Chapter 1.5, Problem 18E

(a)

To determine

To calculate: To find $\left(f+g\right)\left(x\right)$

Answer to Problem 18E

  $\left(f+g\right)\left(x\right)=\frac{x^4+x+1}{x^4+x^3}$

Explanation of Solution

Given: $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x^3}$

Formula Used: Addition of two functions $f$ and $g$ is defined as

  $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$

Calculation:

Functions are given as

  $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x^3}$

  $\left(f+g\right)\left(x\right)$ is calculated as follows:

  $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$

Substituting the values:

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

Conclusion:

Hence, $\left(f+g\right)\left(x\right)=\frac{x^4+x+1}{x^4+x^3}$

(b)

To determine

To calculate: To find $\left(f-g\right)\left(x\right)$

Answer to Problem 18E

  $\left(f-g\right)\left(x\right)=\frac{x^4-x-1}{x^4+x^3}$

Explanation of Solution

Given: $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x^3}$

Formula Used: Subtraction of two functions $f$ and $g$ is defined as

  $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$

Calculation:

Functions are given as

  $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x^3}$

  $\left(f-g\right)\left(x\right)$ is calculated as follows:

  $\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)$

Substituting the values:

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

Conclusion:

Hence, $\left(f-g\right)\left(x\right)=\frac{x^4-x-1}{x^4+x^3}$

(c)

To determine

To calculate: To find $\left(fg\right)\left(x\right)$

Answer to Problem 18E

  $\left(fg\right)\left(x\right)=\frac{1}{x^3+x^2}$

Explanation of Solution

Given: $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x^3}$

Formula Used: Product of two functions $f$ and $g$ is defined as

  $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$

Calculation:

Functions are given as

  $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x^3}$

  $\left(fg\right)\left(x\right)$ is calculated as follows:

  $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)$

Substituting the values:

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

Conclusion:

Hence, $\left(fg\right)\left(x\right)=\frac{1}{x^3+x^2}$

(d)

To determine

To calculate: To find $\left(f/g\right)\left(x\right)$

Answer to Problem 18E

  $\left(f/g\right)\left(x\right)=\frac{x^4}{x+1}$

Explanation of Solution

Given: $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x^3}$

Formula Used: Quotient of two functions $f$ and $g$ is defined as

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

Calculation:

Functions are given as

  $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x^3}$

  $\left(f/g\right)\left(x\right)$ is calculated as follows:

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

Substituting the values:

  $\begin{array}{l}\left(f/g\right)\left(x\right)=\frac{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$x+1$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x^3$}\right.}\\ \left(f/g\right)\left(x\right)=\frac{x\times x^3}{x+1}\\ \left(f/g\right)\left(x\right)=\frac{x^4}{x+1}\end{array}$

To calculate domain of $\left(f/g\right)\left(x\right)$ :

Thus, domain of intersection is set of all real numbers except $0$ and $1$

Conclusion:

Hence, $\left(f/g\right)\left(x\right)=\frac{x^4}{x+1}$ and domain is set of all real numbers except $0$ and $1$