(Solved): f(x)=xe2x You can use the following hint in this question: dxndnf(x)=(2xn)(1)n2n1e ...

$f(x)=x{\mathrm{e}}^{-2x}$ You can use the following hint in this question: $\frac{{\mathrm{d}}^n}{\mathrm{d}{x}^n}f(x)=(2x-n)(-1{)}^n{2}^{n-1}{\mathrm{e}}^{-2x}(n\ge 0,n\in \mathbb{Z})$ (a) For $x_0=1$, find the third order Taylor Series (b) According to the results you found in the previous option, find the upper limit of the error that will occur in the range of [0,2]. (c) If we know that the value of this function at a point $x_x$ is $f\left(x_x\right)=0.1210$, use Newton's iteration to find the point $xk$ such that the error is less than 0.0001 . Take $x_0=0.8$ as the initial value.