Question #f729c

Feb 13, 2017

It is neither, but $f \left(x\right) = \ln \left(2 - x\right) - \ln \left(x + 2\right)$ is odd. (Perhaps there's a typo.)

Explanation:

Function $f$ is even if and only if for any $x$ in the domain of $f$, we have $f \left(- x\right) = f \left(x\right)$

Function $f$ is odd if and only if for any $x$ in the domain of $f$, we have $f \left(- x\right) = - f \left(x\right)$

The domain of the function is $\left(2 , \infty\right)$, so for any $x$ in the domain, $- x$ is not in the domain, that is, $f \left(- x\right)$ is not defined.

Bonus Note

$f \left(x\right) = \ln \left(2 - x\right) - \ln \left(x + 2\right)$ is odd.

The domain of $f$ is $\left(- 2 , 2\right)$.

For $x$ in the domain, $- x$ is also in the domain, and

$f \left(- x\right) = \ln \left(2 - \left(- x\right)\right) - \ln \left(\left(- x\right) + 2\right)$

$= \ln \left(2 + x\right) - \ln \left(- x + 2\right)$

$= \ln \left(x + 2\right) - \ln \left(2 - x\right)$

$= - \left(\ln \left(2 - x\right) - \ln \left(x + 2\right)\right)$

$= - f \left(x\right)$