How does one prove the Taylor's Theorem by the Cauchy's Mean Value Theorem? I am trying to prove there is a $c \in (a,b)$ such that
$$
f(b)=f(a)+f'(a)(b-a)+ \frac{1}{2} f''(c)(b-a)^2
$$
which I do believe is a specialized form of the Taylor's Theorem when you plug in $b$ and $a$ into the appropriate spots.  I think you can approach this by proving the Lagrangian remainder is the same, but it doesn't seem all that intuitive.  

Define the function
$$
g(x)=f(x)-\left[f(a)+(x-a)f'(a)\right]-\frac{f(b)-\left[f(a)+(b-a)f'(a)\right]}{(b-a)^2}(x-a)^2
$$
Since $g(a)=0$ and $g(b)=0$, the Mean Value Theorem says there is a $c_1\in(a,b)$ so that $g'(c_1)=0$.
$$
g'(x)=f'(x)-f'(a)-\frac{f(b)-\left[f(a)+(b-a)f'(a)\right]}{(b-a)^2}(x-a)
$$
Since $g'(a)=0$ and $g'(c_1)=0$, the Mean Value Theorem says there is a $c_2\in(a,c_1)$ so that $g''(c_2)=0$. That is,
$$
0=f''(c_2)-2\frac{f(b)-\left[f(a)+(b-a)f'(a)\right]}{(b-a)^2}
$$
this, becomes
$$
f(b)=f(a)+(b-a)f'(a)+\frac12(b-a)^2f''(c_2)
$$
