The number of critical points of the function $f(x)=(x-2)^{2 / 3}(2 x+1)$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}(\mathrm{x})=(\mathrm{x}-2)^{2 / 3}(2 \mathrm{x}+1) \ & \mathrm{f}^{\prime}(\mathrm{x})=\frac{2}{3}(\mathrm{x}-2)^{-1 / 3}(2 \mathrm{x}+1)+(\mathrm{x}-2)^{2 / 3} \ & \mathrm{f}^{\prime}(\mathrm{x})=2 	imes \frac{(2 \mathrm{x}+1)+(\mathrm{x}-2)}{3(\mathrm{x}-2)^{1 / 3}} \ & \frac{3 \mathrm{x}-1}{(\mathrm{x}-2)^{1 / 3}}=0\end{aligned}$ Critical points $\mathrm{x}=\frac{1}{3}$ and $\mathrm{x}=2$

The number of critical points of the function $f(x)=(x-2)^{2 / 3}(2 x+1)$ is

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