Question

In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain.

$$\begin{gathered} \left(a\right)(f+g)\left(x\right)\left(b\right)(f-g)\left(x\right)\left(c\right)(f·g)\left(x\right)\left(d\right)\left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right)(f+g)\left(3\right)\left(f\right)(f-g)\left(4\right)\left(g\right)(f·g)\left(2\right)\left(h\right)\left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

$f\left(x\right)=\left|x\right|;g\left(x\right)=x$

Short answer

The value of $(f+g)\left(x\right)=\left|x\right|+x$and the domain of $f+g$is set of all real numbers.

The value of $(f-g)\left(x\right)=\left|x\right|-x$and the domain of $f-g$is set of all real numbers.

The value of $(f·g)\left(x\right)=x·\left|x\right|$and the domain of $f·g$is set of all real numbers.

The value of $\left(\frac{f}{g}\right)\left(x\right)=\frac{\left|x\right|}{x}$and the domain of $\frac{f}{g}$is $\left\{x\right|x\ne 0\}$.

The value of$(f+g)\left(3\right)=6$

The value of$(f-g)\left(4\right)=0$

The value of$(f·g)\left(2\right)=4$

The value of$\left(\frac{f}{g}\right)\left(1\right)=1$

Step-by-step solution

Step 1. Given Information 

In the given problems we have to solve the given functions f and g, find the following. For parts (a)–(d), we have to also find the domain.

$$\begin{gathered} \left(a\right)(f+g)\left(x\right)\left(b\right)(f-g)\left(x\right)\left(c\right)(f·g)\left(x\right)\left(d\right)\left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right)(f+g)\left(3\right)\left(f\right)(f-g)\left(4\right)\left(g\right)(f·g)\left(2\right)\left(h\right)\left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

The given function is$f\left(x\right)=\left|x\right|;g\left(x\right)=x$

Step 2. The function f  tells us module of a number. The function g tells us a number.

Since these operations can be performed on any real number, we conclude that the domain of $f\mathrm{and}g$is the set of all real numbers.

Part (a) Step 1. We have to find the value of (f+g)(x)

We know that$(f+g)\left(x\right)=f\left(x\right)+g\left(x\right)$

Part (a) Step 2. Putting the value of f(x) and g(x)

$(f+g)\left(x\right)=\left|x\right|+x$

The domain of $f+g$consists of those numbers x that are in the domains of bothfandg. Therefore, the domain of $f+g$is all set of real number.

Part (b) Step 1. We have to find the value of (f-g)(x)We know (f-g)(x)=f(x)-g(x)

Putting the value of $f\left(x\right)\mathrm{and}g\left(x\right)$

$(f-g)\left(x\right)=\left|x\right|-x$

The domain of $f-g$ consists of those numbers x that are in the domains of both . Therefore, the domain of $f-g$is all set of real number.

Part (c) Step 1. We have to find the value of (f·g)(x)We know that (f·g)(x)=f(x)·g(x)

Putting the value of $f\left(x\right)\mathrm{and}g\left(x\right)$

$$\begin{gathered} (f·g)\left(x\right)=\left|x\right|·x \\ (f·g)\left(x\right)=x·\left|x\right| \end{gathered}$$

The domain of $f·g$ consists of those numbers x that are in the domains of both $f\mathrm{and}g$. Therefore, the domain of $f·g$is .

Part (d) Step 1. We have to find the value of fg(x)We know that fg(x)=f(x)g(x)

Putting the value of $f\left(x\right)\mathrm{and}g\left(x\right)$

$\left(\frac{f}{g}\right)\left(x\right)=\frac{\left|x\right|}{x}$

Part (d) Step 2. The domain of fg consists of the numbers x for which g(x)≠0 and that are in the domains of both f and g.  

Since $g\left(x\right)\ne 0$when

$x\ne 0$

The domain of $\frac{f}{g}$is$\left\{x\right|x\ne 0\}$

Part (e) Step 1. We have to find the value of (f+g)(3)From the part (a) we know the value of (f+g)(x)=x+x

Putting $x=3$in the value of $(f+g)\left(x\right)$

$$\begin{gathered} (f+g)\left(3\right)=\left|3\right|+3 \\ (f+g)\left(3\right)=3+3 \\ (f+g)\left(3\right)=6 \end{gathered}$$

Part (f) Step 1. We have to find the value of (f-g)(4)From the part (a) we know the value of (f-g)(x)=x-x

Putting $x=4$in the value of $(f-g)\left(x\right)$

$$\begin{gathered} (f-g)\left(4\right)=\left|4\right|-4 \\ (f-g)\left(4\right)=4-4 \\ (f-g)\left(4\right)=0 \end{gathered}$$

Part (g) Step 1. We have to find the value of (f·g)(2)From the part (a) we know the value of (f·g)(x)=x·x

Putting $x=2$in the value of $(f·g)\left(x\right)$

$$\begin{gathered} (f·g)\left(2\right)=2·\left|2\right| \\ (f·g)\left(2\right)=2·2 \\ (f·g)\left(2\right)=4 \end{gathered}$$

Part (h) Step 1. We have to find the value of fg(1)From the part (a) we know the value of fg(x)=xx

Putting $x=1$in the value of $\left(\frac{f}{g}\right)\left(x\right)$

$$\begin{gathered} \left(\frac{f}{g}\right)\left(1\right)=\frac{\left|1\right|}{1} \\ \left(\frac{f}{g}\right)\left(1\right)=\frac{1}{1} \\ \left(\frac{f}{g}\right)\left(1\right)=1 \end{gathered}$$