# Lagrange mean value theorem

## Statement

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

1.  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 ).
2.  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).

## 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