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Rational Inequalities

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

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Rational Inequalities

Introduction

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

Examples:

  • x+1x−2>0\frac{x + 1}{x - 2} > 0
  • x2−4x+3≤0\frac{x^2 - 4}{x + 3} \leq 0
  • 2xx2−9≥1\frac{2x}{x^2 - 9} \geq 1

Key Differences from Polynomial Inequalities

Critical values come from TWO sources:

  1. Zeros of the numerator (where the expression equals 0)
  2. 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 P(x)Q(x)⋄0\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)=0P(x) = 0 and solve
  • Denominator zeros: Set Q(x)=0Q(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> 0 or ≥0\geq 0: positive intervals
  • For <0< 0 or ≤0\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 (x−a)(x−b)(x−c)\frac{(x - a)(x - b)}{(x - c)} where a<b<ca < b < c:

Interval(x−a)(x-a)(x−b)(x-b)(x−c)(x-c)(x−a)(x−b)(x−c)\frac{(x-a)(x-b)}{(x-c)}
x<ax < a−-−-−-−-
a<x<ba < x < b++−-−-++
b<x<cb < x < c++++−-−-
x>cx > c++++++++

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

Common Mistake to Avoid

WRONG: Multiplying by the denominator without considering its sign

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

Why it's wrong: If x−2<0x - 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 P(x)Q(x)⋄0\frac{P(x)}{Q(x)} \diamond 0, rearrange:

Example: xx−1≥2\frac{x}{x - 1} \geq 2

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

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

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

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

Correct answer: [−1,2)∪(2,∞)[-1, 2) \cup (2, \infty)

Graphical Interpretation

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

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

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

📚 Practice Problems

1Problem 1easy

❓ Question:

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

💡 Show Solution

Solution:

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

Step 1: Find critical values

Numerator zero: x+1=0⇒x=−1x + 1 = 0 \Rightarrow x = -1

Denominator zero: x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2

Critical values: x=−1,2x = -1, 2

Step 2: Create intervals

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

Step 3: Make a sign chart

IntervalTest Point(x+1)(x+1)(x−2)(x-2)x+1x−2\frac{x+1}{x-2}
x<−1x < -1x=−2x = -2−-−-++
−1<x<2-1 < x < 2x=0x = 0++−-−-
x>2x > 2x=3x = 3++++++

Step 4: Identify where expression >0> 0

We need positive intervals:

  • (−∞,−1)(-\infty, -1): positive ✓
  • (2,∞)(2, \infty): positive ✓

Step 5: Check endpoints

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

Answer: (−∞,−1)∪(2,∞)(-\infty, -1) \cup (2, \infty)

Verification:

  • At x=−2x = -2: −2+1−2−2=−1−4=14>0\frac{-2+1}{-2-2} = \frac{-1}{-4} = \frac{1}{4} > 0 ✓
  • At x=0x = 0: 0+10−2=1−2=−12<0\frac{0+1}{0-2} = \frac{1}{-2} = -\frac{1}{2} < 0 ✓
  • At x=3x = 3: 3+13−2=41=4>0\frac{3+1}{3-2} = \frac{4}{1} = 4 > 0 ✓

2Problem 2medium

❓ Question:

Solve x2−4x+3≤0\frac{x^2 - 4}{x + 3} \leq 0 and express the solution in interval notation.

💡 Show Solution

Solution:

Given: x2−4x+3≤0\frac{x^2 - 4}{x + 3} \leq 0

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

Step 2: Find critical values

Numerator zeros:

  • x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2
  • x+2=0⇒x=−2x + 2 = 0 \Rightarrow x = -2

Denominator zero:

  • x+3=0⇒x=−3x + 3 = 0 \Rightarrow x = -3

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

Step 3: Make a sign chart

IntervalTest(x−2)(x-2)(x+2)(x+2)(x+3)(x+3)Expression
x<−3x < -3−4-4−-−-−-−-
−3<x<−2-3 < x < -2−2.5-2.5−-−-++++
−2<x<2-2 < x < 200−-++++−-
x>2x > 233++++++++

Step 4: Identify where expression ≤0\leq 0

We need negative or zero:

  • x<−3x < -3: negative ✓
  • −2<x<2-2 < x < 2: negative ✓

Step 5: Check endpoints

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

Answer: (−∞,−3)∪[−2,2](-\infty, -3) \cup [-2, 2]

Verification:

  • At x=−4x = -4: (−4)2−4−4+3=12−1=−12<0\frac{(-4)^2-4}{-4+3} = \frac{12}{-1} = -12 < 0 ✓
  • At x=−2x = -2: (−2)2−4−2+3=01=0\frac{(-2)^2-4}{-2+3} = \frac{0}{1} = 0 ✓
  • At x=0x = 0: 0−40+3=−43<0\frac{0-4}{0+3} = \frac{-4}{3} < 0 ✓
  • At x=2x = 2: 4−42+3=05=0\frac{4-4}{2+3} = \frac{0}{5} = 0 ✓

3Problem 3hard

❓ Question:

Solve 2xx2−9≥1\frac{2x}{x^2 - 9} \geq 1 and express the solution in interval notation.

💡 Show Solution

Solution:

Given: 2xx2−9≥1\frac{2x}{x^2 - 9} \geq 1

Step 1: Move everything to one side 2xx2−9−1≥0\frac{2x}{x^2 - 9} - 1 \geq 0

Step 2: Get common denominator 2x−(x2−9)x2−9≥0\frac{2x - (x^2 - 9)}{x^2 - 9} \geq 0 2x−x2+9x2−9≥0\frac{2x - x^2 + 9}{x^2 - 9} \geq 0 −x2+2x+9x2−9≥0\frac{-x^2 + 2x + 9}{x^2 - 9} \geq 0

Step 3: Factor

Numerator: −x2+2x+9-x^2 + 2x + 9

Using quadratic formula: x=−2±4+36−2=−2±40−2=−2±210−2=1∓10x = \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=1−10≈−2.16x = 1 - \sqrt{10} \approx -2.16 and x=1+10≈4.16x = 1 + \sqrt{10} \approx 4.16

Denominator: x2−9=(x−3)(x+3)x^2 - 9 = (x - 3)(x + 3)

Denominator zeros: x=−3,3x = -3, 3

Critical values (in order): 1−10,−3,3,1+101 - \sqrt{10}, -3, 3, 1 + \sqrt{10}

Approximately: −2.16,−3,3,4.16-2.16, -3, 3, 4.16

Step 4: Make sign chart

IntervalNumeratorDenominatorExpression
x<1−10x < 1-\sqrt{10}−-++−-
1−10<x<−31-\sqrt{10} < x < -3++++++
−3<x<3-3 < x < 3++−-−-
3<x<1+103 < x < 1+\sqrt{10}++++++
x>1+10x > 1+\sqrt{10}−-++−-

Step 5: Identify where expression ≥0\geq 0

Positive or zero:

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

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

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

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

Verification at x=0x = 0 (should be negative): 2(0)02−9=0−9=0<1\frac{2(0)}{0^2 - 9} = \frac{0}{-9} = 0 < 1 So 2xx2−9−1=−1<0\frac{2x}{x^2-9} - 1 = -1 < 0 ✓

Explain using:

⚠️ Common Mistakes: Rational Inequalities

Avoid these 4 frequent errors

🌍 Real-World Applications: Rational Inequalities

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values

📌 Related Topics in Polynomial and Rational Functions

❓ Frequently Asked Questions

What is Rational Inequalities?▾
Solve rational inequalities by finding critical values from zeros and vertical asymptotes, then using sign analysis.
How can I study Rational Inequalities effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
Is this Rational Inequalities study guide free?▾
Yes — all study notes, flashcards, and practice problems for Rational Inequalities on Study Mondo are free to access. No account is needed.
What course covers Rational Inequalities?▾
Rational Inequalities is part of the AP Precalculus course on Study Mondo, specifically in the Polynomial and Rational Functions section. You can explore the full course for more related topics and practice resources.
Are there practice problems for Rational Inequalities?▾
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.