Answer
$$ L(x) \approx \frac{-1}{16}\left(x-3\right)+\frac{1}{2} $$
Work Step by Step
Given
$$y=(1+x)^{-1 / 2}, \quad a=3$$
Since $y(3)=\frac{1}{2}$ and
\begin{align*}
y'(x)&=\frac{-1}{2}(1+x)^{-3/ 2}\\
y'(3)&=\frac{-1}{16}
\end{align*}
Then the linearization at $a=3 $ is given by
\begin{align*}
L(x)&\approx y^{\prime}(a)(x-a)+y(a)\\
&=y^{\prime}(3)(x-3)+y(3)\\
&=\frac{-1}{16}\left(x-3\right)+\frac{1}{2}
\end{align*}
