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Q.
The critical points of the function $f(x)=(x-2)^{2/3}(2x+1)$ are
Application of Derivatives
Solution:
$f\left(x\right)=\left(x-2\right)^{2/3}\left(2x+1\right)$
$f'\left(x\right)=\frac{2}{3}\left(x-2\right)^{-1/3}\left(2x+1\right)+\left(x-2\right)^{2/3}\cdot 2$
Clearly $f'\left(x\right)$ is not defined at $x = 2$
$\therefore x=2$ is a critical point.
Another critical point is given by $f'\left(x\right)=0$
i.e., $\frac{2}{3} \frac{2x+1}{\left(x-2\right)^{1/3}}+2\left(x-2\right)^{2/3}=0$
$\Rightarrow \frac{2}{3}\left(2x+1\right)+2\left(x-2\right)=0$
$\Rightarrow 4x + 2 + 6x - 12 = 0$
$\Rightarrow 10x-10=0$
$\Rightarrow x=0$
Hence critical points are $1$ and $2$.