (http://planetmath.org/FermatsTheoremStationaryPoints).

Title | proof of Darboux’s theorem |
---|---|

Canonical name | ProofOfDarbouxsTheorem |

Date of creation | 2013-03-22 12:45:04 |

Last modified on | 2013-03-22 12:45:04 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 7 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 26A06 |