Answer
$$0.01544516 ,\ \ 0.015625,\ \ ,\ \ \ \ 0.000179837$$
Work Step by Step
Given $$16.5^{1 / 4}-16^{1 / 4}$$
Consider $f(x)= x^{1/4}$, $a= 16$, $\Delta x= 0.5$, since
\begin{align*}
f'(x) &= \frac{1}{4}x^{-3/4}\\
f'(16)&= 0.03125
\end{align*}
Then the linear approximation is given by
\begin{align*}
\Delta &f \approx f^{\prime}(a) \Delta x\\
&= (0.5)(0.03125)\\
&= 0.015625
\end{align*}
and the actual change is given by
\begin{align*}
\Delta f&=f(a+\Delta x)-f(a)\\
&=f(16.5)-f(16) \\
&\approx 0.01544516
\end{align*}
Hence the error is
$$ | 0.015625- 0.01544516 | \approx 0.000179837 $$
