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

Explanation

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

### Step 1: Check the continuity of \( f(x) \) on \([-1, 1]\)

The logarithmic function is continuous wherever its argument is positive. We need to ensure that \( x^2 + 2 > 0 \) for all \( x \) in the interval \([-1, 1]\).

- For \( x = -1 \):
\[ f(-1) = \log((-1)^2 + 2) - \log(3) = \log(1 + 2) - \log(3) = \log(3) - \log(3) = 0 \]

- For \( x = 0 \):
\[ f(0) = \log(0^2 + 2) - \log(3) = \log(2) - \log(3) \]

- For \( x = 1 \):
\[ f(1) = \log(1^2 + 2) - \log(3) = \log(3) - \log(3) = 0 \]

Since \( x^2 + 2 \) is always greater than 0 for all \( x \), \( f(x) \) is continuous on \([-1, 1]\).

### Step 2: Check the differentiability of \( f(x) \) on \([-1, 1]\)

Next, we need to check if \( f(x) \) is differentiable in the interval \([-1, 1]\). The derivative of \( f(x) \) is given by:

\[ f'(x) = \frac{d}{dx} \left( \log(x^2 + 2) - \log(3) \right) = \frac{2x}{x^2 + 2} \]

The function \( f'(x) \) is defined for all \( x \) in \([-1, 1]\) since \( x^2 + 2 > 0 \).

### Step 3: Evaluate \( f(-1) \) and \( f(1) \)

From our previous calculations:
- \( f(-1) = 0 \)
- \( f(1) = 0 \)

### Step 4: Apply Rolle's Theorem

Since \( f(x) \) is continuous on \([-1, 1]\), differentiable on \((-1, 1)\), and \( f(-1) = f(1) \), we can apply Rolle's theorem. According to Rolle's theorem, there exists at least one \( c \) in the interval \((-1, 1)\) such that:

\[ f'(c) = 0 \]

### Step 5: Solve for \( c \)

Setting the derivative equal to zero:

\[ \frac{2c}{c^2 + 2} = 0 \]

This implies:

\[ 2c = 0 \implies c = 0 \]

### Conclusion

Since \( c = 0 \) lies within the interval \([-1, 1]\), we have verified that Rolle's theorem holds for the function \( f(x) \) on the interval \([-1, 1]\).

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