Suppose is a function defined on a closed interval (with ) satisfying the following three conditions:
- is a continuous function on the closed interval . In particular, is (two-sided) continuous at every point in the open interval , right continuous at , and left continuous at .
- is differentiable on the open interval , i.e., the derivative of exists at all points in the open interval .
Then, there exists in the open interval such that .
- Lagrange mean value theorem
- Bound relating number of zeros of function and number of zeros of its derivative
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||If is zero on all of , then for any choice of||obvious|
|2||must attain both its maximum and its minimum values on||Fact (1)||is continuous on||Given-fact combination direct|
|3||If is not zero on all of , either its absolute maximum value on is positive and attained at a point in the open interval or its absolute minimum value on is negative and attained at a point in the open interval (or possibly both).||Step (2)||[SHOW MORE]|
|4||If is a point in at which attains its maximum value or its minimum value, then .||Fact (2)||is differentiable on||[SHOW MORE]|
|5||There is a point at which .||Steps (1), (3), (4)||[SHOW MORE]|