Why Rational Inequalities Need a Sign Chart
A rational inequality asks where a quotient of polynomials is positive, negative, nonnegative, or nonpositive. For example, solving \(\frac{x-2}{x+1}\geq 0\) requires more than finding where the quotient is zero. The denominator also affects the sign, and its zeros are excluded from the domain.
The useful idea is to divide the real number line at the values where the numerator or denominator is zero. On each resulting interval, the rational expression has a constant sign. A single test value then determines its sign throughout that interval. Finally, the inequality symbol and the original domain determine which boundary values belong in the answer.
Critical Values and Interval Signs
For a nonzero numerator polynomial \(P\), first find the real zeros of \(P\) and the real zeros of \(Q\). These values are called the critical values for the sign chart. Since both polynomials are nonzero, each has only finitely many real zeros. List the distinct critical values in increasing order. They partition the real line into open intervals.
On any one of those open intervals, neither \(P\) nor \(Q\) is zero. Thus the quotient is defined and cannot change sign there. More precisely, it is continuous on that interval. If it took both a positive and a negative value, the Intermediate Value Theorem would imply that it was zero somewhere between them. A quotient with nonzero denominator is zero only where its numerator is zero, and the interval contains no such value. So the sign is constant on the interval.
The endpoint rules are essential. A numerator zero is a point where the quotient equals zero, provided the denominator is nonzero there. It can be included for \(\geq 0\) or \(\leq 0\), but not for \(>0\) or \(<0\). A denominator zero is never included: the expression is undefined there, regardless of the inequality symbol.
| Boundary value | Strict inequality | Non-strict inequality |
|---|---|---|
| Numerator zero, denominator nonzero | Exclude for \(>0\) or \(<0\) | Include for \(\geq 0\) or \(\leq 0\) |
| Denominator zero | Always exclude | Always exclude |
Worked Example: A Nonnegative Rational Expression
Worked Example: A Nonnegative Rational Expression
Solve \(\frac{x-2}{x+1}\geq 0\).
The numerator is zero when \(x-2=0\), so \(x=2\). The denominator is zero when \(x+1=0\), so \(x=-1\), which is excluded from the domain. The critical values in order are \(-1\) and \(2\), giving the intervals \((-\infty,-1)\), \((-1,2)\), and \((2,\infty)\).
Test \(x=-2\) in the first interval: \(\frac{-2-2}{-2+1}=\frac{-4}{-1}=4>0\). Test \(x=0\) in the second: \(\frac{0-2}{0+1}=\frac{-2}{1}=-2<0\). Test \(x=3\) in the third: \(\frac{3-2}{3+1}=\frac{1}{4}>0\).
The quotient is nonnegative on the first and third intervals. The point \(x=2\) is included because the denominator is nonzero there and the quotient equals zero. The point \(x=-1\) is excluded because the expression is undefined there.
As a check, the quotient is positive at \(x=-2\) and \(x=3\), negative at \(x=0\), zero at \(x=2\), and undefined at \(x=-1\), exactly as the solution and endpoint rules require.
Why Testing One Point Is Enough
The sign chart does not assume that a quotient has the same sign everywhere. It establishes that the sign cannot change inside any interval between critical values. There is no numerator zero or denominator zero inside such an interval, and the quotient is continuous there. The test point identifies which of the two possible signs holds throughout that interval.
Find every real zero of the numerator and denominator, keeping denominator zeros marked as excluded.
Order the distinct boundary values and form the open intervals between them, including the two unbounded intervals.
Evaluate the quotient at one convenient point in each interval and record whether the result is positive or negative.
Select intervals with the required sign, include eligible numerator zeros for non-strict inequalities, and exclude all denominator zeros.
Worked Example: A Strictly Negative Inequality
Worked Example: A Strictly Negative Inequality
Solve \(\frac{(x+3)(x-1)}{x-2}<0\).
The numerator zeros are \(x=-3\) and \(x=1\). The denominator zero is \(x=2\), which is not in the domain. The intervals are \((-\infty,-3)\), \((-3,1)\), \((1,2)\), and \((2,\infty)\).
For \(x=-4\), the quotient is \(\frac{(-4+3)(-4-1)}{-4-2}=\frac{(-1)(-5)}{-6}=-\frac{5}{6}<0\). For \(x=0\), it is \(\frac{(3)(-1)}{-2}=\frac{3}{2}>0\). For \(x=\frac32\), it is \(\frac{(\frac92)(\frac12)}{-\frac12}=-\frac92<0\). For \(x=3\), it is \(\frac{(6)(2)}{1}=12>0\).
The negative intervals are \((-\infty,-3)\) and \((1,2)\). Because the inequality is strict, the numerator zeros \(x=-3\) and \(x=1\) are not included. The denominator zero \(x=2\) is also excluded.
Checking the chosen test values confirms that each listed interval has a negative value. The boundary points are correctly omitted: the quotient is zero at \(-3\) and \(1\), and undefined at \(2\).
Repeated Zeros and Intervals with the Same Sign
Not every critical value causes the sign to change. In particular, when a numerator has a repeated zero, the sign on the intervals immediately to its left and right may be the same. The sign-chart method remains reliable because it tests each interval rather than assuming that signs alternate. The earlier material on polynomial sign charts and multiplicity explains this behavior.
Worked Example: A Nonpositive Quotient with a Repeated Zero
Solve \(\frac{(x-1)^2}{(x+2)(x-3)}\leq 0\).
The numerator is zero at \(x=1\). The denominator is zero at \(x=-2\) and \(x=3\), so those two values are excluded. The intervals are \((-\infty,-2)\), \((-2,1)\), \((1,3)\), and \((3,\infty)\).
At \(x=-3\), the quotient is \(\frac{(-4)^2}{(-1)(-6)}=\frac{16}{6}=\frac83>0\). At \(x=0\), it is \(\frac{(-1)^2}{(2)(-3)}=-\frac16<0\). At \(x=2\), it is \(\frac{1^2}{(4)(-1)}=-\frac14<0\). At \(x=4\), it is \(\frac{3^2}{(6)(1)}=\frac32>0\).
The quotient is negative on both middle intervals. At \(x=1\), its numerator is zero and its denominator is \((3)(-2)=-6\ne0\), so the quotient equals zero and is included for a nonpositive inequality. The denominator zeros remain excluded.
The two intervals adjacent to \(x=1\) both have negative signs; including \(x=1\) joins them into the interval \((-2,3)\). The excluded endpoints \(-2\) and \(3\) are denominator zeros.
A Special Case: The Numerator Is Identically Zero
The sign-chart procedure above assumes that \(P\) is a nonzero polynomial, so that its real zeros form a finite list. If \(P\) is identically zero, every point in the domain makes the quotient equal to zero. Handle this case directly instead of trying to list numerator zeros as critical values.
For instance, \(\frac{0}{x-1}\geq0\) is true for every \(x\ne1\), so its solution is \((-\infty,1)\cup(1,\infty)\). The point \(1\) is still excluded because it makes the denominator zero. If \(Q\equiv0\), the quotient is nowhere defined and there are no solutions.
A Common Pitfall: Cancellation Does Not Restore Domain Values
Factoring and canceling can simplify a quotient, but the original denominator still determines which inputs are allowed. Consider \(\frac{x-1}{(x-1)(x+2)}>0\). Its original domain excludes both \(x=1\) and \(x=-2\).
For \(x\ne1\), cancellation gives \(\frac{1}{x+2}\), but this equality holds only on the original domain. The simplified expression is positive when \(x>-2\). The value \(x=1\), although it satisfies \(1>-2\), remains excluded from the original expression.
Check Your Understanding
For each problem, identify the critical values, determine the interval signs, and apply the endpoint rules.
- Solve \(\frac{x+1}{x-3}\leq0\).
- Solve \(\frac{(x-2)(x+4)}{x+1}>0\).
- Why must a zero of the original denominator be excluded even if its factor cancels?
- What is the solution of \(\frac{0}{x-5}\leq0\)?
- Can a rational expression keep the same sign on both sides of a numerator zero? Explain how the sign-chart method handles this case.
