Chapter 1.4, Problem 7E

Solution

To find: The formula and domain for functions $f/g$and $g/f$ .

The required functions are: $\left(f/g\right)x=\frac{x^2}{\sqrt{1-x^2}}$ and $\left(g/f\right)x=\frac{\sqrt{1-x^2}}{x^2}$ . The domain of the function $\left(f/g\right)x$ is $\left(-1,1\right)$ and that of $\left(g/f\right)x$ is the interval $\left(-1,1\right)-\left\{0\right\}$ .

Given data: The functions $f\left(x\right)$ and $g\left(x\right)$ are $f\left(x\right)=x^2$ and $g\left(x\right)=\sqrt{1-x^2}$ .

Method/Formula used:

The division of the two functions $f\left(x\right)$ and $g\left(x\right)$ is also a function given by $f\left(x\right)/g\left(x\right)=\left(f/g\right)x$ , and $g\left(x\right)/f\left(x\right)=\left(g/f\right)x$ . The domain of resulting function $\left(f/g\right)x$ and $\left(g/f\right)x$ are the values of values of x that define $\left(f/g\right)x$ and $\left(g/f\right)x$ .

Calculation: The given functions are $f\left(x\right)=x^2$ ; $g\left(x\right)=\sqrt{1-x^2}$ . Therefore, using the definition of $f\left(x\right)/g\left(x\right)$ and $g\left(x\right)/f\left(x\right)$ are:

  $\begin{array}{c}\left(f/g\right)x=\frac{x^2}{\sqrt{1-x^2}}\\ \left(g/f\right)x=\frac{\sqrt{1-x^2}}{x^2}\end{array}$

The graphs of functions $\left(f/g\right)x$ and $\left(g/f\right)x$ are shown in Fig 1(a) and Fig 1(b) respectively.

  

  

The function $\left(f/g\right)x=\frac{x^2}{\sqrt{1-x^2}}$ does not exists when $1-x^2\le 0$ or $x^2\ge 1$ , i.e., when $x\ge \pm 1$ .

The function $\left(g/f\right)x=\frac{\sqrt{1-x^2}}{x^2}$ does not exists when $1-x^2\le 0$ or $x^2\ge 1$ , i.e., when $x\ge \pm 1$ and when $x=0$ .

Therefore, the domain of the function $\left(f/g\right)x$ is the interval $\left(-1,1\right)$ , and the domain of the function $\left(g/f\right)x=\frac{\sqrt{1-x^2}}{x^2}$ is $\left(-1,1\right)-\left\{0\right\}$ .