Let `f(x)=(4-x^(2))^(2//3)`, then f has a

Options

A. local minima at x = 0
B. local maxima at x = 2
C. local maxima at x = -2
D. none of these

Answer

To solve the problem, we need to analyze the function \( f(x) = (4 - x^2)^{\frac{2}{3}} \) and find its critical points to determine where it has maxima and minima.

### Step-by-Step Solution:

1. **Find the derivative \( f'(x) \)**:
We start by applying the chain rule to differentiate \( f(x) \).
\[ f'(x) = \frac{d}{dx} \left( (4 - x^2)^{\frac{2}{3}} \right) = \frac{2}{3} (4 - x^2)^{-\frac{1}{3}} \cdot (-2x) \]
Simplifying this gives:
\[ f'(x) = -\frac{4x}{3(4 - x^2)^{\frac{1}{3}}} \]

2. **Set the derivative to zero to find critical points**:
We set \( f'(x) = 0 \):
\[ -\frac{4x}{3(4 - x^2)^{\frac{1}{3}}} = 0 \]
This implies \( 4x = 0 \) or \( x = 0 \).

3. **Determine the intervals for testing**:
The critical points divide the number line into intervals. We need to check the sign of \( f'(x) \) in the intervals:
- \( (-\infty, -2) \)
- \( (-2, 0) \)
- \( (0, 2) \)
- \( (2, \infty) \)

4. **Test the intervals**:
- For \( x < -2 \) (e.g., \( x = -3 \)):
\[ f'(-3) = -\frac{4(-3)}{3(4 - 9)^{\frac{1}{3}}} \text{ (undefined as denominator is negative)} \]
- For \( -2 < x < 0 \) (e.g., \( x = -1 \)):
\[ f'(-1) = -\frac{4(-1)}{3(4 - 1)^{\frac{1}{3}}} > 0 \text{ (positive)} \]
- For \( 0 < x < 2 \) (e.g., \( x = 1 \)):
\[ f'(1) = -\frac{4(1)}{3(4 - 1)^{\frac{1}{3}}} < 0 \text{ (negative)} \]
- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ f'(3) = -\frac{4(3)}{3(4 - 9)^{\frac{1}{3}}} \text{ (undefined as denominator is negative)} \]

5. **Analyze the sign changes**:
- From \( (-\infty, -2) \): undefined
- From \( (-2, 0) \): \( f' > 0 \) (increasing)
- At \( x = 0 \): \( f' = 0 \) (critical point)
- From \( (0, 2) \): \( f' < 0 \) (decreasing)
- From \( (2, \infty) \): undefined

6. **Conclusion about maxima and minima**:
Since \( f(x) \) increases on \( (-2, 0) \) and decreases on \( (0, 2) \), we conclude that there is a local maximum at \( x = 0 \).

### Final Answer:
The function \( f(x) \) has a local maximum at \( x = 0 \).