Chapter 1, Problem 22SGA

To determine

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

Answer to Problem 22SGA

  $\begin{array}{c}\left(f+g\right)\left(x\right)=x^2+x\\ \left(f-g\right)\left(x\right)=x^2-x-2\\ \left(f\cdot g\right)\left(x\right)=x^3+x^2-x-1\\ \left(\frac{f}{g}\right)\left(x\right)=x-1\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

  $\begin{array}{l}f\left(x\right)=x^2-1\\ g\left(x\right)=x+1\end{array}$

Calculation:

Put the values of $f\left(x\right)$ and $g\left(x\right)$ in the equation below and simplify as follows:

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\left(x^2-1\right)+\left(x+1\right)\\ =x^2+x\end{array}$

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

Put the values of $f\left(x\right)$ and $g\left(x\right)$ in the equation below and simplify as follows:

  $\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =\left(x^2-1\right)-\left(x+1\right)\\ =x^2+x-2\end{array}$

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

Put the values of $f\left(x\right)$ and $g\left(x\right)$ in the equation below and simplify as follows:

  $\begin{array}{c}\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)\\ =\left(x^2-1\right)\cdot \left(x+1\right)\\ =x^3+x^2-x-1\end{array}$

Thus, the value of $\left(f\cdot g\right)\left(x\right)$ is $x^3+x^2-x-1$ .

Put the values of $f\left(x\right)$ and $g\left(x\right)$ in the equation below and simplify as follows:

  $\begin{array}{c}\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{x^2-1}{x+1}\\ =\frac{\left(x+1\right)\left(x-1\right)}{x+1}\\ =x-1\end{array}$

Thus, the value of $\left(\frac{f}{g}\right)\left(x\right)$ is $x-1$ .