Sorry for the vague title, unsure of how to phrase a more specific one in relation to the question's context.
Anyways, here is my question:
- Suppose $f'$ exists and is continuous on $[a,b]$ and $f''$ exists on $(a,b)$, where $a < b$. Show that there exists a number $c \in (a,b)$ such that $$f(b) = f(a) + (b-a) f'(a) + f''(c) \frac{(b-a)^2}{2}$$
I know you have to use MVT, basically since $f$ is continuous and differentiable on $(a,b)$, there exists a point $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{(b-a)}$. Let $g(x) = f(x) - [f'(c)(x-a) + f(a)]$. Since $g(a) = g(b) = 0$, there exists a $d \in\ (a,b)$ such that $g'(d) = 0 = f'(x) - f'(c)$.
However, I'm unsure of how to proceed after this part. May I know if there are any areas in my thought-process that are wrong? Or did I miss anything out? Thank you!
Let $$ g(x)=f(x) - [f'(a)(x-a) + f(a)], h(x)=(x-a)^2. $$ Then $$ g'(x)=f'(x)-f'(a), h'(x)=2(x-a), g''(x)=f''(x),h''(x)=2. $$ By Cauchy's MVT, you have that, there is $c\in(a,b)$ such that $$ \frac{g(b)-g(a)}{h(b)-h(a)}=\frac{g'(c)}{h'(c)}. \tag{1}$$ Again by Cauchy's MVT, you have that, there is $c_1\in(a,c)$ such that $$ \frac{g'(c)}{h'(c)}=\frac{g'(c)-g'(a)}{h'(c)-h'(a)}=\frac{g''(c_1)}{h''(c_1)}=\frac{f''(c_1)}{2}. \tag{2} $$ Combining (1) and (2) gives $$ \frac{g(b)-g(a)}{h(b)-h(a)}=\frac{f''(c_1)}{2} $$ or $$ f(b)=f(a)+f'(a)(b-a)+\frac{f''(c_1)}{2}(b-a)^2. $$