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 continuity of \( f(x) \) on \([-2, 2]\)

The function \( f(x) = \sqrt{4 - x^2} \) is defined for all \( x \) such that \( 4 - x^2 \geq 0 \). This inequality holds true when \( -2 \leq x \leq 2 \). Since \( f(x) \) is a composition of continuous functions (the square root function and the polynomial), it is continuous on the closed interval \([-2, 2]\).

### Step 2: Check the differentiability of \( f(x) \) on \((-2, 2)\)

The function \( f(x) \) is differentiable wherever its derivative exists. We can find the derivative of \( f(x) \) using the chain rule. The function is differentiable in the open interval \((-2, 2)\) since it is continuous and does not have any points of non-differentiability in this interval.

### Step 3: 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 4: Find \( c \) such that \( f'(c) = 0 \)

Now we differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx} \left( \sqrt{4 - x^2} \right) = \frac{1}{2\sqrt{4 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{4 - x^2}} \]

Setting the derivative equal to zero to find \( c \):

\[ \frac{-x}{\sqrt{4 - x^2}} = 0 \]

The numerator must be zero:

\[ -x = 0 \implies x = 0 \]

### Conclusion

The value \( c = 0 \) lies within the interval \([-2, 2]\). Therefore, by Rolle's Theorem, there exists at least one point \( c \) in the interval such that \( f'(c) = 0 \).

### Summary

1. \( f(x) \) is continuous on \([-2, 2]\).
2. \( f(x) \) is differentiable on \((-2, 2)\).
3. \( f(-2) = f(2) = 0 \).
4. \( c = 0 \) satisfies \( f'(c) = 0 \).

Thus, we have verified Rolle's Theorem for the function \( f(x) = \sqrt{4 - x^2} \) on the interval \([-2, 2]\).