1. The function
$$
f(x)=|x+1|+|x+3|+\mathrm{e}^{-x}(x \in \mathbf{R})
$$

has a minimum value of $\qquad$ .
$-1.6-2 \ln 2$
When $x \leqslant-3$, $f(x)=-2 x-4+\mathrm{e}^{-x}$, $f(x)$ is monotonically decreasing.

When $-3 \leqslant x \leqslant-1$, $f(x)=2+\mathrm{e}^{-x}$, at this time, $f(x)$ is also monotonically decreasing.
When $x \geqslant-1$,
$$
\begin{array}{l}
f(x)=2 x+4+\mathrm{e}^{-x}, f^{\prime}(x)=2-\mathrm{e}^{-x} . \\
\text { Let } f^{\prime}(x)=0 \Rightarrow x=-\ln 2 .
\end{array}
$$

Therefore, $f(x)$ reaches its minimum value at $x=-\ln 2$
$$
6-2 \ln 2
$$