SOLUTION: For the following equation, find the interval(s) where {{{f(x)<0}}}. {{{f(x)=1/(x^2-2x-8)}}}

Algebra ->  Inequalities -> SOLUTION: For the following equation, find the interval(s) where {{{f(x)<0}}}. {{{f(x)=1/(x^2-2x-8)}}}      Log On


   

Question 92932This question is from textbook Algebra and Trigonometry
: For the following equation, find the interval(s) where f%28x%29%3C0.
f%28x%29=1%2F%28x%5E2-2x-8%29 This question is from textbook Algebra and Trigonometry

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
For the following equation, find the interval(s) where f%28x%29%3C0.
f%28x%29=1%2F%28x%5E2-2x-8%29

First let's learn what we want to find.  Let's look at 
the graph of
y=f%28x%29=1%2F%28x%5E2-2x-8%29 which can be found by plotting
points and drawing dot-to-dot:



The two green lines are not part of the graph of f(x).  Only 
the blue curves make up the graph of f%28x%29=1%2F%28x%5E2-2x-8%29. 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%28x%29%3C0 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%28x%29=1%2F%28x%5E2-2x-8%29 and we want f%28x%29%3C0, so

we want:

1%2F%28x%5E2-2x-8%29%3C0

We factor the denominator

1%2F%28%28x-4%29%28x%2B2%29%29%3C0

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%2F%28%28x-4%29%28x%2B2%29%29%3C0

1%2F%28%28%28-3%29-4%29%28%28-3%29%2B2%29%29%3C0 

1%2F%28%28-7%29%28-1%29%29%3C0

1%2F7%3C0

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%2F%28%28x-4%29%28x%2B2%29%29%3C0

1%2F%28%280-4%29%280%2B2%29%29%3C0 

1%2F%28%28-4%29%282%29%29%3C0

-1%2F8%3C0

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%2F%28%28x-4%29%28x%2B2%29%29%3C0

1%2F%28%285-4%29%285%2B2%29%29%3C0 

1%2F%28%281%29%287%29%29%3C0

1%2F7%3C0

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