Suppose I have a $C^\infty$ smooth function $f$ defined on the reals.


I can apply Taylor's formula and get the local expression


$$
f(x) = \sum\_{i=0}^l\frac{f^{(i)}(0)}{l!}x^i+ f^{(l+1)}(\xi(x))x^{l+1}.
$$


Question: Is the function $\xi $ smooth? The function $f$ can in principle be as nice as you want.



Taylor Series Remainder
Note that the point $\xi$ in the expression of the remainder is not unique in general (as it is clear already for $l=0$). According to a common phenomenon, lack of unicity may cause a lack of continuity. For an example where there is no continuous $\xi$ (again for $l=0$) think of a smooth function $f$ which is positive and concave on $I:=(0,1)$; with $f(0)=f(1)=0$, with $f'(1) < 1$, and which is flat on an interval $J:=\{f'(x)=0\}\subset I$. Crossing the point $x\_0=1$, the point $\xi(x)$ has to jump the interval $J$, causing a discontinuity at $x=1$. Note that in this example $f^{l+2}(\xi(x\_0))=0$. 


On the other hand, going back to the general situation, if you have a point $\xi\_0$ for the expression of the remainder corresponding to $x\_0\neq0$, and if $f^{(l+2)}(\xi\_0)\neq0$, then the implicit function theorem applies, giving a smooth function $\xi$ in a nbd of $x\_0$.


Finally note that any such function $\xi$ is certainly continuous at $x=0$, but it may have discontinuities in any nbd of $0$ even if $f$ is smooth (think of a proper version of the first example, with flat intervals accumulating at $0$). 


