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 determine the local maxima and minima of the function \( f(x) = (4 - x^2)^{\frac{2}{3}} \), we will follow these steps:

### Step 1: Find the First Derivative
We start by finding the first derivative \( f'(x) \).

Using the chain rule:
\[ f'(x) = \frac{d}{dx} \left( (4 - x^2)^{\frac{2}{3}} \right) = \frac{2}{3} (4 - x^2)^{-\frac{1}{3}} \cdot (-2x) \]
This simplifies to:
\[ f'(x) = -\frac{4x}{3(4 - x^2)^{\frac{1}{3}}} \]

### Step 2: Set the First Derivative to Zero
To find critical points, we set \( f'(x) = 0 \):
\[ -\frac{4x}{3(4 - x^2)^{\frac{1}{3}}} = 0 \]
This implies:
\[ 4x = 0 \quad \Rightarrow \quad x = 0 \]

### Step 3: Check Points of Non-Differentiability
Next, we check where the derivative does not exist. The denominator \( 3(4 - x^2)^{\frac{1}{3}} \) is undefined when \( 4 - x^2 = 0 \):
\[ 4 - x^2 = 0 \quad \Rightarrow \quad x^2 = 4 \quad \Rightarrow \quad x = \pm 2 \]
Thus, \( x = 2 \) and \( x = -2 \) are points where the function is not differentiable.

### Step 4: Find the Second Derivative
Now we find the second derivative \( f''(x) \) to determine the nature of the critical point \( x = 0 \):
\[ f''(x) = \frac{d}{dx} \left( -\frac{4x}{3(4 - x^2)^{\frac{1}{3}}} \right) \]
Using the quotient rule:
\[ f''(x) = \frac{(3(4 - x^2)^{\frac{1}{3}})(-4) - (-4x)(-\frac{1}{3}(4 - x^2)^{-\frac{2}{3}}(-2x))}{(3(4 - x^2)^{\frac{1}{3}})^2} \]
This simplifies to:
\[ f''(x) = \frac{-12(4 - x^2)^{\frac{1}{3}} + \frac{8x^2}{3(4 - x^2)^{\frac{2}{3}}}}{9(4 - x^2)^{\frac{2}{3}}} \]

### Step 5: Evaluate the Second Derivative at the Critical Point
Now we evaluate \( f''(0) \):
\[ f''(0) = \frac{-12(4)^{\frac{1}{3}} + \frac{8(0)^2}{3(4)^{\frac{2}{3}}}}{9(4)^{\frac{2}{3}}} = \frac{-12 \cdot 2 + 0}{9 \cdot 4} = \frac{-24}{36} = -\frac{2}{3} \]
Since \( f''(0) < 0 \), this indicates that \( x = 0 \) is a local maximum.

### Conclusion
Thus, the function \( f(x) = (4 - x^2)^{\frac{2}{3}} \) has a local maximum at \( x = 0 \).

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