Rational Inequalities

Main idea

A fraction can change sign where its top is zero or where its bottom is zero, so both kinds of points go on the sign chart. The bottom's zeros are never part of the answer, because the fraction doesn't exist there.

\frac{p(x)}{{\htmlData{sym=0}{q(x)}}} \ge 0: \text{ critical points are zeros of } p \text{ and of } q

What the symbols mean

$q(x)$
Denominator
Its zeros split intervals too, but are always excluded.

Explanation

A rational inequality is simply an inequality with a fraction of polynomials in it, like "when is $\frac{x - 1}{x + 2}$ at least zero?" The formula up top just lists where to look: the zeros of $p$ (where the top is zero, so the fraction is zero) and the zeros of $q$ (where the bottom is zero, so the fraction blows up). Those are the only places the sign can switch. It's the same sign chart as last lesson with one extra kind of point.

Why the bottom's zeros count. Last lesson, a polynomial could only change sign by passing through zero. A fraction has a second way: near a zero of the denominator it shoots off to $+\infty$ on one side and can come back from $-\infty$ on the other, skipping zero entirely. So those points split the number line too, even though the fraction is undefined there.

The method. Move everything to one side, combine into a single fraction, find the zeros of the top and bottom, and make a sign chart. Zeros of the numerator are included for $\ge$ and $\le$ ; zeros of the denominator never are.

A common mistake. It's tempting to multiply both sides by the denominator to clear the fraction, the way you would in an equation. But the denominator contains $x$ , so you don't know if it's positive or negative, and you don't know whether to flip the sign. Cross-multiplying quietly assumes it's positive and can throw away half the answer.

See it

This solves (x − 1)/(x + 2) ≥ 0. The sign can only switch at x = 1, where the top is zero, or at x = −2, where the bottom is. Notice that at −2 the curve doesn't pass through zero at all; it jumps from far above to far below. The shaded solution is x < −2 or x ≥ 1, with −2 itself left out.

Worked examples

Example 1

Solve $\dfrac{x - 1}{x + 2} \ge 0$ .

1
The bottom is zero at $-2$ , which is always excluded. The top is zero at $1$ , which is included because the inequality allows equality.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Solve $\dfrac{x - 1}{x + 2} \ge 0$ .

Problem 2+5 XP

Solve $\dfrac{3}{x - 4} < 0$ .

Problem 3+5 XP

Solve $\dfrac{x}{x - 3} > 2$ .

Problem 4+5 XP

Is $x = -2$ part of the solution of $\dfrac{x - 1}{x + 2} \ge 0$ ?

Problem 5+5 XP

Solve $\dfrac{x + 1}{x^2 + 1} \le 0$ .

Problem 6+5 XP

A factory's average cost per unit is $\frac{500 + 3q}{q}$ . What is the smallest production level $q$ for which the average cost is at most 8?

Problem 7+5 XP

Solve $\dfrac{x + 3}{x - 1} \le 0$ . Enter the endpoints of the solution interval.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $\dfrac{1}{x - 2} > \dfrac{1}{x + 1}$ .

Where it's used

Average cost per unit (a rational function) must stay below a price for profit; concentration limits in pharmacology lead to rational inequalities.

When Does Making More Get Cheap Enough?Algebra · Economics & Finance

Polynomial Inequalities Absolute Value Equations

Rational Inequalities

Main idea

\frac{p(x)}{{\htmlData{sym=0}{q(x)}}} \ge 0: \text{ critical points are zeros of } p \text{ and of } q

What the symbols mean

$q(x)$
Denominator
Its zeros split intervals too, but are always excluded.

Explanation

See it

Worked examples

Example 1

Solve $\dfrac{x - 1}{x + 2} \ge 0$ .

1
The bottom is zero at $-2$ , which is always excluded. The top is zero at $1$ , which is included because the inequality allows equality.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Solve $\dfrac{x - 1}{x + 2} \ge 0$ .

Problem 2+5 XP

Solve $\dfrac{3}{x - 4} < 0$ .

Problem 3+5 XP

Solve $\dfrac{x}{x - 3} > 2$ .

Problem 4+5 XP

Is $x = -2$ part of the solution of $\dfrac{x - 1}{x + 2} \ge 0$ ?

Problem 5+5 XP

Solve $\dfrac{x + 1}{x^2 + 1} \le 0$ .

Problem 6+5 XP

A factory's average cost per unit is $\frac{500 + 3q}{q}$ . What is the smallest production level $q$ for which the average cost is at most 8?

Problem 7+5 XP

Solve $\dfrac{x + 3}{x - 1} \le 0$ . Enter the endpoints of the solution interval.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Solve $\dfrac{1}{x - 2} > \dfrac{1}{x + 1}$ .

Where it's used

Average cost per unit (a rational function) must stay below a price for profit; concentration limits in pharmacology lead to rational inequalities.

When Does Making More Get Cheap Enough?Algebra · Economics & Finance