Answer
$$L(x)\approx -\frac{1}{128} x+\frac{3}{8}
$$
$$ 0.14 4\%~error$$
Work Step by Step
Given $$\frac{1}{\sqrt 17} $$
Consider $f(x)=x^{-1 / 2}, a=16,$ and $\Delta x=1$, since
\begin{align*}
f^{\prime}(x)&=-\frac{1}{2} x^{-3 / 2}\\
f^{\prime}(16)&=-\frac{1}{128}
\end{align*}
Then the linearization to $f (x)$ is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&=-\frac{1}{128}(x-16)+\frac{1}{4}\\
&=-\frac{1}{128} x+\frac{3}{8}
\end{align*}
Since
\begin{align*}
L(17)&= f(17)\\
&=\frac{1}{\sqrt{17}}\\
&\approx -\frac{1}{128} (17)+\frac{3}{8}\\
&\approx 0.24219
\end{align*}
Hence the error is given by
$$ | \frac{1}{\sqrt{17}}-0.24219|=4.35\times 10^{-4}$$
and the error percentage is
$$\frac{4.35\times 10^{-4}}{1/\sqrt{17}}\times 100 \% \approx 0.14 4\%$$
