Rolle's theorem

Contents

Statement

Suppose $f$ is a function defined on a closed interval $[a,b]$ (with $a < b$) satisfying the following three conditions:

1. $f$ is a continuous function on the closed interval $[a,b]$. In particular, $f$ is (two-sided) continuous at every point in the open interval $(a,b)$, right continuous at $a$, and left continuous at $b$.
2. $f$ is differentiable on the open interval $(a,b)$, i.e., the derivative of $f$ exists at all points in the open interval $(a,b)$.
3. $f(a) = f(b) = 0$.

Then, there exists $c$ in the open interval $(a,b)$ such that $f'(c) = 0$.

Related facts

Applications

• Lagrange mean value theorem
• Bound relating number of zeros of function and number of zeros of its derivative

Facts used

1. Extreme value theorem
2. Point of local extremum implies critical point

Proof

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	If $f$ is zero on all of $[a,b]$, then $f'(c) = 0$ for any choice of $c\in (a,b)$				obvious
2	$f$ must attain both its maximum and its minimum values on $[a,b]$	Fact (1)	$f$ is continuous on $[a,b]$		Given-fact combination direct
3	If $f$ is not zero on all of $[a,b]$, either its absolute maximum value on $[a,b]$ is positive and attained at a point in the open interval $(a,b)$ or its absolute minimum value on $[a,b]$ is negative and attained at a point in the open interval $(a,b)$ (or possibly both).		$f(a) = f(b) = 0$	Step (2)	[SHOW MORE] If $f$ is not everywhere zero, it must either attain positive values or negative values somewhere in $(a,b)$. By Step (2), the function attains both its maximum and minimum values. If the function attains positive values, then the maximum value is positive, hence must be attained at some point in the interior. If the function attains negative values, then the minimum value is negative, hence must be attained at some point in the interior.
4	If $c$ is a point in $(a,b)$ at which $f$ attains its maximum value or its minimum value, then $\! f'(c) = 0$.	Fact (2)	$f$ is differentiable on $(a,b)$		[SHOW MORE] A maximum value (respectively minimum value) in the interior is also a local maximum value (respectively, local minimum value) for the function, so by Fact (2), $c$ is a critical point for $f$. Thus, either $f'(c) = 0$ or $f'(c)$ does not exist. The does not exist case cannot arise because $f$ is given to be differentiable on $(a,b)$. Thus, we are forced to have $f'(c) = 0$.
5	There is a point $c\in (a,b)$ at which $\! f'(c) = 0$.			Steps (1), (3), (4)	[SHOW MORE] Step (1) settles the case of the zero function. If $f$ is not the zero function, Step (3) says that $f$ attains either its maximum value or its minimum value at some interior point. Step (4) now tells us that the derivative at that point is zero, completing the proof.