#### Answer

$$2.0125,\ \ \ \ 2.0123466,\ \ \ 1.533\times 10^{-4}$$

#### Work Step by Step

Given $$\sqrt[5]{33}$$
Consider $f(x)=x^{1/5}, a=32,$ and $\Delta x=1$, since
\begin{align*}
f^{\prime}(x)&=\frac{1}{5}x^{-4/5} \\
f^{\prime}(32)&=\frac{1}{80}
\end{align*}
Then the linear approximation is given by
\begin{align*}
L(x)&=f^{\prime}(a)(x-a)+f(a)\\
&= \frac{1}{80}(x-32)+ 2\\
L(33)&\approx\frac{1}{80}(33-32)+ 2= 2.0125
\end{align*}
Since by using a calculator, $\sqrt[5]{33} =2.0123466$ and the error in the linear approximation is$$
|2.0125- 2.012346|=1.533\times 10^{-4}
$$