Proof of Mean Value Theorem with Rolle's Theorem

Theorem: if $f$ is continuous on $[a,b]$ and $f$ is differentiable on $(a,b)$, then $\frac{f(b)-f(a)}{b-a}=f^\prime(c)$ for some $c$ in $(a,b)$.

Rolle's Theorem: if $g$ is continuous on $[a,b]$ , $g$ is differentiable on $(a,b)$ and $g(b)=g(a)$ , then there is $c$ in $(a,b)$ such that $g^\prime(c)=0$ .

Proof:

$\text{Let }g(x)=f(x)-\left(\frac{f(b)-f(a)}{b-a}\right)x$

$\begin{aligned} g(a)&=f(a)-\left(\frac{f(b)-f(a)}{b-a}\right)a\\ &=f(a)-\left(\frac{a}{b-a}\right)f(b)+\left(\frac{a}{b-a}\right)f(a)\\ &=\left(1+\frac{a}{b-a}\right)f(a)-\left(\frac{a}{b-a}\right)f(b)\\ &=\left(\frac{b}{b-a}\right)f(a)-\left(\frac{a}{b-a}\right)f(b) \end{aligned}$

$\begin{aligned} g(b)&=f(b)-\left(\frac{f(b)-f(a)}{b-a}\right)b\\ &=f(b)-\left(\frac{b}{b-a}\right)f(b)+\left(\frac{b}{b-a}\right)f(a)\\ &=\left(1-\frac{b}{b-a}\right)f(b)+\left(\frac{b}{b-a}\right)f(a)\\ &=-\left(\frac{a}{a-b}\right)f(b)+\left(\frac{b}{b-a}\right)f(a) \end{aligned}$

$g(b)=g(a)\Rightarrow g^\prime(c)=0$

$\begin{aligned} g(x)&=f(x)-\left(\frac{f(b)-f(a)}{b-a}\right)x\\ g^\prime(x)&=f^\prime(x)-\frac{f(b)-f(a)}{b-a}\\ g^\prime(c)&=f^\prime(c)-\frac{f(b)-f(a)}{b-a}\\ 0&=f^\prime(c)-\frac{f(b)-f(a)}{b-a}\\ \frac{f(b)-f(a)}{b-a}&=f^\prime(c)\quad\blacksquare \end{aligned}$

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Proof of Mean Value Theorem with Rolle's Theorem

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