Q. 69 - 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. 

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

$f\left(x\right)=1+\frac{1}{x}; g\left(x\right)=\frac{1}{x}$

Step-by-Step Solution

Answer

The value of $(f+g)\left(x\right)=\frac{x+2}{x}$ and the domain of $f+g$ is $\left\{x\right|x\ne 0\}$

The value of $(f-g)\left(x\right)=1$ and the domain of $f-g$ is $\left\{x\right|x\ne 0\}$

The value of $(f·g)\left(x\right)=\frac{x+1}{x^2}$ and the domain of $f·g$ is $\left\{x\right|x\ne 0\}$.

The value of $\left(\frac{f}{g}\right)\left(x\right)=x+1$ and the domain of $\frac{f}{g}$ is $\left\{x\right|x\ne 0\}$.

The value of $(f+g)\left(3\right)=\frac{5}{3}$

The value of $(f-g)\left(4\right)=1$

The value of $(f·g)\left(2\right)=\frac{3}{4}$

The value of $\left(\frac{f}{g}\right)\left(1\right)=2$

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 $f\left(x\right)=1+\frac{1}{x}; g\left(x\right)=\frac{1}{x}$

Step 2. The function f tells us one divided by a number then add one. The function g tells us one divided by a number.

This requires that:

$x\ne 0$

So the domain of both $f \mathrm{and} g$are $\left\{x\right|x\ne 0\}$

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 )

$$\begin{gathered} (f+g)\left(x\right)=1+\frac{1}{x}+\frac{1}{x} \\ (f+g)\left(x\right)=1·\frac{x}{x}+\frac{1}{x}+\frac{1}{x} \\ (f+g)\left(x\right)=\frac{x+1+1}{x} \\ (f+g)\left(x\right)=\frac{x+2}{x} \end{gathered}$$

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 $\left\{x\right|x\ne 0\}$.  

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 $f \mathrm{and} g$

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

The domain of $f-g$ consists of those numbers x that are in the domains of both . Therefore, the domain of $f-g$ is $\left\{x\right|x\ne 0\}$.

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 $f\left(x\right) \mathrm{and} g\left(x\right)$

$$\begin{gathered} (f·g)\left(x\right)=\left(1+\frac{1}{x}\right)·\frac{1}{x} \\ (f·g)\left(x\right)=1·\frac{1}{x}+\frac{1}{x}·\frac{1}{x} \\ (f·g)\left(x\right)=\frac{1}{x}·\frac{x}{x}+\frac{1}{x^2} \\ (f·g)\left(x\right)=\frac{x}{x^2}+\frac{1}{x^2} \\ (f·g)\left(x\right)=\frac{x+1}{x^2} \end{gathered}$$

Using the commutative property 

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 $\left\{x\right|x\ne 0\}$.

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

Firstly solving $f\left(x\right)=1+\frac{1}{x}$

$$\begin{gathered} f\left(x\right)=1·\frac{x}{x}+\frac{1}{x} \\ f\left(x\right)=\frac{x+1}{x} \end{gathered}$$

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

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

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 

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 + 2 x

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

$$\begin{gathered} (f+g)\left(3\right)=\frac{3+2}{3} \\ (f+g)\left(3\right)=\frac{5}{3} \end{gathered}$$

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 ) = 1

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

$(f-g)\left(4\right)=1$

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 + 1 x 2

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

$$\begin{gathered} (f·g)\left(2\right)=\frac{2+1}{{\left(2\right)}^2} \\ (f·g)\left(2\right)=\frac{3}{4} \end{gathered}$$

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 + 1

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

$$\begin{gathered} \left(\frac{f}{g}\right)\left(1\right)=1+1 \\ \left(\frac{f}{g}\right)\left(1\right)=2 \end{gathered}$$