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 $f'(x) = 0$:

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