Chapter 1.5, Problem 11E

(a)

To determine

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

Answer to Problem 11E

  $\left(f+g\right)\left(x\right)=2x$

Explanation of Solution

Given: $f\left(x\right)=x+3$ and $g\left(x\right)=x-3$

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

  $f\left(x\right)=x+3$ and $g\left(x\right)=x-3$

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

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

Substituting the values:

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

Conclusion:

Hence, $\left(f+g\right)\left(x\right)=2x$

(b)

To determine

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

Answer to Problem 11E

  $\left(f-g\right)\left(x\right)=6$

Explanation of Solution

Given: $f\left(x\right)=x+3$ and $g\left(x\right)=x-3$

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

  $f\left(x\right)=x+3$ and $g\left(x\right)=x-3$

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

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

Substituting the values:

  $\begin{array}{l}\left(f-g\right)\left(x\right)=\left(x+3\right)-\left(x-3\right)\\ \left(f-g\right)\left(x\right)=x+3-x+3\\ \left(f-g\right)\left(x\right)=6\end{array}$

Conclusion:

Hence, $\left(f-g\right)\left(x\right)=6$

(c)

To determine

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

Answer to Problem 11E

  $\left(fg\right)\left(x\right)=x^2-9$

Explanation of Solution

Given: $f\left(x\right)=x+3$ and $g\left(x\right)=x-3$

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

  $f\left(x\right)=x+3$ and $g\left(x\right)=x-3$

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

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

Substituting the values:

  $\begin{array}{l}\left(fg\right)\left(x\right)=\left(x+3\right)\left(x-3\right)\\ \left(fg\right)\left(x\right)={\left(x\right)}^2-{\left(3\right)}^2\\ \left(fg\right)\left(x\right)=x^2-9\end{array}$

Conclusion:

Hence, $\left(fg\right)\left(x\right)=x^2-9$

(d)

To determine

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

Answer to Problem 11E

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

Explanation of Solution

Given: $f\left(x\right)=x+3$ and $g\left(x\right)=x-3$

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

  $f\left(x\right)=x+3$ and $g\left(x\right)=x-3$

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

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

Substituting the values:

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

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

Denominator can never be zero

  $\begin{array}{l}x-3\ne 0\\ x\ne 3\end{array}$

Thus, domain is $\left(-\infty ,\infty \right)-\left\{3\right\}$

Conclusion:

Hence, $\left(f/g\right)\left(x\right)=\frac{x+3}{x-3}$ and domain is all set of real numbers except $3$