# Calculus/Rolle's Theorem

 ← Extreme Value Theorem Calculus Mean Value Theorem for Functions → Rolle's Theorem
Rolle's Theorem

If a function, ${\displaystyle f(x)\ }$, is continuous on the closed interval ${\displaystyle [a,b]\ }$, is differentiable on the open interval ${\displaystyle (a,b)\ }$, and ${\displaystyle f(a)=f(b)\ }$, then there exists at least one number c, in the interval ${\displaystyle (a,b)\ }$ such that ${\displaystyle f'(c)=0\ .}$

Rolle's Theorem is important in proving the Mean Value Theorem.

## Examples

Example:

${\displaystyle f(x)=x^{2}-3x}$. Show that Rolle's Theorem holds true somewhere within this function. To do so, evaluate the x-intercepts and use those points as your interval.

Solution:

1: The question wishes for us to use the x-intercepts as the endpoints of our interval.

Factor the expression to obtain ${\displaystyle x(x-3)=0}$. x = 0 and x = 3 are our two endpoints. We know that f(0) and f(3) are the same, thus that satisfies the first part of Rolle's theorem (f(a) = f(b)).

2: Now by Rolle's Theorem, we know that somewhere between these points, the slope will be zero. Where? Easy: Take the derivative.

${\displaystyle dy \over dx}$ ${\displaystyle =2x-3}$

Thus, at ${\displaystyle x=3/2}$, we have a spot with a slope of zero. We know that ${\displaystyle 3/2}$ (or 1.5) is between 0 and 3. Thus, Rolle's Theorem is true for this (as it is for all cases). This was merely a demonstration.