Let `f(x) = 2 + 2x -3x^(2//3) " on " [-1, 10//3]`. Then f has

Options

A. Absolute maximum at an end point
B. Absolute minimum at an interior point
C. Absolute minimum is `f(10//3)`
D. Absolute minimum is `f(-1)`

Answer

To find the absolute maximum and minimum values of the function \( f(x) = 2 + 2x - 3x^{2/3} \) on the interval \([-1, \frac{10}{3}]\), we will follow these steps:

### Step 1: Find the derivative of \( f(x) \)

We start by differentiating the function \( f(x) \):

\[ f'(x) = \frac{d}{dx}(2 + 2x - 3x^{2/3}) \]

Using the power rule for differentiation, we get:

\[ f'(x) = 0 + 2 - 3 \cdot \frac{2}{3} x^{-1/3} = 2 - 2x^{-1/3} \]

Thus,

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

### Step 2: Set the derivative to zero to find critical points

To find the critical points, we set the derivative equal to zero:

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

Solving for \( x \):

\[ \frac{2}{x^{1/3}} = 2 \implies x^{1/3} = 1 \implies x = 1 \]

### Step 3: Evaluate the function at the critical point and endpoints

Now we evaluate \( f(x) \) at the critical point \( x = 1 \) and the endpoints of the interval \( x = -1 \) and \( x = \frac{10}{3} \).

1. **At \( x = -1 \)**:
\[ f(-1) = 2 + 2(-1) - 3(-1)^{2/3} = 2 - 2 - 3(1) = -3 \]

2. **At \( x = 1 \)**:
\[ f(1) = 2 + 2(1) - 3(1)^{2/3} = 2 + 2 - 3 = 1 \]

3. **At \( x = \frac{10}{3} \)**:
\[ f\left(\frac{10}{3}\right) = 2 + 2\left(\frac{10}{3}\right) - 3\left(\frac{10}{3}\right)^{2/3} \]
First, calculate \( \left(\frac{10}{3}\right)^{2/3} \):
\[ \left(\frac{10}{3}\right)^{2/3} = \left(\frac{100}{9}\right)^{1/3} = \frac{10^{2/3}}{3^{2/3}} = \frac{10^{2/3}}{3^{2/3}} \approx 2.154 \]
Now substituting back:
\[ f\left(\frac{10}{3}\right) \approx 2 + \frac{20}{3} - 3(2.154) \approx 2 + 6.67 - 6.462 = 2.208 \]

### Step 4: Compare the values

Now we compare the values we calculated:

- \( f(-1) = -3 \)
- \( f(1) = 1 \)
- \( f\left(\frac{10}{3}\right) \approx 2.208 \)

### Conclusion

The absolute maximum value of \( f(x) \) on the interval \([-1, \frac{10}{3}]\) occurs at \( x = \frac{10}{3} \) and is approximately \( 2.208 \).

The absolute minimum value occurs at \( x = -1 \) and is \( -3 \).

### Final Answer

- Absolute Maximum: \( f\left(\frac{10}{3}\right) \approx 2.208 \)
- Absolute Minimum: \( f(-1) = -3 \)

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