## Precalculus (6th Edition) Blitzer

Step I: To check even or odd. In the provided equation, if $f\left( -x \right)=f\left( x \right)$ then it is an even function, and if $f\left( -x \right)=-f\left( x \right)$ , then it is odd. \begin{align} & f\left( x \right)={{x}^{3}}+x \\ & f\left( -x \right)={{\left( -x \right)}^{3}}+\left( -x \right) \\ & =-{{x}^{3}}-x \\ & =-\left( {{x}^{3}}+x \right) \\ & =-f\left( x \right) \end{align} It is an odd function. Step II: To check symmetry about the y-axis: In the provided equation put $x=-x$ , if the equation remains the same, then it has symmetry about the y- axis. \begin{align} & y={{x}^{3}}+x \\ & y={{\left( -x \right)}^{3}}+\left( -x \right) \\ & y=-{{x}^{3}}-x \\ \end{align} It is not the same as the given equation, hence it is not symmetric about the y-axis. Step III: To check symmetry about the x-axis: In the provided equation put $y=-y$ , if the equation remains the same, then it has symmetry about the x-axis. \begin{align} & y={{x}^{3}}+x \\ & \left( -y \right)={{x}^{3}}+x \\ & -y={{x}^{3}}+x \end{align} It is not the same as the given equation, hence it is not symmetric about the x-axis. Step IV: To check symmetry about the origin: In the provided equation put $x=-x,\text{ and }y=-y$ , if the equation remains the same, then it has symmetry about the origin. \begin{align} & y={{x}^{3}}+x \\ & \left( -y \right)={{\left( -x \right)}^{3}}+\left( -x \right) \\ & -y=-{{x}^{3}}-x \\ & y={{x}^{3}}+x \end{align} It is the same as provided equation, hence it has symmetry about the origin. Thus, the given function is an odd function, has symmetry about the origin only.