Lagrange mean value theorem
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:
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).
|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|