# Is the function f(x)=x^2/(x-1) even, odd or neither?

Sep 1, 2015

It is neither odd or even.

#### Explanation:

First of all if you want to discuss if a function is odd or even, the domain of the function must be symetrical to 0 (in other words if $x \in D$ then $- x \in D$).

Function given in this task does not satisfy this condition, because $- 1 \in D$ but $1 \cancel{\in} D$.

Sep 1, 2015

$f$ is neither even nor odd.

#### Explanation:

$f \left(- x\right) = {\left(- x\right)}^{2} / \left(\left(- x\right) - 1\right)$

$= {x}^{2} / \left(- x - 1\right) = {x}^{2} / - \left(x + 1\right) = - {x}^{2} / \left(x + 1\right)$

$f \left(- x\right) \ne f \left(x\right)$ (that is $f \left(- x\right)$ is not always the same as $f \left(x\right)$), so $f$ is not even

$f \left(- x\right) \ne - f \left(x\right)$ (that is $f \left(- x\right)$ is not always the same as $- f \left(x\right)$, so $f$ is not odd.

$f$ is neither even nor odd.