#math/calculus **Mean Value Theorem.** If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in(a,b)$ satisfies $ f'(c)=\frac{f(b)-f(a)}{b-a}. $ Geometrically, at least one tangent line inside the interval is parallel to the secant line through $(a,f(a))$ and $(b,f(b))$. ![[tikz-11c200b731bd.svg]] %% tikz-source ```latex \begin{document} \begin{tikzpicture}[x=1.25cm,y=0.85cm,line cap=round] \draw[->,black!45] (0,0) -- (5.1,0) node[right] {$x$}; \draw[->,black!45] (0,0) -- (0,6.8) node[above] {$y$}; \draw[very thick,blue,domain=0.55:4.45,samples=100] plot (\x,{0.3*\x*\x+0.5}); \draw[red!75!black,thick] (0.7,{0.3*0.7*0.7+0.5}) -- (4.3,{0.3*4.3*4.3+0.5}); \draw[orange!80!black,thick] (1.45,{1.5*(1.45-2.5)+2.375}) -- (3.55,{1.5*(3.55-2.5)+2.375}); \foreach \x/\lab in {0.7/a,2.5/c,4.3/b}{ \draw[dashed,black!30] (\x,0) -- (\x,{0.3*\x*\x+0.5}); \node[below] at (\x,0) {$\lab$}; } \fill (0.7,{0.3*0.7*0.7+0.5}) circle (2pt); \fill (2.5,2.375) circle (2pt); \fill (4.3,{0.3*4.3*4.3+0.5}) circle (2pt); \node[red!75!black,above left] at (4.2,5.9) {secant}; \node[orange!80!black,below right] at (3.45,3.8) {tangent}; \end{tikzpicture} \end{document} ``` %% For a position function $s(t)$, the secant slope is average velocity. The theorem says that the instantaneous velocity equals that average at some time in the interval. The continuity and differentiability assumptions are essential; corners, jumps, or vertical tangents can prevent the required tangent slope.