To find the number of critical points of the function $$f(x)=(x-2)^{2 / 3}(2 x+1)$$, we need to determine where its derivative $$f'(x)$$ is equal to zero or undefined. Critical points occur where the derivative is zero or does not exist.

First, let's find the derivative of the function:

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

We apply the product rule for differentiation, which states that $$(uv)' = u'v + uv'$$, where $$u = (x-2)^{2/3}$$ and $$v = 2x + 1$$.

We need the derivatives of $$u$$ and $$v$$:

For $$u = (x-2)^{2/3}$$, we use the chain rule:

$$u'(x) = \frac{d}{dx}[(x-2)^{2/3}] = \frac{2}{3}(x-2)^{-1/3} \cdot 1 = \frac{2}{3}(x-2)^{-1/3}$$

The derivative of $$v$$ is straightforward, as $$v = 2x + 1$$:

$$v'(x) = 2$$

Now we apply the product rule:

$$f'(x) = u'(x)v(x) + u(x)v'(x)$$

Substituting $$u$$, $$u'$$, and $$v$$, we get:

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

This simplifies to:

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

For critical points, we need to solve $$f'(x) = 0$$ or where it is undefined.

1. Solve for where the derivative is zero:

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

Combining like terms in a common denominator, we get:

$$\frac{2(2x + 1) + 6(x - 2)}{3(x-2)^{1/3}} = 0$$

Simplifying the numerator:

$$2(2x + 1) + 6(x-2) = 4x + 2 + 6x - 12 = 10x - 10 = 10(x-1)$$

So:

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

The numerator is zero when:

$$10(x-1) = 0$$

Therefore:

$$x = 1$$

2. Solve for where the derivative is undefined:

The denominator, $$3(x-2)^{1/3}$$, is undefined when $$(x-2)^{1/3} = 0$$, which happens at:

$$x = 2$$

From the above analysis, the critical points are at $$x = 1$$ and $$x = 2$$. Thus, there are 2 critical points.

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

Option A