# Is the function y=x-sin(x) even, odd or neither?

Jul 20, 2016

The function will be odd.

#### Explanation:

For an even function, $f \left(- x\right) = f \left(x\right)$.
For an odd function, $f \left(- x\right) = - f \left(x\right)$

So we can test this by plugging in $x = - x$:
$- x - \sin \left(x\right) = - x + \sin \left(x\right) = \left(- 1\right) \left(x - \sin \left(x\right)\right)$

This means the function must be odd.

It's not surprising either, since $x$ and $\sin \left(x\right)$ are both odd. In fact, given two functions, $f \left(x\right)$ and $g \left(x\right)$ for which:
$f \left(- x\right) = - f \left(x\right)$
$g \left(- x\right) = - g \left(x\right)$

It is obvious that:
$f \left(- x\right) + g \left(- x\right) = - f \left(x\right) - g \left(x\right) = - \left[f \left(x\right) + g \left(x\right)\right]$

That is, the sum of odd functions is always another odd function.

Jul 20, 2016

$f \left(x\right) = x - \sin x$ is odd

#### Explanation:

A function $f$ is said to be even if $f \left(- x\right) = f \left(x\right)$, and odd if $f \left(- x\right) = - f \left(x\right)$. Then, to check, we will evaluate the function applied to $- x$.

In our case, $f \left(x\right) = x - \sin x$, so

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

$= - x - \left(- \sin x\right)$ (as $\sin x$ is odd)

$= - x + \sin x$

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

#=-f(x)

Thus $f \left(x\right) = x - \sin x$ is odd.