Chapter 1.2, Problem 11E

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 11E

  $\begin{array}{c}f\left(x\right)+g\left(x\right)=x^2-x+9\\ f\left(x\right)-g\left(x\right)=x^2-3x-9\\ f\left(x\right)\cdot g\left(x\right)=x^3+7x^2-18x\\ \frac{f}{g}\left(x\right)=\frac{x^2-2x}{x+9}\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

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

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:

  $\begin{array}{c}f\left(x\right)+g\left(x\right)=\left(x^2-2x\right)+\left(x+9\right)\\ =x^2-x+9\end{array}$

Thus, the value of $f\left(x\right)+g\left(x\right)$ is $x^2-x+9$ .

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:

  $\begin{array}{c}f\left(x\right)-g\left(x\right)=\left(x^2-2x\right)-\left(x+9\right)\\ =x^2-2x-x-9\\ =x^2-3x-9\end{array}$

Thus, the value of $f\left(x\right)-g\left(x\right)$ is $x^2-3x-9$ .

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:

  $\begin{array}{c}f\left(x\right)\cdot g\left(x\right)=\left(x^2-2x\right)\cdot \left(x+9\right)\\ =x^3+9x^2-2x^2-18x\\ =x^3+7x^2-18x\end{array}$

Thus, the value of $f\left(x\right)\cdot g\left(x\right)$ is $x^3+7x^2-18x$ .

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:

  $\frac{f}{g}\left(x\right)=\frac{\left(x^2-2x\right)}{\left(x+9\right)}$

Thus, the value of $\frac{f}{g}\left(x\right)$ is $\frac{x^2-2x}{x+9}$ .