Chapter 1, Problem 23SGA

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 23SGA

  $\begin{array}{c}\left(f+g\right)\left(x\right)=\frac{x^3-8x^2+16x+4}{x-4}\\ \left(f-g\right)\left(x\right)=\frac{x^3-8x^2+16x-4}{x-4}\\ \left(f\cdot g\right)\left(x\right)=4x\\ \left(\frac{f}{g}\right)\left(x\right)=\frac{x{\left(x-4\right)}^2}{4}\end{array}$

Explanation of Solution

Given information: The functions given are as follows:

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

Calculation:

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

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\left(x^2-4x\right)+\left(\frac{4}{x-4}\right)\\ =\left(x^2-4x\right)\left(\frac{x-4}{x-4}\right)+\left(\frac{4}{x-4}\right)\\ =\left(\frac{x^3-8x^2+16x}{x-4}\right)+\left(\frac{4}{x-4}\right)\\ =\frac{x^3-8x^2+16x+4}{x-4}\end{array}$

Thus, the value of $\left(f+g\right)\left(x\right)$ is $\frac{x^3-8x^2+16x+4}{x-4}$ .

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

  $\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =\left(x^2-4x\right)\left(\frac{x-4}{x-4}\right)-\left(\frac{4}{x-4}\right)\\ =\left(\frac{x^3-8x^2+16x}{x-4}\right)-\left(\frac{4}{x-4}\right)\\ =\frac{x^3-8x^2+16x-4}{x-4}\end{array}$

Thus, the value of $\left(f-g\right)\left(x\right)$ is $\frac{x^3-8x^2+16x-4}{x-4}$ .

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

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

Thus, the value of $\left(f\cdot g\right)\left(x\right)$ is $4x$ .

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

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

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