Pre-Calculus Polynomial and rational functions

Rational inequalities (interval notation)

Q: How do you solve a rational inequality?

Move everything to one side, combine into a single fraction, find the zeros of the numerator and denominator as critical values, and use a sign chart.

Q: Why can't you multiply out the denominator?

Because its sign is unknown. Multiplying by a negative quantity would flip the inequality, so cross-multiplying can give wrong answers.

Q: Are the denominator's zeros ever included?

No. The expression is undefined there, so those points are always excluded, even for less-than-or-equal or greater-than-or-equal.

Q: What are the critical values?

The zeros of the numerator and the zeros of the denominator, the only places where the quotient can change sign.

20 practice questions 0 video lessons Theory + worked examples

Rational Inequalities

Texas Precalculus (TEKS) • Standard P.5(L) • Polynomial & Rational Functions

Rational Inequalities is a topic in Polynomial & Rational Functions in the Texas Essential Knowledge and Skills (Precalculus, §111.42). It is aligned to Standard P.5(L), which requires students to solve rational inequalities and express solutions in interval notation.

A rational inequality compares a ratio of polynomials with zero, solved with a sign chart whose critical values include the zeros of both the numerator and the denominator.

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Theory

A rational inequality compares a ratio of polynomials with zero, e.g. \(\dfrac{p(x)}{q(x)}\le 0\). Like a polynomial inequality it is solved with a sign chart, but with one extra rule.

The critical values are the zeros of the numerator and the zeros of the denominator. A denominator zero is always excluded (the expression is undefined there), even for \(\le\) or \(\ge\).

Never multiply both sides by the denominator. Its sign is unknown, so it could flip the inequality. Instead move everything to one side and combine into a single fraction.

Critical values from both numerator (\(x=1\), closed) and denominator (\(x=-2\), open).

A solution such as \([0,3)\): included at the numerator zero, excluded at the pole.

Set up as a single fraction versus zero:

\[\dfrac{p(x)}{q(x)}\ \{<,\le,>,\ge\}\ 0\]

compare p of x over q of x with zero

Sign of a quotient follows the sign of a product: it is positive when top and bottom agree in sign, negative when they differ.

Endpoints: numerator zeros may be included (for \(\le/\ge\)); denominator zeros are always excluded.

How to solve a rational inequality

Move everything to one side so the other is \(0\); combine into one fraction.
Find critical values: zeros of the numerator and of the denominator.
Sign-test each interval between critical values.
Select intervals with the required sign; include numerator zeros if \(\le/\ge\), always exclude denominator zeros.
Write the solution in interval notation.

Example 1 — Basic rational inequality

Solve \(\dfrac{x-1}{x+2}>0\).

Solution

Critical values are the numerator zero \(x=1\) and the denominator zero \(x=-2\) (always excluded). Test each interval.

\(x=-3\)	\(\Rightarrow\)	\(\dfrac{-}{-}=+\)
\(x=0\)	\(\Rightarrow\)	\(\dfrac{-}{+}=-\)
\(x=2\)	\(\Rightarrow\)	\(\dfrac{+}{+}=+\)

We want \(>0\); exclude \(x=-2\):

\[(-\infty,-2)\cup(1,\infty)\]

solution x less than negative 2 or x greater than 1

Example 2 — Inclusive, with an excluded pole

Solve \(\dfrac{x}{x-3}\le 0\).

Solution

Numerator zero \(x=0\) (included by \(\le\)); denominator zero \(x=3\) (always excluded). Sign-test the intervals.

\(x=-1\)	\(\Rightarrow\)	\(\dfrac{-}{-}=+\)
\(x=1\)	\(\Rightarrow\)	\(\dfrac{+}{-}=-\)
\(x=4\)	\(\Rightarrow\)	\(\dfrac{+}{+}=+\)

We want \(\le 0\): the middle interval, including \(0\) but not \(3\).

\[[0,\ 3)\]

solution is the interval from 0 included to 3 excluded

Example 3 — Combine into one fraction first

Solve \(\dfrac{2}{x-1}\ge 1\).

Solution

Move everything to one side and combine over a common denominator — never multiply out by \(x-1\), whose sign is unknown.

\(\dfrac{2}{x-1}-1\)	\(\ge\)	\(0\)
\(\dfrac{2-(x-1)}{x-1}\)	\(\ge\)	\(0\)
\(\dfrac{3-x}{x-1}\)	\(\ge\)	\(0\)

Critical values \(x=3\) (included) and \(x=1\) (excluded). The quotient is \(\ge 0\) between them:

\[(1,\ 3]\]

solution is the interval from 1 excluded to 3 included

Example 4 — Reading the sign chart

From the chart for \(\dfrac{(x+1)}{(x-2)}\), where is the expression negative?

Solution

Critical values \(x=-1\) (numerator) and \(x=2\) (denominator). The quotient is negative where numerator and denominator have opposite signs.

\(-1<x<2\)

\(\Rightarrow\)

\(\dfrac{+}{-}=-\)

\[(-1,\ 2)\]

expression is negative on the open interval from negative 1 to 2

Common pitfalls

Don't cross-multiply. Multiplying by \(q(x)\) can flip the inequality because you don't know its sign.

Denominator zeros are always excluded. Use an open endpoint there even for \(\le\) or \(\ge\).

Combine before testing. Get a single fraction \(=0\) form; testing a difference of fractions is error-prone.

Frequently asked questions

How do you solve a rational inequality?

Move everything to one side, combine into a single fraction, find the zeros of the numerator and denominator as critical values, and use a sign chart.

Why can't you multiply out the denominator?

Because its sign is unknown. Multiplying by a negative quantity would flip the inequality, so cross-multiplying can give wrong answers.

Are the denominator's zeros ever included?

No. The expression is undefined there, so those points are always excluded, even for \(\le\) or \(\ge\).

What are the critical values?

The zeros of the numerator and the zeros of the denominator — the only places where the quotient can change sign.

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