Taylor's expansion and remainder of $f(x)=0, -1\le x\le0$ and $f(x)=x^4, 0<x\le 1$

Let $f(x)=0, -1\le x\le0$ and $f(x)=x^4, 0<x\le 1$

If $$f(x)=\sum_{k=0}^n\frac{f^{(k)}(0)x^k}{k!}+\frac{f^{(n+1)}(\xi)x^k}{(n+1)!}$$ is the Taylor's formula for $f$ about $x=0$ with maximum possible value of $n$, then find the value of $\xi$ for $0<x\le 1$.

I know how to calculate remainder and I can do that. My question is what is the maximum possible value of $n$ and how to find it.

• The function is $3$ times differentiable at $x=0$, and no more. Feb 6 '16 at 3:42
• Can you please provide some explanation
– Mix
Feb 6 '16 at 3:43
• When you differentiate three times, you get the function which is $0$ to the left of $0$ and $24x$ to the right of $0$. (It is not hard to check by using the definition of the derivative that the first three derivatives are $0$ at $0$.) But the function which is $0$ to the left of $0$ and $24x$ for $x\ge 0$ is not differentiable at $0$. (It has a sharp kink at $0$.) Feb 6 '16 at 3:49