## Calculus (3rd Edition)

$$0.01544516 ,\ \ 0.015625,\ \ ,\ \ \ \ 0.000179837$$
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 $$| 0.015625- 0.01544516 | \approx 0.000179837$$