Chapter 0.6, Problem 9QR

To determine

To show: A function as an odd function.

Explanation of Solution

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

Method/Formula used:

A function $f\left(x\right)$ is an even function if $f\left(-x\right)=f\left(x\right)$ and odd function if $f\left(-x\right)=-f\left(x\right)$ .

The graph of a function is symmetrical about origin if both the points $\left(a,b\right)$ and $\left(-a,-b\right)$ lie on it.

Proof:

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

Now, replace x by − x ,

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

Since, for the given function $f\left(x\right)=x^3-3x$ , $f\left(-x\right)=-f\left(x\right)$ , therefore, $f\left(x\right)=x^3-3x$ is an odd function.

Now, write $y=x^3-3x$

Suppose the point $\left(a,b\right)$ lies on the graph of the function $y=x^3-3x$ . Then, $b=a^3-3a$ . Check whether point $\left(-a,-b\right)$ lies on the graph of $y=x^3-3x$ as:

  $\begin{array}{l}\left(-b\right)={\left(-a\right)}^3-3\left(-a\right)\\ -b=-a^3+3a\\ b=a^3-3a\end{array}$

Thus, the points $\left(a,b\right)$ and $\left(-a,-b\right)$ both lie on the graph of the function $y=x^3-3x$ that shows, its graph is symmetrical about the origin.

The graph of the function $f\left(x\right)=x^3-3x$ is shown in Fig.1.