Verify Rolles theorem for function `f(x)=log(x^2+2)-log3` on `[-1,\ 1]`

Explanation

To verify Rolle's Theorem for the function \( f(x) = \log(x^2 + 2) - \log(3) \) on the interval \([-1, 1]\), we will follow the steps outlined by the theorem.

### Step 1: Check the conditions of Rolle's Theorem

**Condition 1:** The function \( f(x) \) must be continuous on the closed interval \([-1, 1]\).

- The function \( f(x) = \log(x^2 + 2) - \log(3) \) is a logarithmic function.
- The argument of the logarithm, \( x^2 + 2 \), is always positive for all real \( x \) because \( x^2 \) is non-negative and adding 2 ensures it is greater than 0.
- Therefore, \( f(x) \) is continuous on \([-1, 1]\).

**Condition 2:** The function \( f(x) \) must be differentiable on the open interval \((-1, 1)\).

- The function \( f(x) \) is differentiable wherever its argument is positive. Since \( x^2 + 2 > 0 \) for all \( x \), \( f(x) \) is differentiable on \((-1, 1)\).

**Condition 3:** The function values at the endpoints must be equal, i.e., \( f(-1) = f(1) \).

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

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

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

### Step 2: Find \( c \) in \((-1, 1)\) such that \( f'(c) = 0 \)

- Differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( \log(x^2 + 2) \right) - 0 = \frac{1}{x^2 + 2} \cdot (2x) = \frac{2x}{x^2 + 2} \]

- Set the derivative equal to zero:
\[ \frac{2x}{x^2 + 2} = 0 \]

- The numerator must be zero:
\[ 2x = 0 \implies x = 0 \]

### Conclusion

The value \( c = 0 \) lies in the interval \((-1, 1)\). Therefore, by Rolle's Theorem, since all conditions are satisfied, we conclude that there exists at least one \( c \) in \((-1, 1)\) such that \( f'(c) = 0 \).

Thus, Rolle's Theorem is verified for the function \( f(x) = \log(x^2 + 2) - \log(3) \) on the interval \([-1, 1]\).

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