Question

Is $f\left(x\right)=\frac{x}{x^2+2\left|x\right|+1}$ is an odd function

Answer 1

Let $f\left(x\right)=\frac{x}{x^2+2\left|x\right|+1}$

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

As ${(-x)}^2=x^2$ and $|-x|=\left|x\right|$ we get

$f(-x)=\frac{-x}{x^2+2\left|x\right|+1}$

$\Rightarrow f(-x)=-\left(\frac{x}{x^2+2\left|x\right|+1}\right)$

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

So, $f\left(x\right)$ is an odd function