# proof of Rolle’s theorem

Because $f$ is continuous^{} on a compact (closed and bounded) interval $I=[a,b]$, it attains its
maximum and minimum values. In case $f(a)=f(b)$ is both the maximum and
the minimum, then there is nothing more to say, for then $f$ is a constant function and
${f}^{\prime}\equiv 0$ on the whole interval $I$. So suppose otherwise, and $f$ attains an extremum^{}
in the open interval
$(a,b)$, and without loss of generality, let this extremum be a maximum, considering $-f$ in
lieu of $f$ as necessary. We claim that at this extremum $f(c)$ we have ${f}^{\prime}(c)=0$, with $$.

To show this, note that $f(x)-f(c)\le 0$ for all
$x\in I$, because $f(c)$ is the maximum. By definition of the derivative^{}, we have that

$${f}^{\prime}(c)=\underset{x\to c}{lim}\frac{f(x)-f(c)}{x-c}.$$ |

Looking at the one-sided limits, we note that

$$R=\underset{x\to {c}^{+}}{lim}\frac{f(x)-f(c)}{x-c}\le 0$$ |

because the numerator in the limit is nonpositive in the interval $I$, yet $x-c>0$, as $x$ approaches $c$ from the right. Similarly,

$$L=\underset{x\to {c}^{-}}{lim}\frac{f(x)-f(c)}{x-c}\ge 0.$$ |

Since $f$ is differentiable^{} at $c$, the left and right limits must coincide, so
$0\le L=R\le 0$, that is to say, ${f}^{\prime}(c)=0$.

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

Canonical name | ProofOfRollesTheorem |

Date of creation | 2013-03-22 12:40:19 |

Last modified on | 2013-03-22 12:40:19 |

Owner | rmilson (146) |

Last modified by | rmilson (146) |

Numerical id | 5 |

Author | rmilson (146) |

Entry type | Proof |

Classification | msc 26A06 |