2. Given a function $$ f(x)=\left\{\begin{array}{rll}x^{3}-\frac{x}{3}, & 0 \leq x \leq 1 \ x^{2}+x-\frac{4}{3}, & 1

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To determine the extreme values of the function $$ f(x) $$ defined on the interval $$[0, 2]$$:

$$
f(x) = 
\begin{cases} 
x^3 - \frac{x}{3}, & 0 \leq x \leq 1 \ 
x^2 + x - \frac{4}{3}, & 1 < x \leq 2 
\end{cases}
$$

**1. Evaluate at Endpoints and Critical Points:**

- **At $$ x = 0 $$:**
  $$
  f(0) = 0^3 - \frac{0}{3} = 0
  $$

- **At $$ x = 1 $$:**
  $$
  f(1) = 1^3 - \frac{1}{3} = \frac{2}{3}
  $$

- **At $$ x = 2 $$:**
  $$
  f(2) = 2^2 + 2 - \frac{4}{3} = 4 + 2 - \frac{4}{3} = \frac{14}{3} \approx 4.67
  $$

**2. Critical Points in Each Piece:**

- **For $$ 0 \leq x \leq 1 $$:**
  $$
  f'(x) = 3x^2 - \frac{1}{3}
  $$
  Setting $$ f'(x) = 0 $$:
  $$
  3x^2 - \frac{1}{3} = 0 \implies x = \frac{1}{3}
  $$
  Evaluating $$ f(x) $$ at $$ x = \frac{1}{3} $$:
  $$
  f\left(\frac{1}{3}\right) = \left(\frac{1}{3}\right)^3 - \frac{\frac{1}{3}}{3} = \frac{1}{27} - \frac{1}{9} = -\frac{2}{27} \approx -0.074
  $$

- **For $$ 1 < x \leq 2 $$:**
  $$
  f'(x) = 2x + 1
  $$
  Setting $$ f'(x) = 0 $$:
  $$
  2x + 1 = 0 \implies x = -\frac{1}{2} \quad (\text{Not within the interval})
  $$
  Hence, no critical points in this interval.

**3. Summary of Extreme Values:**

- **Maximum Value:** $$ f(2) = \frac{14}{3} \approx 4.67 $$
- **Minimum Value:** $$ f\left(\frac{1}{3}\right) = -\frac{2}{27} \approx -0.074 $$

**Correct Option:**

**D.** $$ 4.67, -0.074 $$

**Answer:**  
**D.** $$ 4.67, -0.074 $$
The extreme values of the function are $$ 4.67 $$ (maximum) and $$ -0.074 $$ (minimum).

Given a function What is the extrime value?

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