To find the critical numbers of the function $f(x) = x(x-3)^{1/3}$, here’s a step-by-step, simplified approach:

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

We use the product rule since the function is a product of two terms, $x$ and $(x-3)^{1/3}$: $f(x) = x \cdot (x-3)^{1/3}.$ The product rule states: $f'(x) = u'(v) + u(v'),$ where $u = x$ and $v = (x-3)^{1/3}$.

• Derivative of $u = x$ is $u' = 1$.
• Derivative of $v = (x-3)^{1/3}$ is: $v' = \frac{1}{3}(x-3)^{-2/3} \cdot 1.$

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

Step 2: Simplify the derivative

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

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

Simplify the terms inside the brackets: $f'(x) = (x-3)^{-2/3} \left[\frac{3(x-3) + x}{3}\right].$

Distribute $3(x-3)$: $f'(x) = (x-3)^{-2/3} \cdot \frac{3x - 9 + x}{3}.$

Combine like terms: $f'(x) = (x-3)^{-2/3} \cdot \frac{4x - 9}{3}.$

Thus: $f'(x) = \frac{(4x - 9)}{3(x-3)^{2/3}}.$

Step 3: Solve for critical points

Critical points occur when $f'(x) = 0$ or $f'(x)$ is undefined.

1. When $f'(x) = 0$: The numerator $4x - 9 = 0$: $4x = 9 \implies x = \frac{9}{4}.$

2. When $f'(x)$ is undefined: The denominator $(x-3)^{2/3} = 0$, which happens when $x = 3$.

Step 4: Final Answer

The critical numbers are: $x = \frac{9}{4} \quad \text{and} \quad x = 3.$

Would you like a detailed explanation of any step? Let me know!

Related Questions

1. What are critical numbers, and why do they matter in calculus?
2. How does the product rule help in differentiation?
3. What is the significance of the $(x-3)^{2/3}$ term in this function?
4. Could we simplify $f'(x)$ further for easier computation?
5. How would this process change for a different function, like $g(x) = x(x-3)^n$ for general $n$?

Tip

Always check both when $f'(x) = 0$ and when $f'(x)$ is undefined to ensure you capture all critical numbers!