Chapter 1, Problem 11RE

Solution

To find: The domain and range of a function.

The domain D and the range R of function $g\left(x\right)=x^3$ in interval notation are $D\text{=}\left(-\infty ,\infty \right)$ and $R\text{=}\left(-\infty ,\infty \right)$ respectively.

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

Method/Formula used:

Suppose $y=f\left(x\right)$ be a function $f:A\to B$ , $x\in A,y\in B$ the set A is called the domain of function f and set B the co-domain. Thus, the domain of a function $f\left(x\right)$ are the values of x for which the function $f\left(x\right)$ is defined. The range of the function is the set of points which are images of points of domain of function f . Thus, the range is a subset of the codomain of the function f . The figure below explains domain, co-domain and the range of f .

  

Calculations:

The given function is $g\left(x\right)=x^3$ . Let $g:A\to B$

Domain:

This function $g\left(x\right)=x^3$ is defined for all real numbers, therefore, the set of all real numbers $ℝ\text{or}\left(-\infty ,\infty \right)$ is the domain of function $g\left(x\right)=x^3$ .

Range:

Suppose $y=g\left(x\right)$ , then the set of all y for which there exists an ‘ x’ in the domain of the function of $g\left(x\right)$ or set of points y for which the inverse function $g^{-1}\left(y\right)=x$ is defined.

Here,

  $\left.\begin{array}{l}y=x^3\\ \Rightarrow x={\left(y\right)}^{\frac{1}{3}}\end{array}\right\}$ … (1)

From result (1), it is clear that for every $y\in B$ , there exits cube root of y equals to x in set $A=ℝ$ , therefore, $y\in ℝ$ .

  

Hence, the range of the function $g\left(x\right)=x^3$ is the set of real numbers $ℝ$ .

Thus, the domain and rage of the function $g\left(x\right)=x^3$ are $ℝ\text{or}\left(-\infty ,\infty \right)$ and $ℝ\text{or}\left(-\infty ,\infty \right)$ respectively.