Question Let $f\left(x\right)=\begin{cases} 0 &\text{if } {-}1\leq x\leq0\\ x^4 &\text{if } 0<x\leq1 \end{cases}$. If $f(x) = \sum_{k=0}^{n}$$\frac{f^{(k)}\left(0\right)}{k!}x^{k}+\frac{f^{(n+1)}\left(\xi\right)}{\left(n+1\right)!}x^{n+1}$ is the Taylor's formula for $f$ about $x=0$ with maximum possible value of $n$, then the value of $\xi$ for $0 <x\leq1$ is?
I know 
I don't know how to apply the theorem on this question
Any and all help will be appreciated
First, let us find the value of $n$. Your function $f$ is only 3 times differentiable at $x = 0$, because \begin{align} f'(x) &= \begin{cases} 0 &\text{if } {-}1 \leq x \leq 0 \\ 4x^3 &\text{if } 0 < x < 1 \end{cases} & f''(x) &= \begin{cases} 0 &\text{if } {-}1 \leq x \leq 0 \\ 12x^2 &\text{if } 0 < x < 1 \end{cases} & f'''(x) &= \begin{cases} 0 &\text{if } {-}1 \leq x \leq 0 \\ 24x &\text{if } 0 < x < 1 \end{cases} \end{align} and clearly $f'''$ is not differentiable at $x = 0$. Therefore, $n = 2$ (it is the maximum value for which the Taylor formula in the OP is well-defined). Note that $f(0) = f'(0) = f''(0) = 0$.
So, the Taylor formula for $f$ about $x = 0$ (more precisely, for all $0 < x \leq 1$) is: \begin{align} f(x) = f(0) + \frac{f'(0)}{1}x + \frac{f''(0)}{2}x^2 + \frac{f'''(\xi)}{6}x^3 = 4 \,\xi\, x^3 \end{align} where $0 < \xi < 1$. Since $f(x) = x^4$ for all $ 0 < x \leq 1$, then $$\xi = \frac{x}{4} \ .$$ Note that the condition $0 < \xi < 1$ is fulfilled by the above solution.