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

A critical point occurs where either:

• $f'(x)=0$, or
• $f'(x)$ does not exist, provided $f(x)$ is defined there.

Since $(x-2)^{2/3}$ is defined for all real $x$, $f(x)$ is defined for all real $x$.

2. Differentiate the function using the product rule.

Let $u=(x-2)^{2/3}, \quad v=2x+1.$ Then $u'=\frac{2}{3}(x-2)^{-1/3}, \quad v'=2.$

So, $f'(x)=u'v+uv'$ $f'(x)=\frac{2}{3}(x-2)^{-1/3}(2x+1)+2(x-2)^{2/3}.$

Take common factor $\dfrac{2}{3}(x-2)^{-1/3}$: $f'(x)=\frac{2}{3}(x-2)^{-1/3}\left[(2x+1)+3(x-2)\right].$

Now simplify the bracket: $(2x+1)+3(x-2)=2x+1+3x-6=5x-5=5(x-1).$

Hence, $f'(x)=\frac{2}{3}(x-2)^{-1/3}\cdot 5(x-1)$ $f'(x)=\frac{10}{3}(x-1)(x-2)^{-1/3}.$

3. Find where $f'(x)=0$.

Since $f'(x)=\frac{10}{3}(x-1)(x-2)^{-1/3},$ it is zero when $x-1=0 \implies x=1.$

So one critical point is at $x=1.$

4. Find where $f'(x)$ does not exist.

The derivative contains $(x-2)^{-1/3}$, which is undefined at $x=2.$

But $f(2)$ exists: $f(2)=(2-2)^{2/3}(2\cdot 2+1)=0\cdot 5=0.$

So $x=2$ is also a critical point.

5. Count the critical points.

The critical points occur at:

• $x=1$
• $x=2$

Therefore, the number of critical points is $2.$

6. Check options.

• A: $2$ ✅
• B: $1$ ❌
• C: $0$ ❌
• D: $3$ ❌

So the correct option is A.