Chapter 1, Problem 18RE

Solution

To find: The domain and range of a function.

The domain and the range of the function $k\left(x\right)=\frac{1}{\sqrt{9-x^2}}$ are $\left(-3,3\right)$ and $\left[\frac{1}{3},\infty \right)$ respectively.

Given information: The function is $k\left(x\right)=\frac{1}{\sqrt{9-x^2}}$ .

Method/Formula used:

Suppose $y=f\left(x\right)$ be a function $f:A\to B$ , then the set A is called the domain of function f and set B the co-domain. The range of the function is the set of points which are images of points of domain of function f . Thus, the range is the domain of the function $f^{-1}\left(x\right)$ . The figure below explains domain, co-domain and the range of f .

  

Calculations:

The given function $k:A\to B$ is $k\left(x\right)=\frac{1}{\sqrt{9-x^2}}$ .

(a) Domain:

Clearly, the function $k\left(x\right)=\frac{1}{\sqrt{9-x^2}}$ is not defined when $9-x^2=0\text{or}x=\pm 3$ and $9-x^2<0\text{or}9<x^2$ . Therefore, $k\left(x\right)$ is defined in the interval $\left(-3,3\right)$ . The domain of the function $k\left(x\right)=\frac{1}{\sqrt{9-x^2}}$ is $\left(-3,3\right)$ .

(b) Range:

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

Now let, $y=\frac{1}{\sqrt{9-x^2}}$ . Therefore,

  $\left.\begin{array}{c}{y}^2=\frac{1}{9-x^2}\\ 9-x^2=\frac{1}{y^2}\\ x^2=9-\frac{1}{y^2}\\ x=\frac{\sqrt{9y^2-1}}{y}\\ k^{-1}\left(y\right)=\frac{\sqrt{9y^2-1}}{y}\end{array}\right\}$

The function $k^{-1}\left(y\right)$ is not defined for $y=0$ , and when $9y^2-1<0$ $\text{or}{y}^2<\frac{1}{9}$ .Thus, the function $k^{-1}\left(y\right)$ is defined in the interval $\left[\frac{1}{3},\infty \right)$ .

Hence, the range of function $k\left(x\right)$ is $\left[\frac{1}{3},\infty \right)$ .

The domain and the range of the function $k\left(x\right)=\frac{1}{\sqrt{9-x^2}}$ are $\left(-3,3\right)$ and $\left[\frac{1}{3},\infty \right)$ respectively.