Chapter 1.4, Problem 6E

Solution

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

The required functions are: $\left(f/g\right)x=\frac{\sqrt{x-2}}{\sqrt{x+4}}$ and $\left(g/f\right)x=\frac{\sqrt{x+4}}{\sqrt{x-2}}$ . The domain of each function $\left(f/g\right)x$ and $\left(g/f\right)x$ is the interval $\left(2,\infty \right)$ .

Given data: The functions $f\left(x\right)$ and $g\left(x\right)$ are $f\left(x\right)=\sqrt{x-2}$ and $g\left(x\right)=\sqrt{x+4}$ .

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 $f\left(x\right)=\sqrt{x-2}$ ; $g\left(x\right)=\sqrt{x+4}$ . Therefore, using the definition of $f\left(x\right)/g\left(x\right)$ and $g\left(x\right)/f\left(x\right)$ are:

  $\begin{array}{c}\left(f/g\right)x=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{\sqrt{x-2}}{\sqrt{x+4}}\\ =\sqrt{x-2}\cdot \frac{1}{\sqrt{x+4}}\left\{\text{write}\text{product}\text{of}\text{two}\text{functions}\right\}\end{array}$

The function $\left(\sqrt{x-2}\right)$ is not defined when $x-2$ is negative, i.e., when $x-2<0$ or $x<2$ . The function $\frac{1}{\sqrt{x+4}}$ is not defined $x+4=0$ or $x=-4$ and $x+4$ is negative, i.e., $x+4<0$ or $x<-4$ . Thus the product of functions $\sqrt{x+3}$ and $\frac{1}{x^2}$ is defined only when $x>2$ or $x\in \left(2,\infty \right)$ .

Therefore, the domain of the function $\left(f/g\right)x$ is the interval $\left(2,\infty \right)$ .

  $\begin{array}{c}\left(g/f\right)x=\frac{\sqrt{x+4}}{\sqrt{x-2}}\\ =\frac{\sqrt{x+4}}{\sqrt{x-2}}\\ =\sqrt{x+4}\cdot \frac{1}{\sqrt{x-2}}\left\{\text{write}\text{product}\text{of}\text{two}\text{functions}\right\}\end{array}$

The function $\sqrt{x+4}$ it defined for all $x+4\ge 0$ or $x\ge -4$ . The function $\frac{1}{\sqrt{x-2}}$ is not defined when $x-2$ is negative, i.e., when $x-2<0$ or $x<2$ and $\sqrt{x-2}$ is zero, i.e., when is $x=2$ . Thus, the function $\frac{1}{\sqrt{x-2}}$ is not defined when $x\le 2$ . Thus the product functions $\sqrt{x+4}$ and $\frac{1}{\sqrt{x-2}}$ is defined only when $x\ge 2\text{or}x\in \left(2,\infty \right)$ .

Therefore, the domain of the function $\left(f/g\right)x$ is the interval $\left(2,\infty \right)$ .