Rational inequalities are inequalities in which the unknown appears in the denominator of an algebraic fraction. Solving them relies almost entirely on sign analysis: it is not enough to find where an expression equals zero — one must also determine on which intervals the numerator and denominator share the same sign or have opposite signs.
Table of Contents
- The core idea behind sign analysis
- Domain restrictions
- Critical points and partitioning the number line
- How to build a sign chart
- Multiplicity of zeros
- Fully worked example
- Common mistakes
The core idea behind sign analysis
Consider an inequality of the form:
\[ \frac{P(x)}{Q(x)} > 0 \]
The sign of the fraction depends simultaneously on the sign of the numerator and the sign of the denominator.
A fraction is positive when the numerator and denominator have the same sign; it is negative when they carry opposite signs.
In symbolic form:
\[ \frac{P(x)}{Q(x)}>0 \]
when:
\[ \begin{cases} P(x)>0 \\ Q(x)>0 \end{cases} \quad \text{or} \quad \begin{cases} P(x)<0 \\ Q(x)<0 \end{cases} \]
Likewise:
\[ \frac{P(x)}{Q(x)}<0 \]
when the signs are opposite.
The entire theory of rational inequalities flows directly from this single observation.
Domain restrictions
Before analyzing the sign of the fraction, one must determine for which values the expression is defined.
Since a fraction cannot have a zero denominator, the condition:
\[ Q(x)\neq0 \]
must always hold. The values that make the denominator vanish are variously called:
- excluded values;
- points of exclusion;
- domain restrictions.
These values can never belong to the solution set, even when algebraic cancellation appears to remove them.
For example:
\[ \frac{x^2-4}{x-2} \]
can be rewritten as:
\[ \frac{(x-2)(x+2)}{x-2}=x+2 \]
but:
\[ x=2 \]
remains excluded, because it made the denominator of the original expression zero.
Critical points and partitioning the number line
The values at which a sign change can occur are called critical points.
They consist of:
- the zeros of the numerator;
- the zeros of the denominator.
Consider, for instance:
\[ \frac{(x-1)(x+2)}{x-3} \]
The critical points are:
\[ -2,\quad1,\quad3 \]
These values partition the real number line into the following intervals:
\[ (-\infty,-2),\quad(-2,1),\quad(1,3),\quad(3,+\infty) \]
Within each interval the sign of every factor remains constant. A polynomial can only change sign by passing through one of its zeros.
How to build a sign chart
A sign chart is built directly from the factored form of the fraction.
The first step is to factor both the numerator and the denominator completely. Next, all critical points are identified and arranged on the number line in increasing order.
The sign of each factor is then examined on every interval defined by the critical points.
Consider:
\[ \frac{(x-1)(x+2)}{x-3}>0 \]
| Interval | \((-\infty,-2)\) | \((-2,1)\) | \((1,3)\) | \((3,+\infty)\) |
|---|---|---|---|---|
| \(x+2\) | \(-\) | \(+\) | \(+\) | \(+\) |
| \(x-1\) | \(-\) | \(-\) | \(+\) | \(+\) |
| \(x-3\) | \(-\) | \(-\) | \(-\) | \(+\) |
| Fraction | \(-\) | \(+\) | \(-\) | \(+\) |
Once the sign of each factor is known, the behavior of the fraction follows immediately: factors sharing the same sign yield a positive product, while factors with opposite signs yield a negative product.
Since the inequality requires:
\[ \frac{(x-1)(x+2)}{x-3}>0 \]
we select the intervals in which the bottom row of the chart is positive:
\[ (-2,1)\cup(3,+\infty) \]
Multiplicity of zeros
When a factor appears raised to a power, the sign behavior near its zero depends on the multiplicity of that zero.
If the multiplicity is odd, the factor genuinely crosses zero and changes sign.
For example:
\[ (x-1)^3 \]
is negative for:
\[ x<1 \]
and positive for:
\[ x>1 \]
An even multiplicity, on the other hand, produces no sign change.
Indeed:
\[ (x-1)^2\geq0 \]
for every real value of \(x\).
Geometrically, a zero of odd multiplicity corresponds to the graph crossing the axis, while a zero of even multiplicity corresponds to the graph merely touching it.
Fully worked example
Solve the inequality:
\[ \frac{x^2-4x}{x+2}\geq0 \]
The goal is to find all values of \(x\) for which the fraction is non-negative.
To do so we need to:
- factor the numerator;
- determine the domain restrictions;
- identify the critical points;
- build the sign chart.
Factoring
The numerator has the common factor \(x\):
\[ x^2-4x=x(x-4) \]
The inequality therefore becomes:
\[ \frac{x(x-4)}{x+2}\geq0 \]
In this form the sign of the fraction can be analyzed factor by factor:
\[ x,\qquad x-4,\qquad x+2 \]
Domain restrictions
The denominator cannot be zero:
\[ x+2\neq0 \]
hence:
\[ x\neq-2 \]
This value must be excluded from the final solution regardless of the sign of the fraction.
Critical points
The critical points are the values that make either the numerator or the denominator equal to zero.
In this case:
\[ x=0,\qquad x=4,\qquad x=-2 \]
Arranged in increasing order on the number line:
\[ -2,\quad0,\quad4 \]
These points partition the real line into the following intervals:
\[ (-\infty,-2),\quad(-2,0),\quad(0,4),\quad(4,+\infty) \]
Within each interval the sign of every factor remains constant.
Sign analysis
We now examine the behavior of each factor.
The factor:
\[ x \]
is negative for \(x<0\) and positive for \(x>0\).
The factor:
\[ x-4 \]
is negative for \(x<4\) and positive for \(x>4\).
Finally:
\[ x+2 \]
is negative for \(x<-2\) and positive for \(x>-2\).
Collecting all of this information in the sign chart:
| Interval | \((-\infty,-2)\) | \((-2,0)\) | \((0,4)\) | \((4,+\infty)\) |
|---|---|---|---|---|
| \(x\) | \(-\) | \(-\) | \(+\) | \(+\) |
| \(x-4\) | \(-\) | \(-\) | \(-\) | \(+\) |
| \(x+2\) | \(-\) | \(+\) | \(+\) | \(+\) |
| Fraction | \(-\) | \(+\) | \(-\) | \(+\) |
The bottom row is obtained by multiplying the signs of the individual factors.
For example, on the interval:
\[ (-2,0) \]
we have:
\[ (-)\cdot(-)\cdot(+)=+ \]
so the fraction is positive there.
On the intervals:
\[ (-\infty,-2) \quad\text{and}\quad (0,4) \]
the product of signs is negative.
Reading off the solution
The inequality requires:
\[ \frac{x(x-4)}{x+2}\geq0 \]
so we select the intervals on which the fraction is positive or zero.
From the chart:
\[ (-2,0) \quad\text{and}\quad (4,+\infty) \]
Because the inequality uses:
\[ \geq \]
the zeros of the numerator must also be included:
\[ x=0,\qquad x=4 \]
The value:
\[ x=-2 \]
remains excluded, since it makes the denominator zero.
The solution set is therefore:
\[ (-2,0]\cup[4,+\infty) \]
Common mistakes
Forgetting domain restrictions
This is by far the most frequent error. Zeros of the denominator must always be excluded from the solution.
Reversing the inequality incorrectly
One cannot multiply both sides of an inequality by an expression involving the unknown without first knowing its sign.
Dropping excluded values after cancellation
Even after canceling common factors, any value that made the original denominator zero remains inadmissible.
Overlooking zeros of the numerator
In inequalities involving:
\[ \geq \quad\text{or}\quad \leq \]
zeros of the numerator must be included whenever they belong to the domain.
Rational inequalities illustrate with particular clarity how the behavior of a rational function is governed by the distribution of its zeros and its points of discontinuity.
Sign analysis thus turns an apparently complex problem into a systematic examination of intervals along the real number line.
Careful factoring, combined with a rigorously constructed sign chart, makes it possible to tackle even highly intricate rational inequalities in a methodical, elegant, and reliable way.