Answer
$$L(x) =\frac{8}{3}+\frac{x}{48}$$
$$ 0.000019\%$$
Work Step by Step
Given $$(64.1)^{1 /3}$$
Consider $f(x)=x^{1 / 3}, a=64,$ and $\Delta x=0.1$, since
\begin{align*}
f^{\prime}(x)&=\frac{1}{3}x^{-2/3}\\
f^{\prime}(64)&=\frac{1}{48}
\end{align*}
Then the linearization to $f (x)$ is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&=\frac{1}{48}(x-64)+4\\
&=\frac{8}{3}+\frac{x}{48}
\end{align*}
Since
\begin{align*}
L(64.1) &=\frac{8}{3}+\frac{64}{48}\\
&\approx 4.00208
\end{align*}
Hence the error is given by
$$ | (64.1)^{1 /3}-4.00208|=2.2492\times 10^{-6}$$
and the percentage is
$$\frac{2.2492\times 10^{-6} }{(64.1)^{1 /3}}\times 100 \% \approx 0.000019\%$$
