## Precalculus (6th Edition) Blitzer

The function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function and is symmetric about the origin.
In order to check if the function is even, odd, or neither, substitute $x$ by – $x$ and evaluate the value of $f\left( -x \right)$. \begin{align} & f\left( -x \right)={{\left( -x \right)}^{3}}-5\left( -x \right) \\ & =\left( -x \right)\left( -x \right)\left( -x \right)-5\left( -x \right) \\ & =-{{x}^{3}}+5x \\ & =-\left( {{x}^{3}}-5x \right) \end{align} Since, $f\left( x \right)={{x}^{3}}-5\left( x \right)$, therefore $f\left( -x \right)=-f\left( x \right)$. Now as $f\left( -x \right)=-f\left( x \right)$, it gives that the function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function The function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function, therefore by definition of an odd function, $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is symmetric about the origin. Hence, the function $f\left( x \right)={{x}^{3}}-5\left( x \right)$ is an odd function and is symmetric about the origin.