Chapter 6.1, Problem 58PPS

(a).

To determine

Copy and complete the table.

Explanation of Solution

Given:

You will explore $\left(f\cdot g\right)\text{ and }\left(\frac{f}{g}\right)$ if $f(x)=x^2+1\text{ and }g(x)=x-3$ .

Calculation:

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

x	$\left(f\cdot g\right)(x)=\left(x^2+1\right)\left(x-3\right)$	$\left(\frac{f}{g}\right)(x)=\frac{x^2+1}{x-3}$
1	-4	-1
2	-5	-5
3	0	$\frac{10}{0}=+\infty$
4	17	17
5	52	13

(b).

To determine

Graph $\left(f\cdot g\right)\text{ and }\left(\frac{f}{g}\right)$ on the same coordinate grid.

Explanation of Solution

Given:

You will explore $\left(f\cdot g\right)\text{ and }\left(\frac{f}{g}\right)$ if $f(x)=x^2+1\text{ and }g(x)=x-3$ .

Calculation:

From part (a), we have:

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

x	$\left(f\cdot g\right)(x)=\left(x^2+1\right)\left(x-3\right)$	$\left(\frac{f}{g}\right)(x)=\frac{x^2+1}{x-3}$
1	-4	-1
2	-5	-5
3	0	$\frac{10}{0}=+\infty$
4	17	17
5	52	13

Plot the points on the coordinate plane :

  

(c).

To determine

Explain the relationship between $\left(f\cdot g\right)\text{ and }\left(\frac{f}{g}\right)$ .

Explanation of Solution

Given:

You will explore $\left(f\cdot g\right)\text{ and }\left(\frac{f}{g}\right)$ if $f(x)=x^2+1\text{ and }g(x)=x-3$ .

Calculation:

From part (b), we have:

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

The graphs intersect at the points (4,17) and (2,-5). They do not have the same domain , since $\left(f\cdot g\right)(x)=\left(x^2+1\right)\left(x-3\right)=x^3-2x^2-3$ is cubic equation , its domain is whole real number ; $ℝ$ and $\left(\frac{f}{g}\right)(x)=\frac{x^2+1}{x-3}$ has domain real numbers except 3; $ℝ\backslash \left\{3\right\}$ s, since $g(x)=x-3$ is 0 at x=3.