#math/calculus #math/flashcards
**Horserace Theorem.** Suppose $f_{1}$ and $f_{2}$ are continuous on $[a,b]$, differentiable on $(a,b)$, and start and finish together:
$f_{1}(a)=f_{2}(a)$ and $f_{1}(b)=f_{2}(b)$. Then some $c\in(a,b)$ satisfies $f_{1}'(c)=f_{2}'(c)$.
Apply [[3C3 - Rolle's Theorem]] to $g=f_{1}-f_{2}$. Since $g(a)=g(b)=0$, some $c$ has $g'(c)=0$, which is exactly $f_{1}'(c)=f_{2}'(c)$.
![[tikz-76df740358d4.svg]]
%% tikz-source
```latex
\begin{document}
\begin{tikzpicture}[x=1.15cm,y=0.9cm,line cap=round]
\begin{scope}
\draw[->,black!45] (-0.2,0) -- (4.7,0) node[right] {$x$};
\draw[->,black!45] (0,-0.2) -- (0,3.4) node[above] {$y$};
\draw[very thick,blue,domain=0.35:4.35,samples=100] plot (\x,{0.4*\x+1.3*sin(45*(\x-0.35))});
\draw[very thick,red!70!black] (0.35,0.14) -- (4.35,1.74);
\draw[orange!80!black,thick] (1.45,1.88) -- (3.25,2.60);
\fill (0.35,0.14) circle (2pt);
\fill (2.35,2.24) circle (2pt);
\fill (4.35,1.74) circle (2pt);
\node[blue,above] at (3.7,2.45) {$f_{1}$};
\node[red!70!black,below] at (3.7,1.48) {$f_{2}$};
\node[below] at (0.35,0) {$a$};
\node[below] at (2.35,0) {$c$};
\node[below] at (4.35,0) {$b$};
\end{scope}
\begin{scope}[xshift=6.2cm]
\draw[->,black!45] (-0.2,0) -- (4.7,0) node[right] {$x$};
\draw[->,black!45] (0,-0.2) -- (0,2.1) node[above] {$g$};
\draw[very thick,teal!80!black,domain=0.35:4.35,samples=100] plot (\x,{1.3*sin(45*(\x-0.35))});
\draw[orange!80!black,thick] (1.45,1.3) -- (3.25,1.3);
\node[teal!80!black,above] at (3.5,0.8) {$g=f_{1}-f_{2}$};
\node[below] at (0.35,0) {$a$};
\node[below] at (2.35,0) {$c$};
\node[below] at (4.35,0) {$b$};
\end{scope}
\end{tikzpicture}
\end{document}
```
%%
**Mean Value Theorem from the horserace theorem.** Compare $f$ with its secant line
$
\ell(x)=f(a)+(x-a)\frac{f(b)-f(a)}{b-a}.
$
The functions agree at both endpoints, so $f'(c)=\ell'(c)=[f(b)-f(a)]/(b-a)$ for some $c\in(a,b)$.
**Example.** If $f(0)=0$ and $f(1)=1$, compare $f$ with $q(x)=x^{2}$. They agree at both endpoints, so some $c\in(0,1)$ satisfies $f'(c)=q'(c)=2c$.