Q. 68 - Precalculus Enhanced with Graphing Utilities

Question

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

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

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

Step-by-Step Solution

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 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 both f and g. 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 f g ( x ) We know that f g ( x ) = f ( x ) g ( x )

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

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

Part (d) Step 2. The domain of f g 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 f g ( 1 ) From the part (a) we know the value of f g ( x ) = x x

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

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