Verify the conditions of Rolle's Theorem in the following problems. In each case, find a point in the interval, where the derivative vanishes :
`log(x^(2)+2)-log3" on "[-1,1]`

Answer

To verify the conditions of Rolle's Theorem for the function \( f(x) = \log(x^2 + 2) - \log(3) \) on the interval \([-1, 1]\), we will follow these steps:

### Step 1: Check Continuity
We need to check if the function \( f(x) \) is continuous on the closed interval \([-1, 1]\).

1. **Domain of the logarithm**: The function \( \log(x^2 + 2) \) is defined for all \( x \) because \( x^2 + 2 > 0 \) for all real \( x \).
2. **Continuity**: Since \( \log(x^2 + 2) \) is continuous for all \( x \), and \( \log(3) \) is a constant, the function \( f(x) \) is continuous on \([-1, 1]\).

### Step 2: Check Differentiability
Next, we check if \( f(x) \) is differentiable on the open interval \((-1, 1)\).

1. **Differentiability**: The function \( f(x) = \log(x^2 + 2) - \log(3) \) is differentiable wherever \( x^2 + 2 > 0 \). Since \( x^2 + 2 \) is positive for all \( x \), \( f(x) \) is differentiable on \((-1, 1)\).

### Step 3: Check if \( f(a) = f(b) \)
We need to verify that \( f(-1) = f(1) \).

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

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

Since \( f(-1) = f(1) = 0 \), the condition \( f(a) = f(b) \) is satisfied.

### Step 4: Find \( c \) such that \( f'(c) = 0 \)
Now we will find a point \( c \) in the interval \((-1, 1)\) where the derivative \( f'(c) \) vanishes.

1. **Find the derivative**:
\[ f'(x) = \frac{d}{dx} \left( \log(x^2 + 2) - \log(3) \right) = \frac{1}{x^2 + 2} \cdot (2x) = \frac{2x}{x^2 + 2} \]

2. **Set the derivative to zero**:
\[ f'(x) = 0 \implies \frac{2x}{x^2 + 2} = 0 \]
This occurs when \( 2x = 0 \) or \( x = 0 \).

3. **Check if \( c = 0 \) is in the interval**:
Since \( 0 \) is in the interval \((-1, 1)\), we have found our point.

### Conclusion
We have verified all the conditions of Rolle's Theorem:
1. \( f(x) \) is continuous on \([-1, 1]\).
2. \( f(x) \) is differentiable on \((-1, 1)\).
3. \( f(-1) = f(1) = 0 \).

Thus, by Rolle's Theorem, there exists at least one \( c \) in \((-1, 1)\) such that \( f'(c) = 0 \). We found that \( c = 0 \).