Mean Value Theorem Proof

Proof:

1. Assume $f$ is continuous on $[a,b]$
2. Assume $f$ is differentiable on $(a,b)$
3. Let $g(x)=f(x)-\text{secant line}$
4. So $g(x)=f(x)-[\frac{f(b)-f(a)}{b-a}(x-a)+f(a)]$
5. Note $g$ is continuous on $[a,b]$ since $g$ is a difference of polynomials
6. Note $g$ is differentiable on $(a,b)$ as g is difference of polynomials, you can use derivative rules
7. Note $g(a)=f(a)-[\frac{f(b)-f(a)}{b-a}(a-a)+f(a)]=0$
8. Note $g(b)=f(b)-[\frac{f(b)-f(a)}{b-a}(b-a)+f(a)]=0$
9. By Rolle’s Theorem, there exists a $c\in (a,b)$ such that $g^{\prime }(c)=0$
10. So, we have shown there is a $c\in (a,b)$ where $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}-g^{\prime }(c)=0$ this means $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$ $\square$