`f(x) = log(x^(2)+2)-log3`in `[-1,1]`

Explanation

To solve the problem, we need to verify the conditions of Rolle's Theorem for the function \( f(x) = \log(x^2 + 2) - \log(3) \) on the interval \([-1, 1]\).

### Step-by-Step Solution:

1. **Define the Function**:
\[ f(x) = \log(x^2 + 2) - \log(3) \]

2. **Check Continuity**:
The logarithmic function is continuous wherever its argument is positive. Since \( x^2 + 2 \) is always positive for all \( x \) (as \( x^2 \geq 0 \)), \( f(x) \) is continuous on the interval \([-1, 1]\).

3. **Check Differentiability**:
Since \( f(x) \) is composed of continuous and differentiable functions (logarithm and polynomial), \( f(x) \) is differentiable on the interval \([-1, 1]\).

4. **Evaluate the Endpoints**:
- Calculate \( f(-1) \):
\[ f(-1) = \log((-1)^2 + 2) - \log(3) = \log(1 + 2) - \log(3) = \log(3) - \log(3) = 0 \]
- Calculate \( f(1) \):
\[ f(1) = \log(1^2 + 2) - \log(3) = \log(1 + 2) - \log(3) = \log(3) - \log(3) = 0 \]

5. **Check the Conditions for Rolle's Theorem**:
Since \( f(-1) = f(1) = 0 \), the conditions of Rolle's Theorem are satisfied.

6. **Find \( c \) such that \( f'(c) = 0 \)**:
First, we need to find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx} \left( \log(x^2 + 2) \right) - \frac{d}{dx} \left( \log(3) \right) \]
The derivative of \( \log(3) \) is 0, so:
\[ f'(x) = \frac{1}{x^2 + 2} \cdot (2x) = \frac{2x}{x^2 + 2} \]
Set \( f'(c) = 0 \):
\[ \frac{2c}{c^2 + 2} = 0 \]
This implies \( 2c = 0 \), thus \( c = 0 \).

7. **Conclusion**:
Since \( c = 0 \) lies within the interval \([-1, 1]\), we have verified that there exists a \( c \) in \([-1, 1]\) such that \( f'(c) = 0 \), confirming the application of Rolle's Theorem.