Suppose $f : \mathbb R^2 \to \mathbb R$ is a differentiable function at a point $\mathbf a =(a,b) \in \mathbb R^2$. Then for $\mathbf x = (x,y) \in \mathbb R^2$ close to $\mathbf a$, Taylor's theorem tells us $$ f(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) + \frac 1 2 \left(\mathbf x - \mathbf a\right)^t H(a,b)(\mathbf x - \mathbf a) + R_2(x,y), $$ where $R_2(x,y) = o(|x,y|^2)$, and $H(a,b)$ is the Hessian matrix for $f$.
I recently came across a statement claiming $$ f(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) + \frac 1 2 \left(\mathbf x - \mathbf a\right)^t H(\alpha, \beta)(\mathbf x - \mathbf a) $$ for $x > a$ and $y > b$, and for some $\alpha \in [a,x]$, $\beta \in [b,y]$. In particular, they chose $\alpha, \beta$ to eliminate the third-order terms. While I am confident that there are some $\alpha, \beta$ that may achieve this, what allows us to choose $a \leq \alpha \leq x$ and $b \leq \beta \leq y$?
It strikes me as being similar to the intermediate value theorem applied to the function $$ (s,t) \mapsto \frac 1 2 \left(\mathbf x - \mathbf a\right)^t H(s,t)(\mathbf x - \mathbf a),$$ and it would certainly follow if we could show either $$ \frac 1 2 \left(\mathbf x - \mathbf a\right)^t H(a,b)(\mathbf x - \mathbf a) \leq \frac 1 2 \left(\mathbf x - \mathbf a\right)^t H(a,b)(\mathbf x - \mathbf a) + R_3(x,y) \leq \frac 1 2 \left(\mathbf x - \mathbf a\right)^t H(x,y)(\mathbf x - \mathbf a), $$ or the analogous statement with the inequalities reversed, but I couldn't find anything on the statement of Taylor's theorem that guarantees this. Does Taylor's theorem imply the above inequality, or the reverse of the above inequality?