## Calculus (3rd Edition)

$$2.0125,\ \ \ \ 2.0123466,\ \ \ 1.533\times 10^{-4}$$
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$$|2.0125- 2.012346|=1.533\times 10^{-4}$$