The function is, $f\left(x\right)={e}^{x+1}$

The Taylor polynomials of degree 4 at 0 is,

$P}_{4}\left(x\right)=f\left(0\right)+\frac{{f}^{\prime}\left(0\right)}{1!}x+\frac{f{}^{\u2033}\left(0\right)}{2!}{x}^{2}+\frac{{f}^{\left(3\right)}\left(0\right)}{3!}{x}^{3}=\frac{{f}^{\left(4\right)}\left(0\right)}{4!}{x}^{4$

Find the value of $f(x)$ at $x=0$ as,

$f\left(0\right)={e}^{0+1}$

$={e}^{1}$

$=e$

Find the first, second, third and fourth derivatives of $f(x)$ and find its corresponding values at $x=0$ as shown in below table.

$\begin{array}{|ll|}\hline \text{Derivative of function,}f(x)={e}^{x+1}& \text{Value of derivative at}x=0\\ {f}^{\prime}(x)={e}^{x+1}& {f}^{\prime}(0)={e}^{0+1}=e\\ {f}^{\u2033}(x)=\frac{d}{dx}({f}^{\prime}(x))=\frac{d}{dx}({e}^{x+1})={e}^{x+1}& {f}^{\u2033}(0)={e}^{0+1}=e\\ {f}^{(3)}(x)=\frac{d}{dx}({f}^{\u2033}(x))=\frac{d}{dx}({e}^{x+1})={e}^{x+1}& {f}^{(3)}(0)={e}^{0+1}=e\\ {f}^{(4)}(x)=\frac{d}{dx}({f}^{(3)}(x))=\frac{d}{dx}({e}^{x+1})={e}^{x+1}& {f}^{(4)}(0)={e}^{0+1}=e\\ \hline\end{array}$

The Taylor polynomials of degree 4 at 0 is calculated as follows.

$P}_{4}\left(x\right)=f\left(0\right)=\frac{{f}^{\prime}\left(0\right)}{1!}x+\frac{f{}^{\u2033}\left(0\right)}{2!}{x}^{2}+\frac{{f}^{\left(3\right)}\left(0\right)}{3!}{x}^{3}+\frac{{f}^{\left(4\right)}\left(0\right)}{4!}{x}^{4$

$=e+\frac{e}{1}x+\frac{e}{2}{x}^{2}+\frac{e}{6}{x}^{3}+\frac{e}{24}{x}^{4}$