# proof of mean value theorem

Define $h(x)$ on $[a,b]$ by

$$h(x)=f(x)-f(a)-\left(\frac{f(b)-f(a)}{b-a}\right)(x-a)$$ |

Clearly, $h$ is continuous^{} on $[a,b]$, differentiable^{} on $(a,b)$, and

$$\begin{array}{ccc}\hfill h(a)\hfill & \hfill =\hfill & f(a)-f(a)=0\hfill \\ \hfill h(b)\hfill & \hfill =\hfill & f(b)-f(a)-\left(\frac{f(b)-f(a)}{b-a}\right)(b-a)=0\hfill \end{array}$$ |

Notice that $h$ satisfies the conditions of Rolle’s Theorem. Therefore, by Rolle’s Theorem there exists $c\in (a,b)$ such that ${h}^{\prime}(c)=0$.

However, from the definition of $h$ we obtain by differentiation^{} that

$${h}^{\prime}(x)={f}^{\prime}(x)-\frac{f(b)-f(a)}{b-a}$$ |

Since ${h}^{\prime}(c)=0$, we therefore have

$${f}^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$$ |

as required.

## References

- 1 Michael Spivak, Calculus, 3rd ed., Publish or Perish Inc., 1994.

Title | proof of mean value theorem |
---|---|

Canonical name | ProofOfMeanValueTheorem |

Date of creation | 2013-03-22 12:40:57 |

Last modified on | 2013-03-22 12:40:57 |

Owner | Andrea Ambrosio (7332) |

Last modified by | Andrea Ambrosio (7332) |

Numerical id | 5 |

Author | Andrea Ambrosio (7332) |

Entry type | Proof |

Classification | msc 26A06 |