To find the critical numbers of the function $f(x) = x(x-3)^{\frac{1}{3}}$, follow these steps:

Step 1: Find the derivative of $f(x)$

We use the product rule: $\frac{d}{dx}[u \cdot v] = u'v + uv'$ Here:

• $u = x$
• $v = (x-3)^{\frac{1}{3}}$

Derivatives of $u$ and $v$:

• $u' = \frac{d}{dx}[x] = 1$
• $v = (x-3)^{\frac{1}{3}}$, so $v' = \frac{1}{3}(x-3)^{-\frac{2}{3}}$

Now applying the product rule: $f'(x) = u'v + uv' = 1 \cdot (x-3)^{\frac{1}{3}} + x \cdot \frac{1}{3}(x-3)^{-\frac{2}{3}}$ Simplify: $f'(x) = (x-3)^{\frac{1}{3}} + \frac{x}{3}(x-3)^{-\frac{2}{3}}$

Step 2: Set $f'(x) = 0$

To find the critical numbers, solve $f'(x) = 0$: $(x-3)^{\frac{1}{3}} + \frac{x}{3}(x-3)^{-\frac{2}{3}} = 0$

Combine terms:

Factor out $(x-3)^{-\frac{2}{3}}$: $(x-3)^{-\frac{2}{3}} \left[ (x-3) + \frac{x}{3} \right] = 0$

Simplify the term in brackets:

$(x-3) + \frac{x}{3} = \frac{3(x-3) + x}{3} = \frac{3x - 9 + x}{3} = \frac{4x - 9}{3}$

Thus, the equation becomes: $(x-3)^{-\frac{2}{3}} \cdot \frac{4x - 9}{3} = 0$

Step 3: Solve the equation

The product is zero when either factor is zero:

1. $(x-3)^{-\frac{2}{3}} = 0$
   This term is never zero because $(x-3)^{-\frac{2}{3}}$ is undefined when $x = 3$, but it never equals zero.

2. $\frac{4x - 9}{3} = 0$: $4x - 9 = 0 \quad \implies \quad x = \frac{9}{4}$

Step 4: Check the domain

The function $f(x)$ is undefined when $x = 3$, so we exclude this point.

Critical Number:

The critical number is: $x = \frac{9}{4}$

Would you like further clarification or a detailed explanation of any step?

Related Questions:

1. How does the product rule work in derivatives?
2. What are critical points and why are they important?
3. How do you handle fractional exponents in derivatives?
4. Why is $x = 3$ excluded in the domain of $f(x)$?
5. What is the significance of the derivative being undefined?

Tip: Always verify the domain of a function before solving for critical points to avoid including undefined values.