Chapter 2.1, Problem 72AYU

a.

To determine

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

Answer to Problem 72AYU

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

Domain of $\left(f+g\right)\left(x\right)$ is $\left\{x|x\ge -1,x\ne 0\right\}$

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\frac{2}{x}$ .

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

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

Substitute $\sqrt{x+1}$ for $f\left(x\right)$ and $\frac{2}{x}$ for $g\left(x\right)$ in the sum formula.

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

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

Now, check the domain of $f\left(x\right)=\sqrt{x+1}$

  $\begin{array}{c}x+1\ge 0\\ x\ge -1\end{array}$

Domain of $f\left(x\right)=\sqrt{x+1}$ is set of all real numbers greater than equal to $-1$ .

Now, check the domain of $g\left(x\right)=\frac{2}{x}$

  $x\ne 0$

Domain of $g\left(x\right)=\frac{2}{x}$ is set of all real numbers except 0.

So, domain of $f+g$ can be found by

  $\begin{array}{c}\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)=\left\{x|x\ge -1\right\}\cap \left\{x|x\ne 0\right\}\\ =\left\{x|x\ge -1,x\ne 0\right\}\end{array}$

So, domain of $f+g$ is $\left\{x|x\ge -1,x\ne 0\right\}$ .

b.

To determine

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

Answer to Problem 72AYU

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

Domain of $\left(f-g\right)\left(x\right)$ is $\left\{x|x\ge -1,x\ne 0\right\}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\frac{2}{x}$ .

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

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

Substitute $\sqrt{x+1}$ for $f\left(x\right)$ and $\frac{2}{x}$ for $g\left(x\right)$ in the difference formula.

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

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

Now, check the domain of $f\left(x\right)=\sqrt{x+1}$

  $\begin{array}{c}x+1\ge 0\\ x\ge -1\end{array}$

Domain of $f\left(x\right)=\sqrt{x+1}$ is set of all real numbers greater than equal to $-1$ .

Now, check the domain of $g\left(x\right)=\frac{2}{x}$

  $x\ne 0$

Domain of $g\left(x\right)=\frac{2}{x}$ is set of all real numbers except 0.

So, domain of $f-g$ can be found by

  $\begin{array}{c}\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)=\left\{x|x\ge -1\right\}\cap \left\{x|x\ne 0\right\}\\ =\left\{x|x\ge -1,x\ne 0\right\}\end{array}$

So, domain of $\left(f-g\right)\left(x\right)$ is $\left\{x|x\ge -1,x\ne 0\right\}$ .

c.

To determine

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

Answer to Problem 72AYU

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

Domain of $\left(fg\right)\left(x\right)$ is $\left\{x|x\ge -1,x\ne 0\right\}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\frac{2}{x}$ .

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

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

Substitute $\sqrt{x+1}$ for $f\left(x\right)$ and $\frac{2}{x}$ for $g\left(x\right)$ in the product formula.

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

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

Now, check the domain of $f\left(x\right)=\sqrt{x+1}$

  $\begin{array}{c}x+1\ge 0\\ x\ge -1\end{array}$

Domain of $f\left(x\right)=\sqrt{x+1}$ is set of all real numbers greater than equal to $-1$ .

Now, check the domain of $g\left(x\right)=\frac{2}{x}$

  $x\ne 0$

Domain of $g\left(x\right)=\frac{2}{x}$ is set of all real numbers except 0.

So, domain of $fg$ can be found by

  $\begin{array}{c}\left(\text{Domain}\text{of}f\right)\cap \left(\text{Domain}\text{of}g\right)=\left\{x|x\ge -1\right\}\cap \left\{x|x\ne 0\right\}\\ =\left\{x|x\ge -1,x\ne 0\right\}\end{array}$

So, domain of $\left(f-g\right)\left(x\right)$ is $\left\{x|x\ge -1,x\ne 0\right\}$ .

So, domain of $\left(fg\right)\left(x\right)$ is $\left\{x|x\ge -1,x\ne 0\right\}$ .

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 72AYU

The value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{x\sqrt{x+1}}{2}$ and the domain is $\left\{x|x\ge -1\right\}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\frac{2}{x}$ .

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

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

Substitute $\sqrt{x+1}$ for $f\left(x\right)$ and $\frac{2}{x}$ for $g\left(x\right)$ in the division formula.

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

Now, check the domain of $f\left(x\right)=\sqrt{x+1}$

  $\begin{array}{c}x+1\ge 0\\ x\ge -1\end{array}$

Domain of $f\left(x\right)=\sqrt{x+1}$ is set of all real numbers greater than equal to $-1$ .

So, domain of $\left(\frac{f}{g}\right)$ is $\left\{x|x\ge -1\right\}$ .

e)

To determine

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

Answer to Problem 72AYU

The value of $\left(f+g\right)\left(3\right)$ is $\frac{8}{3}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\frac{2}{x}$ .

Calculation:

Now, from part (a)

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

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

  $\begin{array}{c}\left(f+g\right)\left(x\right)=\sqrt{x+1}+\frac{2}{x}\\ \left(f+g\right)\left(3\right)=\sqrt{3+1}+\frac{2}{3}\\ =\sqrt{4}+\frac{2}{3}\\ =2+\frac{2}{3}\\ =\frac{8}{3}\end{array}$

Hence, the value of $\left(f+g\right)\left(3\right)$ is $\frac{8}{3}$ .

f)

To determine

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

Answer to Problem 72AYU

The value of $\left(f-g\right)\left(4\right)$ is $\sqrt{5}-\frac{1}{2}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\frac{2}{x}$ .

Calculation:

Now, from part (b)

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

Substitute $4$ for $x$ in the composite function $\left(f-g\right)\left(x\right)=\sqrt{x+1}-\frac{2}{x}$ .

  $\begin{array}{c}\left(f-g\right)\left(x\right)=\sqrt{x+1}-\frac{2}{x}\\ \left(f-g\right)\left(4\right)=\sqrt{4+1}-\frac{2}{4}\\ =\sqrt{5}-\frac{1}{2}\end{array}$

Hence, the value of $\left(f-g\right)\left(4\right)$ is $\sqrt{5}-\frac{1}{2}$ .

g)

To determine

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

Answer to Problem 72AYU

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\frac{2}{x}$ .

Calculation:

Now, from part (c)

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

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

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

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

h)

To determine

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

Answer to Problem 72AYU

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x+1}$ and $g\left(x\right)=\frac{2}{x}$ .

Calculation:

Now, from part (d)

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

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

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

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