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

1 and 2

We have,

$f(x) =(x-2)^{2 / 3}(2 x+1)$

$\Rightarrow f'(x) =\frac{2}{3}(x-2)^{-1 / 3}(2 x+1)+2(x-2)^{2 / 3}$

Clearly, f'(x) is not defined at x = 2.

So, x = 2 is a critical point. Another critical point is given by

$f'(x)=0$

$\Rightarrow (2 / 3)(x-2)^{-1 / 3}(2 x+1)+2(x-2)^{2 / 3}=0$

$\Rightarrow 10(x-1)(x-2)^{\frac{-1}{3}}=0 \Rightarrow x=1$.

Hence, 1 and 2 are two critical points of f(x).