Chapter 1.2, Problem 13E

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 13E

  $\begin{array}{c}f\left(x\right)+g\left(x\right)=\frac{x^3-2x^2-35x+3}{x-7},x\ne 7\\ f\left(x\right)-g\left(x\right)=\frac{-x^3+2x^2+35x+3}{x-7},x\ne 7\\ f\left(x\right)\cdot g\left(x\right)=\frac{3x^2+15x}{x-7},x\ne 7\\ \frac{f}{g}\left(x\right)=\frac{3}{\left(x-7\right)\left(x^2+5x\right)},x\ne -5,0\text{ or 7}\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

  $\begin{array}{l}f\left(x\right)=\frac{3}{x-7}\\ g\left(x\right)=x^2+5x\end{array}$

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:

  $\begin{array}{c}f\left(x\right)+g\left(x\right)=\left(\frac{3}{x-7}\right)+\left(x^2+5x\right)\\ =\left(\frac{3}{x-7}\right)+\left(x^2+5x\right)\left(\frac{x-7}{x-7}\right)\\ =\frac{3+x^3+5x^2-7x^2-35x}{x-7}\\ =\frac{x^3-2x^2-35x+3}{x-7}\end{array}$

Thus, the value of $f\left(x\right)+g\left(x\right)$ is $\frac{x^3-2x^2-35x+3}{x-7}$ , where $x\ne 7$ .

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:

  $\begin{array}{c}f\left(x\right)-g\left(x\right)=\left(\frac{3}{x-7}\right)-\left(x^2+5x\right)\\ =\left(\frac{3}{x-7}\right)-\left(x^2+5x\right)\left(\frac{x-7}{x-7}\right)\\ =\frac{3-x^3-5x^2+7x^2+35x}{x-7}\\ =\frac{-x^3+2x^2+35x+3}{x-7}\end{array}$

Thus, the value of $f\left(x\right)-g\left(x\right)$ is $\frac{-x^3+2x^2+35x+3}{x-7}$ , where $x\ne 7$ .

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:

  $\begin{array}{c}f\left(x\right)\cdot g\left(x\right)=\left(\frac{3}{x-7}\right)\cdot \left(x^2+5x\right)\\ =\frac{3x^2+15x}{x-7}\end{array}$

Thus, the value of $f\left(x\right)\cdot g\left(x\right)$ is $\frac{3x^2+15x}{x-7}$ , where $x\ne 7$ .

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:

  $\begin{array}{c}\frac{f}{g}\left(x\right)=\frac{\left(\frac{3}{x-7}\right)}{\left(x^2+5x\right)}\\ =\frac{3}{\left(x-7\right)\left(x^2+5x\right)}\end{array}$

The denominator should not be zero, so check the values of x which will not be included. This can be evaluated as follows:

  $\begin{array}{c}x-7=0\\ x=7\end{array}$

Similarly, evaluate another value of x.

  $\begin{array}{c}{x}^2+5x=0\\ x\left(x+5\right)=0\\ x=0,-5\end{array}$

Thus, the value of $\frac{f}{g}\left(x\right)$ is $\frac{3}{\left(x-7\right)\left(x^2+5x\right)}$ , where $x\ne -5,0\text{ or 7}$ .