2.7 Solving Rational
Inequalities
In addition to the equal sign, there are
five unequal signs which stand for the different ways two quantities may
be unequal. The five signs are
<
less than
< less than or equal to
>
greater than
> greater than or equal to
±
not equal to
An inequality is a statement that two
quantities are unequal in one of the five ways listed above. A solution of an inequality is a
description of the set of all values of the variable for which the inequality
is true.
A rational inequality is in inequality in
which the two expressions are rational expressions. That is, polynomials or ratios of
polynomials.
Example:
Solve the inequality
2 x + 3 > 9
As in an equation, we may subtract three
from both sides is the inequality to get
2 x > 6
Then we may divide by 2 to get
x > 3
The solution is the set of all numbers
larger than 3, or the interval (3, ¥ ).
Although the steps are the same steps one
would take in solving 2 x + 3 = 9, there is one important difference when one
is solving inequalities:
Multiplication or
division of both sides of an inequality by a negative number reverses the
direction of the inequality.
Example:
Solve the inequality
1 - x < 5
Subtract 1 from each side to get
- x < 4
Then divide both sides by - 1,
remembering to reverse the direction of the inequality.
x > - 4
So the solution is ( - 4, ¥ ).
To see that this is the correct solution,
we can work the problem an alternate way which avoids division by a negative
number.
1 - x < 5
Add x to both sides to get
1 <
x + 5
Then subtract 5 from both sides to get
- 4 <
x
Now first, we got x > - 4 and now we
get - 4 < x. Do you see that these
two statements describe the same relationship between - 4 and x ? They mean the same thing. - 4 < x is equivalent to x > - 4. You can always turn an inequality around so
long as you keep the 'pointy end' pointed at the same expression.
In any event we see that x > - 4 is
the correct solution to the inequality.
Exercise 2.7.1
Solve the linear inequality
3 x - 7 > 5
We are going to look at a general method
of solving an inequality in which one side of the inequality is zero.
Requiring that one side of the inequality
equal zero is not a difficult requirement, since we can always achieve that
state in one step by subtracting the expression on one side of the inequality
from both sides of the inequality.
Consider the first inequality we looked
at above
2 x + 3 > 9
Subtract 9 from each side to get
2 x - 6 > 0
Now factor the left side
2 ( x - 3 ) > 0
Now sketch the graph of y = 2 ( x - 3
). It crosses the x-axis at 3 since 3
is a transitive intercept. The
rightmost part of the graph is above the x-axis since the leading coefficient,
2, is positive. So the graph lies above
the x-axis to the right of 3 and the graph lies below the x-axis to the left of
3. Thus, we can see that y > 0 when
x > 3. Thus the solution of the
inequality is
( 3, ¥ ).
So the method we are going to learn for
solving rational inequalities will involve getting zero on one side of the
inequality and a rational expression on the other side. Then we graph the rational expression and
see which portions of the graph satisfy the requirements of the inequality. The domain of those portions is the solution
to the inequality.
Exercise 2.7.2
Sketch the graph of y = 2 - x. Use the graph to solve the inequality 2 - x
< 0.
Now it is not actually necessary to
sketch the entire graph. All we need to
know is the following.
(1) What are the critical numbers of
the rational expression? That is, what
values of the variable cause the expression to equal 0 or to be undefined? That is, what are the zeros of the numerator
and what are the zeros of the denominator?
(2)
Which of the critical numbers are transitive and which are
intransitive? Remember that the
critical number c is transitive if
( x - c ) is a factor of
odd multiplicity and intransitive if ( x - c ) is a factor of
even multiplicity. This is true whether
( x - c ) is a factor of the numerator or a factor of the
denominator. The expression changes
signs at transitive critical numbers, but not at intransitive critical numbers.
(3)
Is the expression positive or is it negative to the right of the largest
critical number? To find out, take the
ratio of the leading coefficient of the numerator to the leading coefficient of
the denominator. If the ratio is
positive, then the expression is positive to the right of the largest critical
number. If the ratio is negative, then
the expression is negative to the right of the largest critical number.
For example, solve
2 x ( 1 - x ) / ( x
+ 2 )2 >
0
The critical numbers are - 2, 0 and
1. - 2 is intransitive and 0 and 1 are
transitive. The leading coefficient of
the numerator is - 2 and the leading coefficient of the denominator is 1. So
the ratio of the leading coefficients is a negative number. Thus the expression is negative to the right
of 1, zero at 1, positive between 0 and 1, zero at 0, negative between - 2 and
0, undefined at -2 and negative to the left of - 2. So the only place the expression is greater than or equal to zero
is on the interval [ 0, 1 ].
Exercise 2.7.3
Solve the inequality
( x2 + 1 )( 2 x - 1 )
/ [ ( x + 1 )3( x - 1 )2 ] < 0
When solving rational inequalities, be
sure NOT to divide both sides by any expression containing the variable. This can result in an inequality which is
not equivalent to the original, that is, an inequality which does not have the
same solution as the original.
For example, solve
x / ( x + 1 ) <
x / ( x + 2 )
Tempting though it is, we cannot divide
both sides by x. Instead, we subtract
one of the two expressions from both sides, find a common denominator and
factor the expression.
x / (x + 1 ) - x
/ ( x + 2 ) < 0
x / [ ( x + 1 ) ( x
+ 2 ) ] < 0
The critical numbers are - 2, - 1, and 0,
all transitive. The expression is zero
at 0 and undefined at - 1 and - 2. The expression
is positive to the right of 0, zero at 0, negative between - 1 and 0, positive
between - 2 and - 1 and negative to the left of - 2. Thus the expression is less than or equal to zero on the interval
( - ¥,
-2 ) and ( - 1, 0 ]. So the solution is
the set union of those two intervals, or
( - ¥, - 2 ) U ( - 1, 0 ].
Exercise 2.7.4
Solve the inequality
x ( x - 1 ) > (
x - 1 ) / x