> [!NOTE] Theorem (Rolle) > Given a [[Continuous Real Function|continuous function]] $f:[a,b]\to \mathbb{R}$ that is [[Fréchet Differentiation|differentiable]] on the open interval $(a,b)$ and $f(a)=f(b)$. Then there is a point $c\in(a,b)$ where $f'(c)= 0$ **Proof**: If $f$ is constant on the interval then by [[Real Function with Zero Derivative is Constant]], its derivative is zero everywhere. If not, it takes values different from $f(a)=f(b)$. WLOG suppose it is somewhere larger than $f(a).$ Since $f$ is continuous on the closed interval, it follows from [[Extreme Value Theorem]] that $f$ attains its maximum value at some point $c\in [a,b]$ and this cannot be $a$ or $b$: so $c\in(a,b)$. If $x>c$, then $f(x)-f(c)\leq 0$ while $x-c>0$ so $\frac{f(x)-f(c)}{x-c} \leq 0.$So $f'(c)$ is a limit of non-positive values so is not positive. On the other hand, if $x<c$ then $f(x)-f(c)\leq 0$ while $x-c<0$ so the ratio $\frac{f(x)-f(c)}{x-c}\geq 0.$So $f'(c)$ is a limit of non-negative values and so is not negative. Therefore $f'(c)=0$.