Chapter 2.1, Problem 67AYU

a.

To determine

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

Answer to Problem 67AYU

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

Domain of $\left(f+g\right)\left(x\right)$ is set of all real numbers greater than equal to zero.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

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 $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

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

Substitute $\sqrt{x}$ for $f\left(x\right)$ and $3x-5$ for $g\left(x\right)$ in the sum formula.

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\sqrt{x}+3x-5\end{array}$

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

As domain of $f\left(x\right)=\sqrt{x}$ is set of all real numbers greater than equal to zero and domain of $g\left(x\right)=3x-5$ is set of all real numbers.

So, domain of $f+g$ is set of all real numbers greater than equal to zero.

b.

To determine

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

Answer to Problem 67AYU

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

Domain of $\left(f-g\right)\left(x\right)$ is set of all real numbers greater than equal to zero.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

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 $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

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

Substitute $\sqrt{x}$ for $f\left(x\right)$ and $3x-5$ for $g\left(x\right)$ in the difference formula.

  $\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =\sqrt{x}-3x+5\end{array}$

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

As domain of $f\left(x\right)=\sqrt{x}$ is set of all real numbers greater than equal to zero and domain of $g\left(x\right)=3x-5$ is set of all real numbers.

So, domain of $f-g$ is set of all real numbers greater than equal to zero.

c.

To determine

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

Answer to Problem 67AYU

The value of $\left(fg\right)\left(x\right)$ is $\sqrt{x}\left(3x-5\right)$ .

Domain of $\left(fg\right)\left(x\right)$ is set of all real numbers greater than equal to zero.

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

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 $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

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

Substitute $\sqrt{x}$ for $f\left(x\right)$ and $3x-5$ for $g\left(x\right)$ in the product formula.

  $\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\sqrt{x}\left(3x-5\right)\end{array}$

Hence, the value of $\left(fg\right)\left(x\right)$ is $\sqrt{x}\left(3x-5\right)$ .

As domain of $f\left(x\right)=\sqrt{x}$ is set of all real numbers greater than equal to zero and domain of $g\left(x\right)=3x-5$ is set of all real numbers.

So, domain of $\left(fg\right)\left(x\right)$ is set of all real numbers greater than equal to 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 67AYU

The value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{\sqrt{x}}{3x-5}$ and the domain is all the real numbers $x\ge 0$ except $x=\frac{5}{3}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

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 $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

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

Substitute $\sqrt{x}$ for $f\left(x\right)$ and $3x-5$ for $g\left(x\right)$ in the division formula.

  $\begin{array}{c}\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{\sqrt{x}}{3x-5}\end{array}$

Find the domain of the function.

  $\begin{array}{c}x\le 0\\ 3x-5=0\\ x=\frac{5}{3}\end{array}$

The domain is all the real numbers $x\ge 0$ except $x=\frac{5}{3}$ because the denominator becomes zero at $x=\frac{5}{3}$ .

Hence, the value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{\sqrt{x}}{3x-5}$ and the domain is all the real numbers $x\ge 0$ except $x=\frac{5}{3}$ .

e)

To determine

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

Answer to Problem 67AYU

The value of $\left(f+g\right)\left(3\right)$ is $\sqrt{3}+4$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

Calculation:

Now, from part (a)

  $\left(f+g\right)\left(x\right)=\sqrt{x}+3x-5$

Substitute $3$ for $x$ in the composite function $\left(f+g\right)\left(x\right)=\sqrt{x}+3x-5$ .

  $\begin{array}{c}\left(f+g\right)\left(x\right)=\sqrt{x}+3x-5\\ \left(f+g\right)\left(3\right)=\sqrt{3}+3\left(3\right)-5\\ =\sqrt{3}+9-5\\ =\sqrt{3}+4\end{array}$

Hence, the value of $\left(f+g\right)\left(3\right)$ is $\sqrt{3}+4$ .

f)

To determine

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

Answer to Problem 67AYU

The value of $\left(f-g\right)\left(4\right)$ is $-5$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

Calculation:

Now, from part (b)

  $\left(f-g\right)\left(x\right)=\sqrt{x}-3x+5$

Substitute $4$ for $x$ in the composite function $\left(f-g\right)\left(x\right)=\sqrt{x}-3x+5$ .

  $\begin{array}{c}\left(f-g\right)\left(x\right)=\sqrt{x}-3x+5\\ \left(f-g\right)\left(4\right)=\sqrt{4}-3\left(4\right)+5\\ =2-12+5\\ =-5\end{array}$

Hence, the value of $\left(f-g\right)\left(4\right)$ is $-5$ .

g)

To determine

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

Answer to Problem 67AYU

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

Calculation:

Now, from part (c)

  $\left(fg\right)\left(x\right)=\sqrt{x}\left(3x-5\right)$

Substitute $2$ for $x$ in the composite function $\left(fg\right)\left(x\right)=\sqrt{x}\left(3x-5\right)$ .

  $\begin{array}{c}\left(fg\right)\left(x\right)=\sqrt{x}\left(3x-5\right)\\ \left(fg\right)\left(2\right)=\sqrt{2}\left(3\cdot 2-5\right)\\ =\sqrt{2}\end{array}$

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

h)

To determine

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

Answer to Problem 67AYU

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=3x-5$ .

Calculation:

Now, from part (d)

  $\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x}}{3x-5}$

Substitute $1$ for $x$ in the composite function $\left(\frac{f}{g}\right)\left(x\right)=\frac{\sqrt{x}}{3x-5}$ .

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

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