Verify Rolles theorem for the function `f(x)=sqrt(4-x^2)` on `[-2,\ 2]` .

Explanation

To verify Rolle's theorem for the function \( f(x) = \sqrt{4 - x^2} \) on the interval \([-2, 2]\), we will follow these steps:

### Step 1: Check the conditions of Rolle's Theorem
Rolle's theorem states that if a function \( f \) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \( f(a) = f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f'(c) = 0 \).

### Step 2: Check continuity
The function \( f(x) = \sqrt{4 - x^2} \) is defined for \( 4 - x^2 \geq 0 \). This gives us the domain \( -2 \leq x \leq 2 \).

- Since \( f(x) \) is a square root function and is defined on the interval \([-2, 2]\), it is continuous on this interval.

### Step 3: Check differentiability
Next, we need to check if \( f(x) \) is differentiable on the open interval \((-2, 2)\).

- The derivative \( f'(x) \) can be found using the chain rule:
\[ f'(x) = \frac{d}{dx}(\sqrt{4 - x^2}) = \frac{1}{2\sqrt{4 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{4 - x^2}} \]
- The derivative \( f'(x) \) is defined for \( x \in (-2, 2) \) except at \( x = \pm 2 \). Thus, \( f(x) \) is differentiable on \((-2, 2)\).

### Step 4: Check if \( f(-2) = f(2) \)
Now we need to evaluate \( f(-2) \) and \( f(2) \):
- \( f(-2) = \sqrt{4 - (-2)^2} = \sqrt{4 - 4} = \sqrt{0} = 0 \)
- \( f(2) = \sqrt{4 - (2)^2} = \sqrt{4 - 4} = \sqrt{0} = 0 \)

Since \( f(-2) = f(2) = 0 \), the condition \( f(a) = f(b) \) is satisfied.

### Step 5: Find \( c \) such that \( f'(c) = 0 \)
Now we need to find \( c \) in \((-2, 2)\) such that \( f'(c) = 0 \):
\[ f'(c) = \frac{-c}{\sqrt{4 - c^2}} = 0 \]
This implies that \( -c = 0 \), hence \( c = 0 \).

### Step 6: Conclusion
The value \( c = 0 \) lies in the interval \((-2, 2)\). Therefore, we have verified that all conditions of Rolle's theorem are satisfied.

### Final Answer
Rolle's theorem is verified for the function \( f(x) = \sqrt{4 - x^2} \) on the interval \([-2, 2]\).

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