# Lagrange mean value theorem

## Contents

## Statement

Suppose is a function defined on a closed interval (with ) such that the following two conditions hold:

- is a continuous function on the closed interval (i.e., it is right continuous at , left continuous at , and two-sided continuous at all points in the open interval ).
- is a differentiable function on the open interval , i.e., the derivative exists at all points in . Note that we
*do not*require the derivative of to be a continuous function.

Then, there exists in the open interval such that the derivative of at equals the difference quotient . More explicitly:

Geometrically, this is equivalent to stating that the tangent line to the graph of at is parallel to the chord joining the points and .

Note that the theorem simply guarantees the existence of , and does not give a formula for finding such a (which may or may not be unique).

## Related facts

## Facts used

## Proof

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|

1 | Consider the function . Then, is a linear (and hence a continuous and differentiable) function with and |
Just plug in and check. Secretly, we obtained by trying to write the equation of the line joining the points and . | |||

2 | Define on , i.e., . | ||||

3 | is continuous on | Fact (1) | is continuous on | Steps (1), (2) | [SHOW MORE] |

4 | is differentiable on | Fact (2) | is differentiable on | Steps (1), (2) | [SHOW MORE] |

5 | Steps (1), (2) | [SHOW MORE] | |||

6 | There exists such that . | Fact (3) | Steps (3), (4), (5) | [SHOW MORE] | |

7 | For the obtained in step (6), | Fact (4) | Steps (2), (6) | [SHOW MORE] | |

8 | for all . In particular, . | Step (1) | Differentiate the expression for from Step (1). | ||

9 | Steps (7), (8) | Step-combination direct |