Chapter 1.2, Problem 14E

To determine

To find: $f\left(x\right)+g\left(x\right),f\left(x\right)-g\left(x\right),f\left(x\right)\cdot g\left(x\right)\text{ and }\frac{f}{g}\left(x\right)$ .

Answer to Problem 14E

  $\begin{array}{c}f\left(x\right)+g\left(x\right)=\frac{x^2-15}{x-5},x\ne 5\\ f\left(x\right)-g\left(x\right)=\frac{x^2-4x-15}{x-5},x\ne 5\\ f\left(x\right)\cdot g\left(x\right)=\frac{2x^2-6x}{x-5},x\ne 5\\ \frac{f}{g}\left(x\right)=\frac{\left(x+3\right)\left(x-5\right)}{2x},x\ne 0\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

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

Calculation:

To find $f\left(x\right)+g\left(x\right)$ , put the values of $f\left(x\right)$ and $g\left(x\right)$ in $f\left(x\right)+g\left(x\right)$ and simplify as follows:

  $\begin{array}{c}f\left(x\right)+g\left(x\right)=\left(x+3\right)+\left(\frac{2x}{x-5}\right)\\ =\left(x+3\right)\left(\frac{x-5}{x-5}\right)+\left(\frac{2x}{x-5}\right)\\ =\left(\frac{x^2+3x-5x-15}{x-5}\right)+\left(\frac{2x}{x-5}\right)\\ \frac{x^2-15}{x-5}\end{array}$

Thus, the value of $f\left(x\right)+g\left(x\right)$ is $\frac{x^2-15}{x-5}$ , where $x\ne 5$ .

Similarly, put the values of $f\left(x\right)$ and $g\left(x\right)$ in $f\left(x\right)-g\left(x\right)$ and simplify as follows:

  $\begin{array}{c}f\left(x\right)-g\left(x\right)=\left(x+3\right)-\left(\frac{2x}{x-5}\right)\\ =\left(x+3\right)\left(\frac{x-5}{x-5}\right)-\left(\frac{2x}{x-5}\right)\\ =\left(\frac{x^2+3x-5x-15}{x-5}\right)-\left(\frac{2x}{x-5}\right)\\ =\frac{x^2-4x-15}{x-5}\end{array}$

Thus, the value of $f\left(x\right)-g\left(x\right)$ is $\frac{x^2-4x-15}{x-5}$ , where $x\ne 5$ .

Similarly, put the values of $f\left(x\right)$ and $g\left(x\right)$ in $f\left(x\right)\cdot g\left(x\right)$ and simplify as follows:

  $\begin{array}{c}f\left(x\right)\cdot g\left(x\right)=\left(x+3\right)\cdot \left(\frac{2x}{x-5}\right)\\ =\frac{2x^2-6x}{x-5}\end{array}$

Thus, the value of $f\left(x\right)\cdot g\left(x\right)$ is $\frac{2x^2-6x}{x-5}$ , where $x\ne 5$ .

Similarly, put the values of $f\left(x\right)$ and $g\left(x\right)$ in $\frac{f}{g}\left(x\right)$ and simplify as follows:

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

Thus, the value of $\frac{f}{g}\left(x\right)$ is $\frac{\left(x+3\right)\left(x-5\right)}{2x}$ , where $x\ne 0$ .