Chapter 2, Problem 20RE

a.

To determine

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

Answer to Problem 20RE

The value of $\left(f+g\right)\left(x\right)$ is $x^2+4x+1$ .

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ .

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 $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ .

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

Substitute $3x$ for $f\left(x\right)$ and $1+x+x^2$ for $g\left(x\right)$ in the sum formula.

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

Hence, the value of $\left(f+g\right)\left(x\right)$ is $x^2+4x+1$ .

As domain of $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ is set of real numbers.

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

b.

To determine

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

Answer to Problem 20RE

The value of $\left(f-g\right)\left(x\right)$ is $-x^2+2x-1$ .

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ .

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 $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ .

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

Substitute $3x$ for $f\left(x\right)$ and $1+x+x^2$ for $g\left(x\right)$ in the difference formula.

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

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

As domain of $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ is set of real numbers.

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

c.

To determine

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

Answer to Problem 20RE

The value of $\left(fg\right)\left(x\right)$ is $3x^3+3x^2+3x$ .

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ .

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 $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ .

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

Substitute $3x$ for $f\left(x\right)$ and $1+x+x^2$ for $g\left(x\right)$ in the product formula.

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

Hence, the value of $\left(fg\right)\left(x\right)$ is $3x^3+3x^2+3x$ .

As domain of $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ is set of real numbers.

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

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 20RE

The value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{3x}{x^2+x+1}$ and the domain is all the real numbers $x$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ .

Formula Used:

Let f and g be two functions.

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

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 $f\left(x\right)=3x$ and $g\left(x\right)=1+x+x^2$ .

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

Substitute $3x$ for $f\left(x\right)$ and $1+x+x^2$ for $g\left(x\right)$ in the division formula.

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

Find the domain of the function.

  $x^2+x+1=0$

But for all real number x , $x^2+x+1\ne 0$

The domain is all the real numbers $x$ .

Hence, the value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{3x}{x^2+x+1}$ and the domain is all the real numbers $x$ .