Stuck on Taylor expansion question I was given this question by a friend:

Let $f(x)$ be 2 times differentiable in $R$ and $x_{0}$ a local extremum.
Show that there are $a, b, c \in R$ such that a function $g(x)=f(x+a)+b$ exists. And $g(x)=cx^2+h(x)$ where $\lim_{x\to 0}\frac{h(x)}{x^2}=0$.

I began by writing a first degree Taylor expansion of $f(x)$ around $x_{0}=a$ using Lagrange's remainder:
$$f(x)=f(a)+\frac{f''(\xi)}{2}(x-a)^2$$
From the above we see that $f(x+a)=f(a)+\frac{f''(\xi)}{2}x^2$. Now comes the part I'm not sure of: Can I say that $g(x)=\frac{f''(\xi)}{2}x^2 \implies g(x)=f(x+a)-f(a)$? Is $g(x)$ defined properly? The reason I'm not sure is because the point $\xi$ is different for every $x$.
 A: The key here is to look at the condition
$$g(x) = cx^2 + h(x)\qquad\mbox{with $\lim\limits_{x\to 0}\frac{h(x)}{x^2} = 0$.}$$
If you think about a potential Taylor expansion of $g(x)$ of degree $2$, this tells you that $h(x)$ has to consists of the degree $3$ and above terms only; that means that the Taylor expansion of $g(x)$ around $0$ cannot have a linear or constant terms, so $g(0)=0$ and $g'(0)=0$.
Now, since we will want $g(x) = f(x+a)+b$, then $g'(x) = f'(x+a)$, so in order to get $g'(0)=0$, we need $f'(a)=0$. The only point where we know $f'(a)=0$ is at $a=x_0$, so we'll pick $a=x_0$. Thus, $g(x) = f(x+x_0) + b$. In order to also get $g(0) = 0$, we need $f(x_0) + b = 0$, so $b=-f(x_0)$. Thus, the function we expect will work is
$$g(x) = f(x+x_0) - f(x_0).$$
Now, to see this works, we proceed as you did: taking the Taylor expansion for $f$ around $x_0$ up to degree $2$, we get
$$f(x) = f(x_0) + \frac{f''(x_0)}{2}(x-x_0)^2 + R_3(x).$$
So 
$$g(x) = f(x+x_0) - f(x_0) = \frac{f''(x_0)}{2}x^2 + R_3(x+x_0).$$
Thus, $c=\frac{f''(x_0)}{2}$, and $h(x) = R_3(x+x_0)$. We just need to show that $h(x)/x^2 \to 0$ as $x\to 0$, but this is equivalent to $R_3(x+x_0)/x^2\to 0$ as $x\to 0$, which in turn is equivalent to 
$$\lim_{x\to x_0}\frac{R_3(x)}{(x-x_0)^2} = 0.$$
This hold precisely because $R_3$ is the degree $3$ remainder at $x_0$. 
So, in summary, we set $a=x_0$, $b=-f(x_0)$, $c=\frac{f''(x_0)}{2}$, $g(x) = f(x+x_0) - f(x_0)$, and $h(x) = R_3(f,x_0,x)$ (the Remainder of degree $3$ in the Taylor expansion of $f$ around $x_0$, evaluated at $x$). 
