Answer
$$L(x)=\frac{5}{3}(x-1)+1 $$
$$1.608\%$$
Work Step by Step
Given $$(1.2)^{5 / 3}$$
Consider $f(x)=x^{5 / 3}, a=1,$ and $\Delta x=0.2$, since
\begin{align*}
f^{\prime}(x)&=\frac{5}{3}x^{2/3}\\
f^{\prime}(1)&=\frac{5}{3}
\end{align*}
Then the linearization to $f (x)$ is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&=\frac{5}{3}(x-1)+1
\end{align*}
Since
\begin{align*}
L(1.2) &=\frac{5}{3}(1.2-1)+1 \\
&\approx 1.3333
\end{align*}
Hence the error given
$$ | (1.2)^{5 /3}-1.3333|=0.02179$$
and the percentage is
$$\frac{0.02179 }{(1.2)^{5 /3}}\times 100 \% \approx 1.608\%$$
