To determine the absolute extrema of $f(x) = 3x^{2/3} - 2x$ on the interval $[-1, 1]$, we'll follow these steps:

1. Find the derivative of $f(x)$ to locate any critical points within the interval.
2. Evaluate the function at the critical points and endpoints to determine the absolute maximum and minimum values.

Step 1: Differentiate $f(x)$

Given: $f(x) = 3x^{2/3} - 2x$

The derivative $f'(x)$ is: $f'(x) = 3 \cdot \frac{2}{3} x^{-1/3} - 2 = 2x^{-1/3} - 2$ Simplifying, we get: $f'(x) = \frac{2}{x^{1/3}} - 2$

Step 2: Set $f'(x) = 0$ to find critical points

$\frac{2}{x^{1/3}} - 2 = 0$ Solving for $x$: $\frac{2}{x^{1/3}} = 2$ $x^{1/3} = 1 \Rightarrow x = 1$

So, $x = 1$ is a critical point. Note also that $f'(x)$ is undefined at $x = 0$, so $x = 0$ is another point to check.

Step 3: Evaluate $f(x)$ at the endpoints and critical points

We need to evaluate $f(x)$ at $x = -1$, $x = 0$, and $x = 1$.

1. At $x = -1$: $f(-1) = 3(-1)^{2/3} - 2(-1) = 3 \cdot 1 + 2 = 5$

2. At $x = 0$: $f(0) = 3 \cdot 0^{2/3} - 2 \cdot 0 = 0$

3. At $x = 1$: $f(1) = 3 \cdot 1^{2/3} - 2 \cdot 1 = 3 - 2 = 1$

Conclusion

The absolute maximum value of $f(x)$ on $[-1, 1]$ is $5$ at $x = -1$.
The absolute minimum value of $f(x)$ on $[-1, 1]$ is $0$ at $x = 0$.

Answers:

1. The absolute maximum occurs at $x = -1$.
2. The absolute minimum occurs at $x = 0$.

Would you like further clarification or details?

Here are five related questions for further practice:

1. How do you find critical points for functions with fractional exponents?
2. Why does the function have an undefined derivative at $x = 0$?
3. What is the significance of evaluating at endpoints for absolute extrema?
4. How can we use the second derivative to classify critical points?
5. How would the process change if the interval were open instead of closed?

Tip: Always check endpoints in a closed interval when finding absolute extrema, as they can often hold the maximum or minimum values.