#math/calculus **Rolle's Theorem.** If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then some $c\in(a,b)$ satisfies $f'(c)=0$. ![[tikz-9d2df7c48938.svg]] %% tikz-source ```latex \begin{document} \begin{tikzpicture}[x=1.25cm,y=1cm,line cap=round] \draw[->,black!45] (-0.2,0) -- (4.8,0) node[right] {$x$}; \draw[->,black!45] (0,-0.2) -- (0,2.8) node[above] {$y$}; \draw[very thick,blue,domain=0.4:4.4,samples=100] plot (\x,{0.6+1.5*sin(45*(\x-0.4))}); \draw[orange!80!black,thick] (1.45,2.1) -- (3.35,2.1); \fill (0.4,0.6) circle (2pt); \fill (2.4,2.1) circle (2pt); \fill (4.4,0.6) circle (2pt); \node[below] at (0.4,0) {$a$}; \node[below] at (2.4,0) {$c$}; \node[below] at (4.4,0) {$b$}; \draw[dashed,black!30] (0.4,0) -- (0.4,0.6); \draw[dashed,black!30] (2.4,0) -- (2.4,2.1); \draw[dashed,black!30] (4.4,0) -- (4.4,0.6); \end{tikzpicture} \end{document} ``` %% **Proof.** By the [[3B3 - Extreme Value Theorem]], $f$ attains a maximum and a minimum on $[a,b]$. If they are equal, $f$ is constant and every interior point works. Otherwise, because the endpoint values agree, at least one nontrivial extremum occurs at an interior point $c$. Differentiability then gives $f'(c)=0$. $\blacksquare$ **Example.** Let $f(x)=x^{4}-9x^{3}+26x^{2}-24x$. Since $f(0)=f(2)=0$, Rolle's theorem guarantees some $c\in(0,2)$ for which $f'(c)=4c^{3}-27c^{2}+52c-24=0$. This proves the cubic has a root in $(0,2)$ without locating it.