Question 92932
For the following equation, find the interval(s) where {{{f(x)<0}}}.
{{{f(x)=1/(x^2-2x-8)}}}
<pre><font size = 4 color = "darkblue"><b>
First let's learn what we want to find.  Let's look at 
the graph of
{{{y=f(x)=1/(x^2-2x-8)}}} which can be found by plotting
points and drawing dot-to-dot:

{{{drawing(400,200,-5,7,-3,3,

graph(400,200,-5,7,-3,3,0,0,(1/(x^2-2x-8)) ( sqrt(x+1.9)/sqrt(x+1.9) )( sqrt(4-x)/sqrt(4-x) ) ),

graph( 400,200,-5,7,-3,3,0,0,(1/(x^2-2x-8)) ( sqrt(-2-x)/sqrt(-2-x) )  ), 

graph( 400,200,-5,7,-3,3,0,0,(1/(x^2-2x-8)) ( sqrt(x-4.1)/sqrt(x-4.1) )  ),

graph( 400,200,-5,7,-3,3,0,999(x+2)), graph( 400,200,-5,7,-3,3,0,999(x-4))
)}}}

The two green lines are not part of the graph of f(x).  Only 
the blue curves make up the graph of {{{f(x)=1/(x^2-2x-8)}}}. The blue 
curves approach but never reach the green lines.  The green 
lines are called "asymptotes". But that's beside the point.

The question we are asking is this:

What part of the x-axis is the blue graph BELOW the x-axis?
 
Why BELOW the x-axis?  Because the problem is to find
where {{{f(x)<0}}} and a graph is less than zero when it is
BELOW the x-axis.

We can tell that the blue graph is BELOW the x-axis on the
part of the x-axis which is between -2 and 4, which we
can write as  -2 < x < 4 or in interval notation as (-2,4).

That is the correct answer, but we did it graphically. But
you are supposed to do it algebraically. Here is how:

{{{f(x)=1/(x^2-2x-8)}}} and we want {{{f(x)<0}}}, so

we want:

{{{1/(x^2-2x-8)<0}}}

We factor the denominator

{{{1/((x-4)(x+2))<0}}}

The critical values are values that make the
expression either 0 or undefined. So we set
each factor = 0.

x-4=0 gives critical value x=4
x+2=0 gives critical value x=-2

Now we draw a number line and mark these two points:

-----------o-----------------o--------
 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 

(This number line is actually the x-axis of the
graph above without the graph or the y-axis.)

Anyway these two points divide the number line into 
three parts. Let's pick a point in each part.

Pick a point less that (left of) -2, say -3.
Substitute -3 for x in

{{{1/((x-4)(x+2))<0}}}

{{{1/(((-3)-4)((-3)+2))<0}}} 

{{{1/((-7)(-1))<0}}}

{{{1/7<0}}}

This is FALSE! So we DO NOT shade the part of 
the number line to the left of -2. So our number 
line is still as it was:

-----------o-----------------o--------
 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 


Next we pick any point in the middle part,
between -2 and 4. Suppose we choose x=0
Substitute 0 for x in

{{{1/((x-4)(x+2))<0}}}

{{{1/((0-4)(0+2))<0}}} 

{{{1/((-4)(2))<0}}}

{{{-1/8<0}}}

This is TRUE! So we DO shade the part of the 
number line between -2 and 4. So our number 
line now looks like this:

-----------o=================o--------
 -5 -4 -3 -2 -1  0  1  2  3  4  5  6 

Finally we pick a point to the right of 4.
Suppose we choose x=5.

Substitute 5 for x in

{{{1/((x-4)(x+2))<0}}}

{{{1/((5-4)(5+2))<0}}} 

{{{1/((1)(7))<0}}}

{{{1/7<0}}}

This is FALSE! So we do not shade the part to the
right of x=4.  So the answer is this number line:

-----------o=================o--------
 -5 -4 -3 -2 -1  0  1  2  3  4  5  6

which is represented by the interval notation (-2,4),
which is what we observed graphically.

Edwin</pre>