# proof of least and greatest value of function

$f$ is continuous^{}, so it will transform compact sets into compact sets.
Thus since $[a,b]$ is compact, $f([a,b])$ is also compact.
$f$ will thus attain on the interval $[a,b]$ a maximum and a minimum value because real compact sets are closed and bounded^{}.

Consider the maximum and later use the same argument for $-f$ to consider the minimum.

By a known theorem (http://planetmath.org/FermatsTheoremStationaryPoints) if the maximum is attained in the interior of the domain, $c\in ]a,b[$ then $f(c)\text{is a maximum}\u27f9{f}^{\prime}(c)=0$, since $f$ is differentiable^{}.

If the maximum isn’t attained in $]a,b[$ and since it must be attained in $[a,b]$ either $f(a)$ or $f(b)$ is a maximum.

For the minimum consider $-f$ and note that $-f$ will verify all conditions of the theorem and that a maximum of $-f$ corresponds to a minimum of $f$ and that $-{f}^{\prime}(c)=0\iff {f}^{\prime}(c)=0$.

Title | proof of least and greatest value of function |
---|---|

Canonical name | ProofOfLeastAndGreatestValueOfFunction |

Date of creation | 2013-03-22 15:52:09 |

Last modified on | 2013-03-22 15:52:09 |

Owner | cvalente (11260) |

Last modified by | cvalente (11260) |

Numerical id | 5 |

Author | cvalente (11260) |

Entry type | Proof |

Classification | msc 26B12 |

Related topic | FermatsTheoremStationaryPoints |

Related topic | HeineBorelTheorem |

Related topic | CompactnessIsPreservedUnderAContinuousMap |