Chapter 1.4, Problem 8E

Solution

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

The required functions are: $\left(f/g\right)x=\frac{x^3}{\sqrt[3]{1-x^3}}$ and $\left(g/f\right)x=\frac{\sqrt[3]{1-x^3}}{x^3}$ . The domain of the function $\left(f/g\right)x$ is $\left(-\infty ,\infty \right)-\left\{1\right\}$ and that of $\left(g/f\right)x$ is the interval $\left(-\infty ,\infty \right)-\left\{0\right\}$ .

Given data: The functions $f\left(x\right)$ and $g\left(x\right)$ are $f\left(x\right)=x^3$ and $g\left(x\right)=\sqrt[3]{1-x^3}$ .

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 $f\left(x\right)=x^3$ ; $g\left(x\right)=\sqrt[3]{1-x^3}$ . Therefore, using the definition of $f\left(x\right)/g\left(x\right)$ and $g\left(x\right)/f\left(x\right)$ are:

  $\begin{array}{c}\left(f/g\right)x=\frac{x^3}{\sqrt[3]{1-x^3}}\\ \left(g/f\right)x=\frac{\sqrt[3]{1-x^3}}{x^3}\end{array}$

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 $\left(f/g\right)x=\frac{x^3}{\sqrt[3]{1-x^3}}$ does not exists when $1-x^3=0$ or $x=1$ .

The function $\left(g/f\right)x=\frac{\sqrt[3]{1-x^3}}{x^3}$ does not exists when $x^3=0$ or $x=0$ .

Therefore, the domain of the function $\left(f/g\right)x=\frac{x^3}{\sqrt[3]{1-x^3}}$ is the interval $\left(-\infty ,\infty \right)-\left\{1\right\}$ , and the domain of the function $\left(g/f\right)x=\frac{\sqrt{1-x^2}}{x^2}$ is $\left(-\infty ,\infty \right)-\left\{0\right\}$ .