Rational Inequalities

Solve rational inequalities by finding critical values from zeros and vertical asymptotes, then using sign analysis.

Rational Inequalities

Introduction

A rational inequality involves a rational expression (fraction with polynomials) and an inequality sign.

Examples:

$\frac{x + 1}{x - 2} > 0$
$\frac{x^2 - 4}{x + 3} \leq 0$
$\frac{2x}{x^2 - 9} \geq 1$

Key Differences from Polynomial Inequalities

Critical values come from TWO sources:

Zeros of the numerator (where the expression equals 0)
Zeros of the denominator (vertical asymptotes - where expression is undefined)

IMPORTANT: Values that make the denominator zero are NEVER included in the solution, even with $\leq$ or $\geq$ .

Solution Strategy

Step-by-Step Process

Step 1: Move everything to one side

Get the inequality in the form $\frac{P(x)}{Q(x)} \diamond 0$ where $\diamond$ is $<$ , $>$ , $\leq$ , or $\geq$ .

Warning: Never multiply both sides by the denominator (you don't know if it's positive or negative).

Step 2: Find critical values

Numerator zeros: Set $P(x) = 0$ and solve
Denominator zeros: Set $Q(x) = 0$ and solve

Step 3: Create intervals

The critical values divide the number line into regions.

Step 4: Make a sign chart

Test a point from each interval to determine the sign of the rational expression.

Step 5: Identify solution intervals

For $> 0$ or $\geq 0$ : positive intervals
For $< 0$ or $\leq 0$ : negative intervals

Step 6: Check endpoints

Include numerator zeros if using $\leq$ or $\geq$
Never include denominator zeros (always undefined)

Sign Chart for Rational Expressions

For $\frac{(x - a)(x - b)}{(x - c)}$ where $a < b < c$ :

| Interval | $(x-a)$ | $(x-b)$ | $(x-c)$ | $\frac{(x-a)(x-b)}{(x-c)}$ | |----------|---------|---------|---------|---------------------| | $x < a$ | $-$ | $-$ | $-$ | $-$ | | $a < x < b$ | $+$ | $-$ | $-$ | $+$ | | $b < x < c$ | $+$ | $+$ | $-$ | $-$ | | $x > c$ | $+$ | $+$ | $+$ | $+$ |

Note: At $x = c$ , the expression is undefined (vertical asymptote).

Common Mistake to Avoid

WRONG: Multiplying by the denominator without considering its sign

Example of wrong approach: $\frac{x + 1}{x - 2} > 0$ Wrong: Multiply by $(x - 2)$ to get $x + 1 > 0$

Why it's wrong: If $x - 2 < 0$ , multiplying reverses the inequality!

RIGHT: Use sign analysis on the rational expression as-is.

Converting to Standard Form

If the inequality is not in the form $\frac{P(x)}{Q(x)} \diamond 0$ , rearrange:

Example: $\frac{x}{x - 1} \geq 2$

Move everything to one side: $\frac{x}{x - 1} - 2 \geq 0$

Get common denominator: $\frac{x - 2(x - 1)}{x - 1} \geq 0$ $\frac{x - 2x + 2}{x - 1} \geq 0$ $\frac{-x + 2}{x - 1} \geq 0$

Now analyze this rational expression.

Special Considerations

Numerator Zeros vs. Denominator Zeros

Numerator zero: Expression equals 0 (may be included in solution)
Denominator zero: Expression is undefined (never in solution)

Use different notation:

Open circle ○ for denominator zeros (excluded)
Closed circle ● for numerator zeros with $\leq$ or $\geq$ (included)

Multiple Factors

Apply the same multiplicity rules as with polynomials:

Odd multiplicity: sign changes
Even multiplicity: sign stays the same

Simplifying First

Be careful: Canceling common factors can eliminate critical values!

Example: $\frac{(x - 2)(x + 1)}{x - 2} \geq 0$

If you cancel $(x - 2)$ , you get $x + 1 \geq 0$ , which gives $x \geq -1$ .

But: $x = 2$ must be excluded (makes original denominator 0).

Correct answer: $[-1, 2) \cup (2, \infty)$

Graphical Interpretation

The graph of $y = \frac{P(x)}{Q(x)}$ has:

Zeros at numerator zeros (crosses or touches x-axis)
Vertical asymptotes at denominator zeros

Solution to $\frac{P(x)}{Q(x)} > 0$ : where graph is above x-axis Solution to $\frac{P(x)}{Q(x)} < 0$ : where graph is below x-axis

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve $\frac{x + 1}{x - 2} > 0$ and express the solution in interval notation.

💡 Show Solution

Solution:

Given: $\frac{x + 1}{x - 2} > 0$

Step 1: Find critical values

Numerator zero: $x + 1 = 0 \Rightarrow x = -1$

Denominator zero: $x - 2 = 0 \Rightarrow x = 2$

Critical values: $x = -1, 2$

Step 2: Create intervals

The critical values divide the number line: $(-\infty, -1)$ , $(-1, 2)$ , $(2, \infty)$

Step 3: Make a sign chart

| Interval | Test Point | $(x+1)$ | $(x-2)$ | $\frac{x+1}{x-2}$ | |----------|-----------|---------|---------|------------| | $x < -1$ | $x = -2$ | $-$ | $-$ | $+$ | | $-1 < x < 2$ | $x = 0$ | $+$ | $-$ | $-$ | | $x > 2$ | $x = 3$ | $+$ | $+$ | $+$ |

Step 4: Identify where expression $> 0$

We need positive intervals:

$(-\infty, -1)$ : positive ✓
$(2, \infty)$ : positive ✓

Step 5: Check endpoints

At $x = -1$ : Expression equals 0, but we have $>$ (strict), so exclude
At $x = 2$ : Expression is undefined, so exclude

Answer: $(-\infty, -1) \cup (2, \infty)$

Verification:

At $x = -2$ : $\frac{-2+1}{-2-2} = \frac{-1}{-4} = \frac{1}{4} > 0$ ✓
At $x = 0$ : $\frac{0+1}{0-2} = \frac{1}{-2} = -\frac{1}{2} < 0$ ✓
At $x = 3$ : $\frac{3+1}{3-2} = \frac{4}{1} = 4 > 0$ ✓

2Problem 2medium

❓ Question:

Solve $\frac{x^2 - 4}{x + 3} \leq 0$ and express the solution in interval notation.

💡 Show Solution

Solution:

Given: $\frac{x^2 - 4}{x + 3} \leq 0$

Step 1: Factor the numerator $\frac{(x - 2)(x + 2)}{x + 3} \leq 0$

Step 2: Find critical values

Numerator zeros:

$x - 2 = 0 \Rightarrow x = 2$
$x + 2 = 0 \Rightarrow x = -2$

Denominator zero:

$x + 3 = 0 \Rightarrow x = -3$

Critical values: $x = -3, -2, 2$ (in order)

Step 3: Make a sign chart

| Interval | Test | $(x-2)$ | $(x+2)$ | $(x+3)$ | Expression | |----------|------|---------|---------|---------|------------| | $x < -3$ | $-4$ | $-$ | $-$ | $-$ | $-$ | | $-3 < x < -2$ | $-2.5$ | $-$ | $-$ | $+$ | $+$ | | $-2 < x < 2$ | $0$ | $-$ | $+$ | $+$ | $-$ | | $x > 2$ | $3$ | $+$ | $+$ | $+$ | $+$ |

Step 4: Identify where expression $\leq 0$

We need negative or zero:

$x < -3$ : negative ✓
$-2 < x < 2$ : negative ✓

Step 5: Check endpoints

At $x = -3$ : Undefined (denominator zero), exclude
At $x = -2$ : Expression equals 0, and we have $\leq$ , so include
At $x = 2$ : Expression equals 0, and we have $\leq$ , so include

Answer: $(-\infty, -3) \cup [-2, 2]$

Verification:

At $x = -4$ : $\frac{(-4)^2-4}{-4+3} = \frac{12}{-1} = -12 < 0$ ✓
At $x = -2$ : $\frac{(-2)^2-4}{-2+3} = \frac{0}{1} = 0$ ✓
At $x = 0$ : $\frac{0-4}{0+3} = \frac{-4}{3} < 0$ ✓
At $x = 2$ : $\frac{4-4}{2+3} = \frac{0}{5} = 0$ ✓

3Problem 3hard

❓ Question:

Solve $\frac{2x}{x^2 - 9} \geq 1$ and express the solution in interval notation.

💡 Show Solution

Solution:

Given: $\frac{2x}{x^2 - 9} \geq 1$

Step 1: Move everything to one side $\frac{2x}{x^2 - 9} - 1 \geq 0$

Step 2: Get common denominator $\frac{2x - (x^2 - 9)}{x^2 - 9} \geq 0$ $\frac{2x - x^2 + 9}{x^2 - 9} \geq 0$ $\frac{-x^2 + 2x + 9}{x^2 - 9} \geq 0$

Step 3: Factor

Numerator: $-x^2 + 2x + 9$

Using quadratic formula: $x = \frac{-2 \pm \sqrt{4 + 36}}{-2} = \frac{-2 \pm \sqrt{40}}{-2} = \frac{-2 \pm 2\sqrt{10}}{-2} = 1 \mp \sqrt{10}$

So numerator zeros are: $x = 1 - \sqrt{10} \approx -2.16$ and $x = 1 + \sqrt{10} \approx 4.16$

Denominator: $x^2 - 9 = (x - 3)(x + 3)$

Denominator zeros: $x = -3, 3$

Critical values (in order): $1 - \sqrt{10}, -3, 3, 1 + \sqrt{10}$

Approximately: $-2.16, -3, 3, 4.16$

Step 4: Make sign chart

| Interval | Numerator | Denominator | Expression | |----------|-----------|-------------|------------| | $x < 1-\sqrt{10}$ | $-$ | $+$ | $-$ | | $1-\sqrt{10} < x < -3$ | $+$ | $+$ | $+$ | | $-3 < x < 3$ | $+$ | $-$ | $-$ | | $3 < x < 1+\sqrt{10}$ | $+$ | $+$ | $+$ | | $x > 1+\sqrt{10}$ | $-$ | $+$ | $-$ |

Step 5: Identify where expression $\geq 0$

Positive or zero:

$[1 - \sqrt{10}, -3)$ : positive, include left endpoint
$(3, 1 + \sqrt{10}]$ : positive, include right endpoint

Never include $x = -3$ or $x = 3$ (undefined).

Answer: $[1 - \sqrt{10}, -3) \cup (3, 1 + \sqrt{10}]$

Or approximately: $[-2.16, -3) \cup (3, 4.16]$

Verification at $x = 0$ (should be negative): $\frac{2(0)}{0^2 - 9} = \frac{0}{-9} = 0 < 1$ So $\frac{2x}{x^2-9} - 1 = -1 < 0$ ✓

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