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
- Continuous functions form a vector space
- Differentiable functions form a vector space
- Rolle's theorem
- Differentiation is linear
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 ![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935) |
Fact (1) |
is continuous on ![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935) |
Steps (1), (2) |
[SHOW MORE]By Step (2),  . By Step (1),  is linear and hence continuous on ![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935) . Thus, by Fact (1),  , being the difference of two continuous functions, is continuous on ![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935) .
|
4 |
is differentiable on  |
Fact (2) |
is differentiable on  |
Steps (1), (2) |
[SHOW MORE]By Step (2),  . By Step (1),  is linear and hence differentiable on  . Thus, by Fact (1),  , being the difference of two differentiable functions, is differentiable on  .
|
5 |
 |
|
|
Steps (1), (2) |
[SHOW MORE] yields  .  yields  .
|
6 |
There exists such that . |
Fact (3) |
|
Steps (3), (4), (5) |
[SHOW MORE]Steps (3), (4), (5) together show that  satisfies the conditions of Rolle's theorem on the interval ![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935) , hence we get the conclusion of the theorem.
|
7 |
For the obtained in step (6),  |
Fact (4) |
|
Steps (2), (6) |
[SHOW MORE]From Step (2),  . Differentiating both sides by Fact (4), we get  on  . Since  , we obtain that  .
|
8 |
for all . In particular, . |
|
|
Step (1) |
Differentiate the expression for from Step (1).
|
9 |
 |
|
|
Steps (7), (8) |
Step-combination direct
|