#### Answer

$$ \frac{65}{32},\ \ \ 0.0348\%$$

#### Work Step by Step

Given $$(17)^{1 / 4}$$
Consider $f(x)=x^{1 / 4}, a=16,$ and $\Delta x=1$, since
\begin{align*}
f^{\prime}(x)&=\frac{1}{4}x^{-3/4}\\
f^{\prime}(16)&=\frac{1}{32}
\end{align*}
Then the linearization to $f (x)$ is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&=\frac{1}{32}(x-16)+2\\
&=\frac{3}{2}+\frac{x}{32}
\end{align*}
Since
\begin{align*}
L(17) &=\frac{3}{2}+\frac{17}{32}\\
&\approx \frac{65}{32}
\end{align*}
Hence the error is given by
$$ | (17)^{1/4}-\frac{65}{32}|= 0.00070$$
and the percentage is
$$\frac{0.00070}{(17)^{1/4}}\times 100 \% \approx 0.0348\%$$