Solving Rational Inequalities

Strategy: find critical points, build sign chart

A rational inequality has the form $\frac{P(x)}{Q(x)} \geq 0$. The sign can only change at critical points: where the numerator = 0 or the denominator = 0.

Step-by-step: Solve $\dfrac{x - 3}{x + 1} \geq 0$

Step 1 — Find critical points.

Numerator = 0: $x - 3 = 0 \implies x = 3$ (included, since $\geq$).

Denominator = 0: $x + 1 = 0 \implies x = -1$ (excluded — division by zero!).

Step 2 — Mark critical points on number line: $x = -1$ and $x = 3$, creating three intervals: $(-\infty, -1)$, $(-1, 3)$, $(3, \infty)$.

Step 3 — Test one value from each interval:

• $x = -2$: $\frac{-2-3}{-2+1} = \frac{-5}{-1} = +5 > 0$ ✓ positive
• $x = 0$: $\frac{0-3}{0+1} = \frac{-3}{1} = -3 < 0$ ✗ negative
• $x = 4$: $\frac{4-3}{4+1} = \frac{1}{5} > 0$ ✓ positive

Step 4 — Include boundary points where appropriate.

$x = 3$: $\frac{0}{4} = 0 \geq 0$ ✓ (included).

$x = -1$: undefined (excluded — always!).

Step 5 — Solution: $(-\infty, -1) \cup [3, \infty)$.

Key difference from polynomial inequalities

In rational inequalities, denominator zeros are always excluded (even with $\geq$ or $\leq$) because division by zero is undefined. Numerator zeros are included for $\geq$ and $\leq$, excluded for $>$ and $<$.

Try it in the visualization

Adjust the numerator and denominator roots. The sign chart shows $+$, $-$ in each interval. The rational function graph confirms which regions are above/below the x-axis. The vertical asymptote marks the excluded point.