Verify rolle's theorem for f(x)=√(1-x²) in [-1,1]

To verify Rolle's theorem for \( f(x) = \sqrt{1 - x^2} \) on the interval \([-1, 1]\), we need to check three conditions:

1. **Continuity**: \( f(x) \) must be continuous on the closed interval \([-1, 1]\).
2. **Differentiability**: \( f(x) \) must be differentiable on the open interval \((-1, 1)\).
3. **Equal function values at the endpoints**: \( f(-1) = f(1) \).

Let's verify each condition step by step.

### 1. Continuity

The function \( f(x) = \sqrt{1 - x^2} \) is continuous on the interval \([-1, 1]\) because the square root function is continuous wherever its argument is non-negative, and \(1 - x^2 \geq 0\) for \( x \in [-1, 1] \).

### 2. Differentiability

To check differentiability, we need to find the derivative of \( f(x) \) and ensure it exists for \( x \in (-1, 1) \).

\[ f(x) = \sqrt{1 - x^2} \]
Using the chain rule:
\[ f'(x) = \frac{d}{dx} \left( (1 - x^2)^{1/2} \right) \]
\[ f'(x) = \frac{1}{2} (1 - x^2)^{-1/2} \cdot (-2x) \]
\[ f'(x) = \frac{-x}{\sqrt{1 - x^2}} \]

The derivative exists for \( x \in (-1, 1) \), as \( \sqrt{1 - x^2} \neq 0 \) in this interval.

### 3. Equal function values at the endpoints

Calculate \( f(-1) \) and \( f(1) \):

\[ f(-1) = \sqrt{1 - (-1)^2} = \sqrt{1 - 1} = \sqrt{0} = 0 \]
\[ f(1) = \sqrt{1 - 1^2} = \sqrt{0} = 0 \]

Since \( f(-1) = f(1) = 0 \), the function values at the endpoints are equal.

### Applying Rolle's Theorem

Since all three conditions are satisfied, Rolle's theorem guarantees that there exists at least one \( c \in (-1, 1) \) such that \( f'(c) = 0 \).

Set \( f'(x) = 0 \):

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

Learn how to verify Rolle's theorem for the function f(x) = √(1 - x²) on the interval [-1, 1]. This problem covers continuity, differentiability, and equal function values at the endpoints. Suitable for advanced high school students and calculus enthusiasts.

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