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

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

We start by differentiating using the product rule, which gives $$f'(x) = \frac{2}{3}(x-2)^{-1/3}(2x+1) + 2(x-2)^{2/3} = \frac{2(2x+1)}{3(x-2)^{1/3}} + 2(x-2)^{2/3}$$.

Combining these terms over the common denominator $$(x-2)^{1/3}$$ yields $$f'(x) = \frac{2(2x+1) + 6(x-2)}{3(x-2)^{1/3}} = \frac{4x + 2 + 6x - 12}{3(x-2)^{1/3}} = \frac{10x - 10}{3(x-2)^{1/3}} = \frac{10(x-1)}{3(x-2)^{1/3}}$$.

Critical points occur where $$f'(x)=0$$ or where $$f'(x)$$ is undefined. The equation $$10(x-1)=0$$ gives $$x=1$$, and the condition $$(x-2)^{1/3}=0$$ gives $$x=2$$. Since $$f(2)=0$$, $$x=2$$ is also a critical point.

Therefore, the function has exactly two critical points at $$x=1$$ and $$x=2$$. Hence, the number of critical points is 2.