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What is /are the critical points $(\mathrm{s})$ of the function $\mathrm{f}(\mathrm{x})=$ $\mathrm{x}^{2 / 3}(5-2 \mathrm{x})$ on the interval $[-1,2] ?$
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$\mathrm{f}(\mathrm{x})=\mathrm{x}^{2 / 3}(5-2 \mathrm{x})$
or, $f(x)=5 x^{2 / 3}-2 x^{5 / 3}$
differentiating both the sides.
$\mathrm{f}^{\prime}(\mathrm{x})=5 \times \frac{2}{3} \mathrm{x}^{-1 / 3}-2 \times \frac{5}{3} \mathrm{x}^{2 / 3}$
or, $\mathrm{f}^{\prime}(\mathrm{x})=\frac{10}{3}\left(\mathrm{x}^{-1 / 3}-\mathrm{x}^{2 / 3}\right)$
To get the critical value. $\mathrm{f}^{\prime}(\mathrm{x})=0$
so, $\mathrm{x}^{-1 / 3}-\mathrm{x}^{2 / 3}=0 \Rightarrow \mathrm{x}^{-1 / 3}(1-\mathrm{x})=0$
$\Rightarrow 1-x=0$ as $x^{-1 / 3} \neq 0$
or, $\mathrm{x}=1$ is the only value in the interval $[-1,2]$
or, $f(x)=5 x^{2 / 3}-2 x^{5 / 3}$
differentiating both the sides.
$\mathrm{f}^{\prime}(\mathrm{x})=5 \times \frac{2}{3} \mathrm{x}^{-1 / 3}-2 \times \frac{5}{3} \mathrm{x}^{2 / 3}$
or, $\mathrm{f}^{\prime}(\mathrm{x})=\frac{10}{3}\left(\mathrm{x}^{-1 / 3}-\mathrm{x}^{2 / 3}\right)$
To get the critical value. $\mathrm{f}^{\prime}(\mathrm{x})=0$
so, $\mathrm{x}^{-1 / 3}-\mathrm{x}^{2 / 3}=0 \Rightarrow \mathrm{x}^{-1 / 3}(1-\mathrm{x})=0$
$\Rightarrow 1-x=0$ as $x^{-1 / 3} \neq 0$
or, $\mathrm{x}=1$ is the only value in the interval $[-1,2]$
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