Rolle's Theorem says:
Let f be a function that
Then there is some c in the open interval (a, b) with f'(c) = 0.
Sometimes the third condition is stated as f (a) = f (b) = 0, but for the proof, it doesn't matter.
In pictures, we're saying suppose f is a nice smooth function with the same starting and ending height:
If f increases or decreases from its starting height, it needs to turn around and come back in order to end at the same height it started at:
Since f is a nice smooth differentiable function, its derivative at that turn-around point must be 0:
If f doesn't go up or down from its starting point, then f is constant:
In this case, f ' (c) is 0 for every value of c in the interval (a, b).
Rolle's Theorem is reminiscent of the Intermediate Value Theorem. Rolle's Theorem says if f satisfies some assumptions (more mathematically known as hypotheses) then f ' will be zero at some point in (a, b). We could have a constant function, in which case f ' will be 0 infinitely many times:
We could have a function that turns around once:
Or we could have a function that turns around many times:
Rolle's Theorem doesn't tell us where or how many times f' will be zero; it tells us f ' must be zero at least once if the hypotheses are all satisfied.
Suppose f is not continuous on [a, b]. Then there doesn't need to be any c in (a, b) with f ' (c) = 0. Here's an example:
This function is not continuous. At the point of discontinuity, f ' doesn't exist. At all other points in the interval, f ' is positive:
There is no point c in (a, b) where f ' (c) = 0.
We found earlier that the derivative of the absolute value function doesn't exist at 0. When x is negative the slope of the absolute value function is -1; when x is positive the slope of the absolute value function is -1:
There is no value of c anywhere, in any interval (a, b), with f ' (c) = 0. The derivative of the absolute value function isn't 0 anywhere.
This function doesn't have a derivative of 0 anywhere between a and b.
If a function fails any of the hypotheses, we aren't allowed to use Rolle's Theorem.