## Calculus (3rd Edition)

$$L(x) =\frac{8}{3}+\frac{x}{48}$$ $$0.000019\%$$
Given $$(64.1)^{1 /3}$$ Consider $f(x)=x^{1 / 3}, a=64,$ and $\Delta x=0.1$, since \begin{align*} f^{\prime}(x)&=\frac{1}{3}x^{-2/3}\\ f^{\prime}(64)&=\frac{1}{48} \end{align*} Then the linearization to $f (x)$ is given by \begin{align*} L(x)&=f^{\prime}(a)(x-a)+f(a)\\ &=\frac{1}{48}(x-64)+4\\ &=\frac{8}{3}+\frac{x}{48} \end{align*} Since \begin{align*} L(64.1) &=\frac{8}{3}+\frac{64}{48}\\ &\approx 4.00208 \end{align*} Hence the error is given by $$| (64.1)^{1 /3}-4.00208|=2.2492\times 10^{-6}$$ and the percentage is $$\frac{2.2492\times 10^{-6} }{(64.1)^{1 /3}}\times 100 \% \approx 0.000019\%$$