Chapter 1.4, Problem 5E

Solution

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

The required functions are: $\left(f/g\right)x=\frac{\sqrt{x+3}}{x^2}$ and $\left(g/f\right)x=\frac{x^2}{\sqrt{x+3}}$ . The domain of the function $\left(f/g\right)x$ is the interval $\left[-3,0\right)\cup \left(0,\infty \right)$ and that of $\left(g/f\right)x$ , the interval $\left(-3,\infty \right)$ .

Given data: The functions $f\left(x\right)$ and $g\left(x\right)$ are $f\left(x\right)=\sqrt{x+3}$ and $g\left(x\right)=x^2$ .

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 $f\left(x\right)=\sqrt{x+3}$ ; $g\left(x\right)=x^2$ . Therefore, using the definition of $f\left(x\right)/g\left(x\right)$ and $g\left(x\right)/f\left(x\right)$ are:

  $\begin{array}{c}\left(f/g\right)x=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{\sqrt{x+3}}{x^2}\\ =\left(\sqrt{x+3}\right)\cdot \frac{1}{x^2}\left\{\text{write}\text{product}\text{of}\text{two}\text{functions}\right\}\end{array}$

The function $\sqrt{x+3}$ is not defined when $x+3$ is negative, i.e., when $x+3<0$ or $x<-3$ . The function $\frac{1}{x^2}$ is not defined $x=0$ . Thus the product functions $\sqrt{x+3}$ and $\frac{1}{x^2}$ is defined only when $-3\le x<0$ or $x\in \left[-3,0\right)$ , and $x>0$ or $x\in \left(0,\infty \right)$ .

Therefore, the domain of the function $\left(f/g\right)x$ is the interval $\left[-3,0\right)\cup \left(0,\infty \right)$ .

  $\begin{array}{c}\left(g/f\right)x=\frac{g\left(x\right)}{f\left(x\right)}\\ =\frac{x^2}{\sqrt{x+3}}\\ =x^2\cdot \frac{1}{\sqrt{x+3}}\left\{\text{write}\text{product}\text{of}\text{two}\text{functions}\right\}\end{array}$

The function $x^2$ it defined for all $x\in \left(-\infty ,\infty \right)$ . The function $\frac{1}{\sqrt{x+3}}$ is not defined when $x+3$ is negative, i.e., when $x+3<0$ or $x<-3$ and $\sqrt{x+3}$ is zero, i.e., when is $x=-3$ .

Thus the product functions $x^2$ and $\frac{1}{\sqrt{x+3}}$ is defined only when $x\in \left(-3,\infty \right)$ .

Therefore, the domain of the function $\left(f/g\right)x$ is the interval $x\in \left(-3,\infty \right)$ .