# How do you determine if f(x)=x+absx is an even or odd function?

Apr 5, 2016

Relate $f \left(- x\right)$ to $f \left(x\right)$

#### Explanation:

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

$= - x + \left\mid x \right\mid$

Since $f \left(- x\right) \ne f \left(x\right)$, $f \left(x\right)$ is not an even function.
Since $f \left(- x\right) \ne - f \left(x\right)$, $f \left(x\right)$ is not an odd function.

Here is a graph of $y = f \left(x\right)$.
graph{x+abs(x) [-10, 10, -5, 5]}
If $f \left(x\right)$ is an even function, the $y$-axis would be a line of symmetry.

If $f \left(x\right)$ is an odd function, the graph would have rotational symmetry about the origin.

These are graphical methods to check whether a function is odd or even. However neither of the symmetries are present.

Hence, $f \left(x\right)$ is neither an odd function nor an even function.

Apr 5, 2016

Neither.

#### Explanation:

$f \left(- x\right) = - x + | x |$ is neither f(x) nor -f(x).
$| x | = - x , x < 0$ and $| x | = x , x > 0$.
$f \left(x\right) = x - x = 0 , x \le 0$..
$f \left(x\right) = 2 x , x > 0$..
The graph for y = f(x) comprises the negative x-axis continued as the straight line y = 2x in the first quadrant.,