Answer
$$x+1,\ \ 0.9$$
Work Step by Step
Given $$ f(x)= e^x,\ \ \ \ a=0$$
Since $f(0)=1$ and
\begin{align*}
f'(x)&= e^x\\
f'(0)&=1
\end{align*}
Then the linearization is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&=(x-0)+1\\
&=x+1
\end{align*}
Hence
\begin{align*}
e^{-0.1}&= -0.1+1\\
&= 0.9
\end{align*}
