# uc3rmw_0205e04

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Hi there.
In this example, we will solve the
rational inequality 2 over x minus 1
is greater than or equal
to 1 over x minus 2.
The solution to a polynomial inequality
involves a five-step process.
In step 1, we want to arrange the inequality
so that the 0 is on the right-hand side.
So let's subtract 1 over x minus
2 from the right-hand side.
We have that the left-hand side is equal
to 2 over x minus 1 minus 1 over x minus 2.
And that's all now greater
than or equal to 0.
We can simplify the expression on the
left-hand side of the inequality symbol
by finding the common denominator.
We multiply the first fraction, 2 over
x minus 1, by x minus 2 over x minus 2.
And then the second fraction, 1 over
x minus 2, we'll multiply by x minus 1
over x minus 1.
And that's all greater than or equal to 0.
Now we have the common denominator x minus
1 times x minus 2 for both fractions.
And we'll go ahead and
combine the numerators.
We have 2 times x minus 2 minus
x minus 1, in parentheses,
all over that common denominator,
x minus 1 times x minus 2.
This is all still greater
than or equal to 0.
We can simplify the numerator now.
We have to expand everything
out and combine like terms.
We have 2x minus 4 minus x plus 1, all over
our common denominator, x minus 1 times
x minus 2, greater than or equal to 0.
2x minus x is equal to x, and
minus 4 plus 1 is equal to minus 3.
And that's all over, again,
x minus 1 times x minus 2.
That's all greater than or equal to 0.
So now we have a simplified
single fraction expression
on the left-hand side of
the inequality symbol.
We're going to go ahead and assign that
expression the functional assignment
R of x.
In step 2, we're going to factor
R of x and find the real zeros
of the numerator and the denominator.
These numbers will form the boundary points.
So the real zeros of the numerator are
found by setting the numerator equal to 0.
The numerator is x minus 3.
So we're going to set that
equal to 0 and solve for x.
We have that x equals 3.
We'll also find the
zeros of the denominator.
So we're going to set the
denominator equal to 0.
The denominator is our common
denominator x minus 1 times x minus 2.
So we're going to set that
equal to 0 and solve for x.
We set each linear factor equal 0.
We have x minus 1 equals
0 or x minus 2 equals 0.
The solutions are x equal 1 and 2
from both of those linear equations.
And so the boundary points
are the points 1, 2, and 3.
So now we move on to step 3, and we want
to locate the boundary points on the number
line.
These boundary points are going to divide
the number line into separate intervals.
Again, our boundary points are 1, 2, and 3.
So we plot that on the real number line.
Here's the point 1.
Here's the point 2.
And here's the point 3.
That separates the real number
line into four separate intervals.
We have the first interval from minus
infinity to 1, the second one from 1 to 2,
the third from 2 to 3, and
a fourth from 3 to infinity.
We see that the boundary points at
1 and 2 are marked as open circles
to indicate that they do not belong to the
domain of the rational function R of x.
The boundary point at 3 is a solid circle.
Now, in step 4a, we're going
to use the graph method.
We want to determine the
sign of the rational function
R of x equal to 1 N of x over D of x
and form the corresponding polynomial P
of x equal N of x times D of x.
We're going to graph P of
x using the end behavior
and multiplicity where it crosses or
touches the x-axis of the zeros of P
of x and determine the sign of P of x on
the intervals determined by the boundary
points.
Now, the sign of P of x is
the same as the sign of R
of x on the intervals found in step three.
So we want to calculate the
leading term for P of x.
Again, P of x is equal to
the product of the numerator
and denominator of our rational
function, R of x, so that is it's
x minus 3 times x minus 2 times x minus 1.
The leading term is calculated by
just taking the product of the leading
terms of each of these linear factors.
So we have x times x times x equals x cubed.
And therefore, P of x is equal to x cubed
plus, then lower degree powers of x.
And that is that P of x is
a polynomial of degree 3,
and the leading coefficient is a sub 3
equal to 1, where 1 is greater than 0.
That is, as x becomes large positive,
P of x becomes large positive.
And as x becomes large negative, P
of x becomes large negative, as well.
Now, because 1, 2, and 3 are
zeros of odd multiplicity
the graph of P of x crosses
the x-axis at these points.
We're summarizing all that information that
we gathered in the previous slide here.
We see the four different intervals
that the real number line, the x-axis,
was divided into based on those
boundary points, 1, 2, and 3.
So we have the interval minus infinity
to 1, 1 to 2, 2 to 3, and 3 to infinity.
The end behavior is now used.
We see that as x tends towards
large negative numbers,
we have that P of x becomes
large negative in y.
And as x becomes large positive, P
of x becomes large positive, as well.
The odd multiplicity means that the graph
crosses the x-axis at those boundary
points, 1, 2, and 3, as shown.
Therefore, we generate a sign
table at the bottom here.
We indicate that the graph is negative
from the interval from minus infinity to 1,
positive from 1 to 2, negative again from
2 to 3, and positive from 3 to infinity.
In step 4b, we're going to use the
Test Points Method to select points
on each interval determined
by the boundary points
and evaluate R of x at
each of these test points
to find the sign of R of x on that interval.
We're going to choose for the first
interval from minus infinity to 1, a test
0, which is an easy one to evaluate.
We're going to calculate the value
of R of x at that test point.
And the value of R of x at x
equals 0 is given as follows.
We get negative 3/2, which is negative, and
we go ahead and summarize that in the table
here.
For the next interval from 1 to 2,
we're going to choose a test point
1.5 on that interval.
We evaluate R of x at x equal 1.5.
We get a value of 6,
which is positive, and we
go ahead and summarize that in the table.
Next, for the interval from 2 to 3, we
choose a point 2.5 on that interval.
R of x at x equal 2.5 is evaluated
to be minus 2/3, a negative value.
And we go ahead and summarize that there.
And finally, for the interval from 3
to infinity, we choose a test point 4.
We have R of x at x equal 4 evaluated
to be 1/6, which is positive.
And so that's summarized here, as well.
Now, the value of R of x at
each of these test points
dictates the sign of R of x on those
intervals containing those test points.
So we found that at the test point 0 on
the interval from minus infinity to 1
that that value is negative.
Therefore, all the values
of R of x for points
on the interval from minus
infinity to 1 are negative.
For the second interval
from 1 to 2, our test point
gave us a positive value for R of x.
And therefore, all the values of R of x
on the interval from 1 to 2 are positive.
For our test point 2 and 1/2
on the interval from 2 to 3,
we got a negative value
for R of x at that point.
And therefore, all the points for R of x
on that interval from 2 to 3 are negative.
Finally, for 3 to infinity, R of x at
x equal 4 gave us a positive value,
and so R of x is positive everywhere
on that interval from 3 to infinity.
Now, in step 5, we want
to write the solution set
by selecting the intervals in step
4 that satisfy the given inequality.
If the inequality is R of x
greater than or equal to 0
or R of x less than or equal to 0, we
want to include only the boundary points
corresponding to the zeros of the
numerator of the rational function R of x.
Otherwise, we're going to exclude
them if it's just greater than 0
or just less than 0.
Then we're going to graph the solution set.
So in our case, we're solving the inequality
R of x greater than or equal to 0,
and so we're going to find the intervals
on which the function is positive.
Now, since the inequality
involves greater than or equal to,
we're including the zeros of the numerator
of R of x, that is the boundary points,
in the solution set for the numerator.
The only boundary point that
is a 0 of the numerator is 3.
And so the solution set is
the interval from 1 to 2 union
the interval from 3 to
infinity, where, again, we're
including the boundary point 3.
That's the 0 of the numerator.
We show that graphically.
Again, we have parentheses here
to indicate that 1 and 2 are not
included in our solution set.
We've got a bracket here to
indicate that 3 is included.
And the intervals include all the values
from 1 to 2 and all the values from 3
to infinity.