Answer
$$T_3(x) =1+\frac{1}{2} x^{2}-\frac{1}{6} x^{3}$$
Work Step by Step
Given $$f(x) =x+e^{-x},\ \ a=0$$
Since
\begin{align*}
f(x) &=x+e^{-x}\ \ \ \ \ \ \ \ \ \ f(0) =1\\
f'(x) &=1-e^{-x}\ \ \ \ \ \ \ \ \ f'(0 )=0\\
f''(x) &=e^{-x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f''(0 )=1\\
f'''(x) &=-e^{-x}\ \ \ \ \ \ \ \ \ \ \ \ f''(0 )=-1
\end{align*}
Then
\begin{align*}
T_3(x)&= \sum_{n=0}^{3}\frac{f^{(n)}(0)}{n!}(x )^n\\
& =\frac{1}{0 !}+\frac{0}{1 !} x+\frac{1}{2 !} x^{2}-\frac{1}{3 !} x^{3}\\
&=1+\frac{1}{2} x^{2}-\frac{1}{6} x^{3}
\end{align*}
