Chapter 1.4, Problem 4E

Solution

To find: The formula and domain for functions $f+g$ $f-g$and $fg$ .

The required functions are: $\left(f+g\right)=\sqrt{x}+\left|x+3\right|$ , $\left(f-g\right)=\sqrt{x}-\left|x+3\right|$ and $f.g=\sqrt{x}\cdot \left|x+3\right|$ . The domain of each function $\left(f+g\right)$ , $\left(f-g\right)$ and $fg$ is the set of real numbers $x\in \left(0,\infty \right)$ .

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

Method/Formula used:

The sum, difference and product of the two functions $f\left(x\right)$ and $g\left(x\right)$ is also a function given by $f\left(x\right)+g\left(x\right)=\left(f+g\right)x$ , $f\left(x\right)-g\left(x\right)=\left(f-g\right)x$ and $f\left(x\right)\cdot g\left(x\right)=\left(f\cdot g\right)\left(x\right)$ . The domain of resulting functions $\left(f+g\right)x$ , $\left(f-g\right)x$ and $f\left(x\right)\cdot g\left(x\right)$ are the values of x that define these functions $\left(f+g\right)x$ $\left(f-g\right)x$ and $\left(f\cdot g\right)\left(x\right)$

Calculation:

The given functions are $f\left(x\right)=\sqrt{x}$ ; $g\left(x\right)=\left|x+3\right|$ . Therefore, using the definition of $\left(f+g\right)\left(x\right)$ , $\left(f-g\right)\left(x\right)$ and $fg\left(x\right)$ are:

  $\left(f+g\right)\left(x\right)$ :

  $\begin{array}{c}\left(f+g\right)x=f\left(x\right)+g\left(x\right)\\ =\sqrt{x}+\left|x+3\right|\end{array}$

The function $f+g=\sqrt{x}+\left|x+3\right|$ is not defined when $x<0$ .Therefore, the domain of $\left(f+g\right)x$ is $\left[0,\infty \right)$ .

  $\left(f-g\right)\left(x\right)$ :

  $\begin{array}{c}\left(f-g\right)x=f\left(x\right)-g\left(x\right)\\ ==\sqrt{x}-\left|x+3\right|\end{array}$

The function $f-g=\sqrt{x}-\left|x+3\right|$ is not defined when $x<0$ .Therefore, the domain of $\left(f+g\right)x$ is $\left[0,\infty \right)$ .

  $\left(fg\right)\left(x\right)$ :

  $\left(fg\right)x=\sqrt{x}\left|x+3\right|$

The function $f.g=\sqrt{x}\cdot \left|x+3\right|$ is not defined when $x<0$ .Therefore, the domain of $\left(fg\right)x$ is $\left[0,\infty \right)$ .

The graphs of functions $\sqrt{x}+\left|x+3\right|$ , $\sqrt{x}-\left|x+3\right|$ , and $\sqrt{x}\cdot \left|x+3\right|$ are shown in Fig. (1), here that also shows that the functions $\sqrt{x}+\left|x+3\right|$ , $\sqrt{x}-\left|x+3\right|$ , and $\sqrt{x}\cdot \left|x+3\right|$ defined for all $x\ge 0$ , asserting that the domain of each of these function is $\left[0,\infty \right)$ .