Let $\mathrm{a}=64, \mathrm{~h}=2$ and let $\mathrm{f}(\mathrm{x})=\mathrm{x}^{\frac{1}{3}} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{3} \mathrm{x}^{-\frac{2}{3}}=\frac{1}{3 \mathrm{x}^{\frac{2}{3}}}$
$\therefore \mathrm{f}(\mathrm{a})=64^{\frac{1}{3}}=4 \quad$ and $\quad \mathrm{f}^{\prime}(\mathrm{a})=\frac{1}{3(64)^{\frac{2}{3}}}=\frac{1}{3(16)}=\frac{1}{48}$
We have $\mathrm{f}(\mathrm{a}+\mathrm{h}) \neq \mathrm{f}(\mathrm{a})+\mathrm{h} \cdot \mathrm{f}^{\prime}(\mathrm{a})$
$\therefore(66)^{\frac{1}{3}} = 4+(2)\left(\frac{1}{48}\right) = 4+\frac{1}{24} = 4.0416$