Chapter 2.1, Problem 69AYU

a.

To determine

The value of $\left(f+g\right)\left(x\right)$ and the domain.

Answer to Problem 69AYU

The value of $\left(f+g\right)\left(x\right)$ is $1+\frac{2}{x}$ .

Domain of $\left(f+g\right)\left(x\right)$ is set of all real numbers except zero.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Formula Used:

Let f and g be two functions.

Use the sum formula, $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$

For a composite function $f+g$ , the domain of the function is defined by

  $\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)$

If domain is not specified for a function, one should consider the set of largest real number where the function is defined all over.

Calculation:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Use the sum formula, $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ .

Substitute $1+\frac{1}{x}$ for $f\left(x\right)$ and $\frac{1}{x}$ for $g\left(x\right)$ in the sum formula.

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

Hence, the value of $\left(f+g\right)\left(x\right)$ is $1+\frac{2}{x}$ .

As domain of $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ is set of all real numbers except zero.

So, domain of $f+g$ is set of all real numbers except zero.

b.

To determine

The value of $\left(f-g\right)\left(x\right)$ and its domain.

Answer to Problem 69AYU

The value of $\left(f-g\right)\left(x\right)$ is $1$ .

Domain of $\left(f-g\right)\left(x\right)$ is set of all real numbers except zero.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Formula Used:

Let f and g be two functions.

Use the difference formula, $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$ .

For a composite function $f-g$ , the domain of the function is defined by

  $\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)$

If domain is not specified for a function, one should consider the set of largest real number where the function is defined all over.

Calculation:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Use the difference formula, $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$ .

Substitute $1+\frac{1}{x}$ for $f\left(x\right)$ and $\frac{1}{x}$ for $g\left(x\right)$ in the difference formula.

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

Hence, the value of $\left(f-g\right)\left(x\right)$ is $1$ .

As domain of $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ is set of all real numbers except zero.

So, domain of $f-g$ is set of all real numbers except zero.

c.

To determine

The value of $\left(fg\right)\left(x\right)$ and its domain.

Answer to Problem 69AYU

The value of $\left(fg\right)\left(x\right)$ is $\frac{1}{x}+\frac{1}{x^2}$ .

Domain of $\left(fg\right)\left(x\right)$ is set of all real numbers except zero.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Formula Used:

Let f and g be two functions.

Use the product formula, $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$ .

For a composite function $fg$ , the domain of the function is defined by

  $\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)$

If domain is not specified for a function, one should consider the set of largest real number where the function is defined all over.

Calculation:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Use the product formula, $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$ .

Substitute $1+\frac{1}{x}$ for $f\left(x\right)$ and $\frac{1}{x}$ for $g\left(x\right)$ in the product formula.

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\left(1+\frac{1}{x}\right)\frac{1}{x}\\ =\frac{1}{x}+\frac{1}{x^2}\end{array}$

Hence, the value of $\left(fg\right)\left(x\right)$ is $\frac{1}{x}+\frac{1}{x^2}$ .

As domain of $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ is set of all real numbers except zero.

So, domain of $\left(fg\right)\left(x\right)$ is set of all real numbers except zero.

d.

To determine

The value of $\left(\frac{f}{g}\right)\left(x\right)$ . Also, find the domain of the function $\left(\frac{f}{g}\right)$ .

Answer to Problem 69AYU

The value of $\left(\frac{f}{g}\right)\left(x\right)$ is $x+1$ and the domain is all the real numbers.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Formula Used:

Let f and g be two functions.

Use the division formula, $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ .

For a composite function $f-g$ , the domain of the function is defined by

  $\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)$

If domain is not specified for a function, one should consider the set of largest real number where the function is defined all over.

Calculation:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Use the division formula, $\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}$ .

Substitute $1+\frac{1}{x}$ for $f\left(x\right)$ and $\frac{1}{x}$ for $g\left(x\right)$ in the division formula.

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

The domain is all the real numbers

Hence, the value of $\left(\frac{f}{g}\right)\left(x\right)$ is $x+1$ and the domain is all the real numbers.

e)

To determine

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

Answer to Problem 69AYU

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Calculation:

Now, from part (a)

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

Substitute $3$ for $x$ in the composite function $\left(f+g\right)\left(x\right)=1+\frac{2}{x}$ .

  $\begin{array}{c}\left(f+g\right)\left(x\right)=1+\frac{2}{x}\\ \left(f+g\right)\left(3\right)=1+\frac{2}{3}\\ =\frac{5}{3}\end{array}$

Hence, the value of $\left(f+g\right)\left(3\right)$ is $\frac{5}{3}$ .

f)

To determine

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

Answer to Problem 69AYU

The value of $\left(f-g\right)\left(4\right)$ is $1$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Calculation:

Now, from part (b)

  $\left(f-g\right)\left(x\right)=1$

Substitute $4$ for $x$ in the composite function $\left(f-g\right)\left(x\right)=1$ .

  $\begin{array}{c}\left(f-g\right)\left(x\right)=1\\ \left(f-g\right)\left(4\right)=1\end{array}$

Hence, the value of $\left(f-g\right)\left(4\right)$ is $1$ .

g)

To determine

The value of $\left(fg\right)\left(2\right)$ .

Answer to Problem 69AYU

The value of $\left(fg\right)\left(2\right)$ is $\frac{3}{4}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Calculation:

Now, from part (c)

  $\left(fg\right)\left(x\right)=\frac{1}{x}+\frac{1}{x^2}$

Substitute $2$ for $x$ in the composite function $\left(fg\right)\left(x\right)=\frac{1}{x}+\frac{1}{x^2}$ .

  $\begin{array}{c}\left(fg\right)\left(x\right)=\frac{1}{x}+\frac{1}{x^2}\\ \left(fg\right)\left(2\right)=\frac{1}{2}+\frac{1}{4}\\ =\frac{3}{4}\end{array}$

Hence, the value of $\left(fg\right)\left(2\right)$ is $\frac{3}{4}$ .

h)

To determine

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

Answer to Problem 69AYU

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=1+\frac{1}{x}$ and $g\left(x\right)=\frac{1}{x}$ .

Calculation:

Now, from part (d)

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

Substitute $1$ for $x$ in the composite function $\left(\frac{f}{g}\right)\left(x\right)=x+1$ .

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

Hence, the value of $\left(\frac{f}{g}\right)\left(1\right)$ is $2$ .