$f(x) = \log \left( {{x^2} + 2} \right) - \log 3$ in [-1,1]

We have, $f(x) = \log \left( {{x^2} + 2} \right) - \log 3$
(i) Logarithmic functions are continuous in their domain.
Hence, $f(x) = \log \left( {{x^2} + 2} \right) - \log 3$ is continuous in [-1,1] .

(ii) $f(x) = \frac{1}{{{x^2} + 2}} \cdot 2x - 0$
$= \frac{{2x}}{{{x^2} + 2}},$ which exists in (-1,1) .
Hence, $f(x)$ is differentiable in (-1,1) .

(iii) $f( - 1) = \log \left[ {{{( - 1)}^2} + 2} \right] - \log 3 = \log 3 - \log 3 = 0$ and
$f(1) = \log \left( {{1^2} + 2} \right) - \log 3 = \log 3 - \log 3 = 0$

$\Rightarrow$ $f( - 1) = f(1)$

Conditions of Rolle's theorem are satisfied.

Hence, there exists a real number $c$ such that
$f(c) = 0$

$\Rightarrow$ $\frac{{2c}}{{{c^2} + 2}} = 0$

$\Rightarrow$ $c = 0 \in ( - 1,1)$

Hence, Rolle's theorem has been verified.