Chapter 5.6, Problem 4E

Solution

To sketch: the graph of the function $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ and to check the answer with the help of a graphing calculator.

Given:

The function $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ .

Graph:

Identify the zeroes of the function $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ :

Observe that the zeroes of the function $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ are those pairs $\left(x,0\right)$ for which $h\left(x\right)=0$ .

Substitute $h\left(x\right)=0$ in $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ :

  $0=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ .

Solve the equation $0=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ for $x$ :

  $\begin{array}{c}0=\left(x+2\right)\text{or }0=\left(x-3\right)\text{or }0=\left(x+1\right)\text{or }0=\left(x-1\right)\\ -2=x\text{ or 3}=x\text{or}-1=x\text{or}1=x\end{array}$ .

Thus, the zeroes of the function $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ are $\left(-2,0\right),\left(3,0\right),\left(-1,0\right)$ and $\left(1,0\right)$ .

Plot the zeroes $\left(-2,0\right),\left(3,0\right),\left(-1,0\right)$ and $\left(1,0\right)$ of the function $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$

on a coordinate grid:

  

Determine whether $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ is positive or negative on each interval determined by the zeroes $\left(-2,0\right),\left(3,0\right),\left(-1,0\right)$ and $\left(1,0\right)$ :

Observe that the intervals determined by the zeroes $\left(-2,0\right),\left(3,0\right),\left(-1,0\right)$ and $\left(1,0\right)$ are $x<-2,-2<x<-1,-1<x<1,1<x<3$ and $x>3$ .

To determine whether $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ is positive or negative over the interval $x<-2$ , evaluate $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ for some value of $x$ that lies in $x<-2$ , say $x=-3$ , for its sign.

That is, substitute $x=-3$ in $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ and evaluate $h\left(-3\right)$ :

  $\begin{array}{l}h\left(-3\right)=\left(-3+2\right)\left(-3-3\right)\left(-3+1\right)\left(-3-1\right)\\ h\left(-3\right)=48\end{array}$ .

Observe that $h\left(-3\right)=48$ is positive.

Thus, $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ is positive over $x<-2$ .

Similarly, it may be observed that $h\left(-\frac{3}{2}\right)=-\frac{45}{16},h\left(0\right)=6,h\left(2\right)=-12$ and $h\left(4\right)=90$ .

That is, $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ is negative over $-2<x<-1$ , positive over $-1<x<1$ , negative over $1<x<3$ and finally positive over $x>3$ .

Sketch the graph of the function $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$ incorporating all the information gained:

  

Verify the sketch with the help of a graphing calculator:

Graph the function $h\left(x\right)=\left(x+2\right)\left(x-3\right)\left(x+1\right)\left(x-1\right)$

with the help of a graphing calculator on a similar grid:

  

Observe that the graph so obtained looks exactly like the one sketched previously.

Thus, it is verified that the sketch is correct.