Rational Inequalities, precalculus lesson

A print shop charges a 500 dollar setup fee plus 4 dollars per poster, so the average cost of each of $x$ posters is $(500 + 4x)/x$ dollars. Asking how many posters bring that average below 9 dollars is a rational inequality. Here $x$ is positive, so clearing the fraction is safe, and the answer is more than 100 posters. In general a denominator can be negative or zero, the shortcut fails, and a sign chart does the job instead.

Key terms

Rational inequality

Definition A rational inequality is an inequality that can be written in one of the forms

\begin{gathered} r(x) > 0, \qquad r(x) \ge 0, \\[0.4em] r(x) < 0, \qquad r(x) \le 0, \end{gathered}

where $r = p/q$ is a rational function, with $p$ and $q$ polynomials and $q$ not the zero polynomial. Its solution set consists of the real numbers $x$ at which every fraction in the original inequality is defined and the inequality is true. An inequality between rational expressions is brought to this form by subtracting one side from both sides and combining everything over a common denominator. Done without cancelling, this keeps every number that made an original denominator zero as a zero of $q$ , so none of them can slip into the answer. Multiplying both sides by an expression that contains $x$ is not a valid step unless its sign is known, because a negative multiplier would reverse the inequality.

In plain words Get zero on one side and a single fraction on the other. Then the question is only about the sign of that fraction.

Example To solve

\frac{x + 3}{x - 1} \ge 2,

subtract $2$ and combine,

\frac{x + 3 - 2(x - 1)}{x - 1} = \frac{5 - x}{x - 1} \ge 0.

Multiplying the original inequality by $x - 1$ instead would give $x + 3 \ge 2x - 2$ , so $x \le 5$ . That accepts $x = 0$ , where the left side is $3/(-1) = -3$ , which is not at least $2$ . The shortcut kept the direction of the inequality, which is correct only when $x - 1 > 0$ .

Zero points

Definition A zero point of the rational inequality $r(x) > 0$ , or of the same inequality with $\ge$ , $<$ or $\le$ , is a real number $a$ in the domain of $r$ with $r(a) = 0$ . For $r = p/q$ these are the real zeros of $p$ at which $q$ is not zero. Every zero point satisfies a non-strict inequality ( $\ge$ or $\le$ ) and fails a strict one ( $>$ or $<$ ).

In plain words A zero point is where the fraction actually equals zero, because the top is zero and the bottom is not. It belongs to the answer exactly when the inequality allows equality.

Example For $(5 - x)/(x - 1) \ge 0$ the numerator is zero at $x = 5$ , where the denominator is $4$ . So $5$ is a zero point, and since the inequality is $\ge$ , it is a solution. In the original form the check is $(5 + 3)/(5 - 1) = 2$ , which is at least $2$ .

Undefined points

Definition An undefined point of a rational inequality is a real number $a$ that makes a denominator of the original inequality zero, before any cancellation. The expression has no value at $a$ , so an undefined point is never in the solution set, whether the inequality is strict or not. It still counts as a critical value, because the sign of $r$ can change there.

In plain words Undefined points and zero points both go on the number line, but they play different roles. A zero point gets a closed dot when equality is allowed, while an undefined point always gets an open dot.

Example In $(5 - x)/(x - 1) \ge 0$ the denominator is zero at $x = 1$ . So $1$ is an undefined point and is left out of the solution set, even though the inequality is $\ge$ . Its sign still matters, since the fraction is negative just to the left of $1$ and positive just to the right.

Critical intervals

Definition The critical values of a rational inequality are its zero points together with its undefined points. Placed in order on the number line, they cut it into open intervals, called critical intervals. On each critical interval $r$ is defined, continuous and never zero, so it keeps one sign there, and its sign at a single test value is its sign on the whole interval. The solution set is the union of the critical intervals with the right sign, plus the zero points when the inequality is non-strict.

In plain words Mark both kinds of critical values, test one number in each gap between them, then decide about the borders.

Example For $(5 - x)/(x - 1) \ge 0$ the critical values are $1$ and $5$ .

Critical interval	Test value	Value of $(5 - x)/(x - 1)$	Sign
$(-\infty, 1)$	$0$	$-5$	negative
$(1, 5)$	$2$	$3$	positive
$(5, \infty)$	$6$	$-1/5$	negative

The fraction is positive on $(1, 5)$ , the zero point $5$ is included and the undefined point $1$ is not. The solution set is $(1, 5]$ .

Sign chart for a quotient

Definition A quotient of two nonzero numbers has the same sign as their product. So a rational expression written as a constant times factors in the numerator and factors in the denominator has a sign chart built exactly like that of a polynomial, with one row for each factor and a last row for the whole expression. A factor $(x - c)^m$ changes sign at $c$ when $m$ is odd and keeps its sign when $m$ is even, whether it sits in the numerator or the denominator.

In plain words For signs it does not matter whether a factor is on the top or the bottom. Track every factor, count the negative ones, and remember which critical values come from the bottom.

Example Solve

\frac{x - 2}{(x + 1)(x - 4)} \le 0.

The zero point is $2$ and the undefined points are $-1$ and $4$ .

Factor	$(-\infty, -1)$	$(-1, 2)$	$(2, 4)$	$(4, \infty)$
$x - 2$	$-$	$-$	$+$	$+$
$x + 1$	$-$	$+$	$+$	$+$
$x - 4$	$-$	$-$	$-$	$+$
Quotient	$-$	$+$	$-$	$+$

The quotient is negative on $(-\infty, -1)$ and on $(2, 4)$ . Adding the zero point $2$ and leaving out $-1$ and $4$ gives the solution set $(-\infty, -1) \cup [2, 4)$ . The test value $x = 3$ gives $1/\bigl(4 \cdot (-1)\bigr) = -1/4$ , as the chart says.

Cancelled factors

Definition If the numerator and denominator share a factor $x - a$ , cancelling it gives a simpler expression that has the same value as the original at every number where the original is defined. The simpler expression may be defined at $a$ , but the original is not, so $a$ stays an undefined point of the original inequality. Solve the simpler inequality, then remove every undefined point of the original from the answer.

In plain words Cancelling is fine for finding signs, but it cannot bring back a number the original fraction was not allowed to use.

Example Solve

\frac{x^2 - x}{x^2 - 1} > 0.

Factoring and cancelling gives

\begin{gathered} \begin{aligned} \frac{x^2 - x}{x^2 - 1} &= \frac{x(x - 1)}{(x - 1)(x + 1)} \\[0.8em] &= \frac{x}{x + 1} \end{aligned} \\[0.8em] \text{for } x \neq 1 \text{ and } x \neq -1. \end{gathered}

The simpler inequality $x/(x + 1) > 0$ holds on $(-\infty, -1)$ and on $(0, \infty)$ . At $x = 1$ the original fraction is $0/0$ , which has no value, so $1$ must be removed. The solution set is

(-\infty, -1) \cup (0, 1) \cup (1, \infty).

Check yourself

Solve $(x - 4)/(x + 2) < 0$ .

Show the answer

The zero point is $4$ and the undefined point is $-2$ . The test values $-3$ , $0$ and $5$ give $7$ , $-2$ and $1/7$ , so the fraction is negative only on $(-2, 4)$ . The inequality is strict, so neither critical value is included, and the solution set is $(-2, 4)$ .

Solve $1/x \le 2$ .

Show the answer

Subtracting $2$ gives

\frac{1 - 2x}{x} \le 0.

The zero point is $1/2$ and the undefined point is $0$ . The test values $-1$ , $1/4$ and $1$ give $-3$ , $2$ and $-1$ . The solution set is $(-\infty, 0) \cup [1/2, \infty)$ . Multiplying the original inequality by $x$ would have given only $x \ge 1/2$ and lost every negative $x$ .

Solve $(x + 1)/(x - 2)^2 \ge 0$ .

Show the answer

The denominator $(x - 2)^2$ is positive for every $x \neq 2$ , so the fraction has the sign of $x + 1$ . It is positive for $x > -1$ with $x \neq 2$ , zero at the zero point $-1$ , and undefined at $2$ . The solution set is $[-1, 2) \cup (2, \infty)$ .

Rational Inequalities

Key terms

Rational inequality

Zero points

Undefined points

Critical intervals

Sign chart for a quotient

Cancelled factors

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus