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

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

We apply the product rule: [ f'(x) = \frac{d}{dx} \big[x\big] \cdot (x - 3)^{\frac{1}{3}} + x \cdot \frac{d}{dx} \big[(x - 3)^{\frac{1}{3}}\big] ] $f'(x) = 1 \cdot (x - 3)^{\frac{1}{3}} + x \cdot \frac{1}{3}(x - 3)^{-\frac{2}{3}} \cdot 1$ $f'(x) = (x - 3)^{\frac{1}{3}} + \frac{x}{3}(x - 3)^{-\frac{2}{3}}$

Step 2: Simplify $f'(x)$

To simplify, factor out $(x - 3)^{-\frac{2}{3}}$: $f'(x) = (x - 3)^{-\frac{2}{3}} \left[(x - 3) + \frac{x}{3}\right]$ $f'(x) = (x - 3)^{-\frac{2}{3}} \left[\frac{3(x - 3) + x}{3}\right]$ $f'(x) = (x - 3)^{-\frac{2}{3}} \cdot \frac{3x - 9 + x}{3}$ $f'(x) = (x - 3)^{-\frac{2}{3}} \cdot \frac{4x - 9}{3}$

Step 3: Set $f'(x) = 0$ or undefined

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

1. $f'(x) = 0$

For the fraction to equal zero, the numerator must be zero: $4x - 9 = 0 \implies x = \frac{9}{4}$

2. $f'(x)$ undefined

The derivative is undefined where $(x - 3)^{-\frac{2}{3}}$ is undefined, which happens when $x - 3 = 0$: $x = 3$

Step 4: List the critical numbers

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

Would you like further explanations or visualizations of this process?

Related Questions:

1. How do we interpret critical numbers in terms of the graph of $f(x)$?
2. Can we determine whether the critical points are maxima, minima, or neither?
3. How does the fractional exponent $\frac{1}{3}$ affect the domain of $f(x)$?
4. Can we confirm the critical points using a sign chart for $f'(x)$?
5. How would the critical points change if $f(x)$ had a different exponent?

Tip:

Always check where a function's derivative is undefined, especially when fractional powers are involved!