We are tasked with determining the absolute extrema (maximum and minimum) of the function $f(x) = 3x^{2/3} - 2x$ on the interval $[-1, 1]$.

Step-by-step approach:

1. Find the critical points:

   • The first step is to take the derivative of $f(x)$.

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

   Using the power rule for derivatives:

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

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

   $\frac{2}{x^{1/3}} - 2 = 0$

   Solving for $x$:

   $\frac{2}{x^{1/3}} = 2 \quad \Rightarrow \quad x^{1/3} = 1 \quad \Rightarrow \quad x = 1$

2. Check endpoints and critical points: The critical point is $x = 1$. We also need to evaluate the function at the endpoints of the interval $[-1, 1]$, which are $x = -1$ and $x = 1$.

   • At $x = -1$:

     $f(-1) = 3(-1)^{2/3} - 2(-1) = 3(1) + 2 = 5$

   • At $x = 1$:

     $f(1) = 3(1)^{2/3} - 2(1) = 3 - 2 = 1$

   So, the function is evaluated at:

   • $f(-1) = 5$
   • $f(1) = 1$

3. Evaluate the absolute maximum and minimum: Now, compare the values:

   • $f(-1) = 5$
   • $f(1) = 1$

   Since $f(-1) = 5$ is the largest value, the absolute maximum occurs at $x = -1$. Since $f(1) = 1$ is the smallest value, the absolute minimum occurs at $x = 1$.

Final answers:

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

Would you like more details or have any questions?

Here are five questions that build on this topic:

1. How do you find the second derivative to confirm concavity and extrema classification?
2. What happens if the critical point is outside the interval of interest?
3. How do we handle functions with multiple critical points?
4. Can a function have an absolute maximum but no absolute minimum on an open interval?
5. How does the presence of cusps or corners affect critical points?

Tip: Always check both endpoints of the interval along with the critical points when finding absolute extrema on a closed interval.