Rolle's Theorem

Rolle’s theorem states that if the outputs of a differentiable function $\mathit{f}$ are equal at the endpoints of an interval, then there must be an interior point $\mathit{c}$ where ${\mathit{f}}^{\mathit{\prime }}\left(\mathit{c}\right)=0$.

Given figures illustrates this theorem.

If a differentiable function $\mathit{f}$ satisfies $\mathit{f}\left(\mathit{a}\right)=\mathit{f}\left(\mathit{b}\right)$, then its derivative must be zero at some point(s) between $\mathit{a}$ and $\mathit{b}$.

Example: For the function $\mathit{f}\left(\mathit{x}\right)={\mathit{x}}^2+2\mathit{x}$ over $\left[-2,0\right]$. Verify that the function satisfies the criteria stated in Rolle’s theorem and find all values $\mathit{c}$ in the given interval where ${\mathit{f}}^{\mathit{\prime }}\left(\mathit{c}\right)=0$.

Solution: Since $\mathit{f}$ is a polynomial, it is continuous and differentiable everywhere. In addition, $\mathit{f}\left(-2\right)=0=\mathit{f}\left(0\right)$. Therefore, $\mathit{f}$ satisfies the criteria of Rolle’s theorem.

We conclude that there exists at least one value $\mathit{c}\in \left(-2,0\right)$ such that ${\mathit{f}}^{\mathit{\prime }}\left(\mathit{c}\right)=0$. Since ${\mathit{f}}^{\mathit{\prime }}\left(\mathit{x}\right)=2\mathit{x}+2=2\left(\mathit{x}+1\right)$, we see that ${\mathit{f}}^{\mathit{\prime }}\left(\mathit{c}\right)=2\left(\mathit{c}+1\right)=0$ implies $\mathit{c}=-1$ as shown in the following graph.