Chapter 1, Problem 20SGA

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 20SGA

  $\begin{array}{c}\left(f+g\right)\left(x\right)=12x-x^3\\ \left(f-g\right)\left(x\right)=-x^2-3x+4\\ \left(f\cdot g\right)\left(x\right)=12x-3x^3\\ \left(\frac{f}{g}\right)\left(x\right)=\frac{4-x^2}{3x}\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

  $\begin{array}{l}f\left(x\right)=4-x^2\\ g\left(x\right)=3x\end{array}$

Calculation:

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

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\left(4-x^2\right)+\left(3x\right)\\ =12x-x^3\end{array}$

Thus, the value of $\left(f+g\right)\left(x\right)$ is $12x-x^3$ .

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

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

Thus, the value of $\left(f-g\right)\left(x\right)$ is $-x^2-3x+4$ .

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

  $\begin{array}{c}\left(f\cdot g\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)\\ =\left(4-x^2\right)\cdot \left(3x\right)\\ =12x-3x^3\end{array}$

Thus, the value of $\left(f\cdot g\right)\left(x\right)$ is $12x-3x^3$ .

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

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

Thus, the value of $\left(\frac{f}{g}\right)\left(x\right)$ is $\frac{4-x^2}{3x}$ .