Key Concepts and Formulas

• Critical Points: A critical point of a function $f(x)$ is a point $x=c$ in the domain of $f(x)$ where either $f'(c) = 0$ or $f'(c)$ is undefined.
• Product Rule: If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$.
• Chain Rule: $\frac{d}{dx}[g(h(x))] = g'(h(x))h'(x)$.

Step 1: Determine the Domain of the Function $f(x)$

The given function is $f(x)=(x-2)^{2 / 3}(2 x+1)$. We need to find all possible values of $x$ for which $f(x)$ is defined.

• $(x-2)^{2/3}$ can be written as $\sqrt[3]{(x-2)^2}$. Since the cube root is defined for all real numbers, this term is defined for all $x \in \mathbb{R}$.
• $2x+1$ is a linear function, which is defined for all $x \in \mathbb{R}$.

Since both terms are defined for all real numbers, the domain of $f(x)$ is $x \in \mathbb{R}$. This is important because a critical point must be in the domain of the original function.

Step 2: Calculate the First Derivative $f'(x)$ using the Product Rule

We will use the product rule: If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. Let $u(x) = (x-2)^{2/3}$ and $v(x) = (2x+1)$.

First, find $u'(x)$ using the chain rule. Let $g(y) = y^{2/3}$ and $h(x) = x-2$. Then $g'(y) = \frac{2}{3}y^{-1/3}$ and $h'(x) = 1$. So, $u'(x) = \frac{2}{3}(x-2)^{-1/3} \cdot 1 = \frac{2}{3(x-2)^{1/3}}$.

Next, find $v'(x)$: $v'(x) = \frac{d}{dx}(2x+1) = 2$.

Now, substitute into the product rule: $f'(x) = \left( \frac{2}{3(x-2)^{1/3}} \right)(2x+1) + (x-2)^{2/3}(2)$

Step 3: Simplify the Expression for $f'(x)$

To find where $f'(x)$ is zero or undefined, we need to simplify the expression by combining the terms into a single fraction. $f'(x) = \frac{2(2x+1)}{3(x-2)^{1/3}} + 2(x-2)^{2/3}$ Find a common denominator, which is $3(x-2)^{1/3}$. Multiply the second term by $\frac{3(x-2)^{1/3}}{3(x-2)^{1/3}}$: $2(x-2)^{2/3} = 2(x-2)^{2/3} \cdot \frac{3(x-2)^{1/3}}{3(x-2)^{1/3}} = \frac{6(x-2)^{2/3}(x-2)^{1/3}}{3(x-2)^{1/3}}$ Using the property $a^m \cdot a^n = a^{m+n}$: $(x-2)^{2/3}(x-2)^{1/3} = (x-2)^{(2/3)+(1/3)} = (x-2)^1 = x-2$ So, the second term becomes $\frac{6(x-2)}{3(x-2)^{1/3}}$.

Now, combine the terms: $f'(x) = \frac{2(2x+1)}{3(x-2)^{1/3}} + \frac{6(x-2)}{3(x-2)^{1/3}}$ $f'(x) = \frac{2(2x+1) + 6(x-2)}{3(x-2)^{1/3}}$ Simplify the numerator: $2(2x+1) + 6(x-2) = 4x+2 + 6x-12 = 10x - 10 = 10(x-1)$ Thus, the simplified derivative is: $f'(x) = \frac{10(x-1)}{3(x-2)^{1/3}}$

Step 4: Find Critical Points where $f'(x) = 0$

A fraction is equal to zero if and only if its numerator is zero and its denominator is non-zero. Set the numerator to zero: $10(x-1) = 0$ $x-1 = 0$ $x = 1$ Now, we must check if the denominator is non-zero at $x=1$: $3(1-2)^{1/3} = 3(-1)^{1/3} = 3(-1) = -3$. Since $-3 \neq 0$, $f'(1)=0$. Also, $x=1$ is in the domain of $f(x)$. Therefore, $x=1$ is a critical point.

Step 5: Find Critical Points where $f'(x)$ is Undefined

A rational expression (fraction) is undefined when its denominator is zero. Set the denominator to zero: $3(x-2)^{1/3} = 0$ $(x-2)^{1/3} = 0$ Cube both sides: $((x-2)^{1/3})^3 = 0^3$ $x-2 = 0$ $x = 2$ We must check if $x=2$ is in the domain of the original function $f(x)$. As established in Step 1, the domain of $f(x)$ is all real numbers, so $x=2$ is in the domain of $f(x)$. Therefore, $x=2$ is a critical point because $f'(2)$ is undefined, but $f(2)$ is defined.

Step 6: Count the Number of Critical Points

From our analysis, the critical points are $x=1$ and $x=2$. There are a total of 2 critical points.

Common Mistakes & Tips:

• Don't forget the undefined derivative case: Many students only look for $f'(x)=0$. The points where $f'(x)$ is undefined but $f(x)$ is defined are equally important critical points.
• Simplify the derivative fully: Before setting the numerator or denominator to zero, ensure the derivative is combined into a single, simplified fraction. This prevents errors and makes the process clearer.

Summary:

We systematically found the critical points by first determining the domain of the function. Then, we calculated the first derivative using the product and chain rules and simplified it into a single fraction. Finally, we identified points where the derivative was zero (numerator equals zero) and where it was undefined (denominator equals zero), ensuring these points were within the original function's domain. This led us to identify two critical points.

The final answer is $\boxed{2}$, which corresponds to option (A).