Chapter 1.6, Problem 9QR

To determine

To show: Thefunction $f\left(x\right)=x^3-3x$ odd function. Also, explain the reason that the graph is symmetric about the origin.

Explanation of Solution

Given information:The function is $f\left(x\right)=x^3-3x$ .

Calculation:

A function $f$ is said to be an even function if $f\left(-x\right)=f\left(x\right)$ for all $x$ in its domain.

A function $f$ is said to be oddfunction if $f\left(-x\right)=-f\left(x\right)$ for all $x$ in its domain.

Substitute $-x$ for $x$ in the function $f\left(x\right)=2x^2-3$ .

  $\begin{array}{c}f\left(-x\right)={\left(-x\right)}^3-3\left(-x\right)\\ =-x^3+3x\\ =-\left(x^3-3x\right)\\ =-f\left(x\right)\end{array}$

So, the given function is odd function.

Let the point $\left(a,b\right)$ lies on the graph of the given function.

  $f\left(a\right)=b$

Since, $f\left(-x\right)=-f\left(x\right)$ , then $f\left(-a\right)=-f\left(a\right)=-b$ it means that the point $\left(-a,-b\right)$ also lies on the graph. So, the function is symmetric to the origin as for every point $\left(a,b\right)$ on the graph, there is the point $\left(-a,-b\right)$ on the graph.

Hence, it is proved that $f\left(x\right)=2x^2-3$ is odd function.