CSCA MATHEMATICS · MODULE 01 · LESSON 01.6

Rational Inequalities

Domain restrictions, critical values, sign charts, factorization, and full solution sets.

Learning goals

  • Find excluded denominator zeros before solving.
  • Construct and interpret rational sign charts.
  • Handle strict and non-strict inequalities, repeated factors, and canceled factors.
  • Write complete solutions using interval notation and verify domain restrictions.

01.6.1 Introduction to Rational Inequalities

A rational expression is the quotient of two polynomials:

\[R(x)=\frac{P(x)}{Q(x)},\qquad Q(x)\neq0\]

Here, \(P(x)\) is the numerator and \(Q(x)\) is the denominator. The denominator cannot be zero.

Rational inequalities compare rational expressions with another expression using \(>\), \(<\), \(\geq\), or \(\leq\).

\[\frac{P(x)}{Q(x)}>0\qquad \frac{P(x)}{Q(x)}<0\]
\[\frac{P(x)}{Q(x)}\geq0\qquad \frac{P(x)}{Q(x)}\leq0\]

Examples A–D · Recognize rational inequalities

\[\frac{x-3}{x+2}>0\]
\[\frac{2x+1}{x-4}\leq0\]
\[\frac{x^2-9}{x+1}\geq0\]
\[\frac{1}{x-2}>3\]

The last inequality must be rearranged with zero on one side before performing sign analysis.

Rational equations versus inequalities

\[\frac{x-2}{x+3}=0\quad\Rightarrow\quad x=2\]

For an equation, a fraction equals zero when its numerator is zero and the denominator is nonzero. For an inequality, we must identify where the whole fraction has the requested sign.

\[\frac{x-2}{x+3}>0\]
\[x<-3\quad\text{or}\quad x>2\]

Key concept

Rational inequalities typically describe entire intervals, not just isolated roots.

01.6.2 Understanding Domain Restrictions

Non-negotiable rule

Every value that makes an ORIGINAL denominator zero must be excluded, even after simplifying a fraction.

Example 1 · One excluded value

\[\frac{x+4}{x-3}\]

Set the denominator to zero:

\[x-3=0\quad\Rightarrow\quad x=3\]

Domain:

\[(-\infty,3)\cup(3,\infty)\]

Example 2 · Multiple excluded values

\[\frac{x+2}{(x-1)(x+4)}\]

The denominator is zero at \(x=1\) and \(x=-4\).

\[x\neq1,\qquad x\neq-4\]
\[(-\infty,-4)\cup(-4,1)\cup(1,\infty)\]

Zeros versus undefined points

\[R(x)=\frac{x-2}{x+3}\]
Value Reason Status
\(x=2\) Numerator = 0 Fraction equals zero
\(x=-3\) Denominator = 0 Undefined — excluded
\[R(2)=\frac{0}{5}=0\]
\[R(-3)=\frac{-5}{0}\quad\text{undefined}\]

Endpoint rule

A numerator zero may be included when equality is permitted and the denominator is nonzero. A denominator zero is never included.

01.6.3 Understanding the Sign of a Fraction

For \(B
eq0\), the sign of \(A/B\) depends on the signs of numerator and denominator.

Numerator Denominator Fraction
Positive Positive Positive
Negative Negative Positive
Positive Negative Negative
Negative Positive Negative
Zero Nonzero Zero
Any Zero Undefined

Example 3 · Positive fractions

\[\frac63=2>0\]
\[\frac{-6}{-3}=2>0\]

Same signs yield a positive quotient.

Example 4 · Negative fractions

\[\frac{-6}{3}=-2<0\]
\[\frac{6}{-3}=-2<0\]

Opposite signs yield a negative quotient.

Sign test

For \(B
eq0\), \(A/B>0\) requires equal signs, whereas \(A/B<0\) requires opposite signs.

01.6.4 Critical Values and Sign Charts

Critical values are real zeros of the numerator or denominator. They divide the real line into intervals where the rational function has a fixed sign.

Solving workflow

  1. Rewrite with zero on one side.
  2. Identify original denominator restrictions.
  3. Combine into one fraction and factor where possible.
  4. Find numerator zeros and denominator zeros.
  5. Arrange critical values in increasing order.
  6. Test one value in each open interval.
  7. Select the intervals satisfying the inequality.
  8. Check endpoints: permitted numerator zeros may be included; all denominator zeros remain excluded.

Example 5 · Solve a basic rational inequality

\[\frac{x-2}{x+3}>0\]

Numerator zero: \(x=2\). Denominator zero: \(x=-3\) (excluded).

\[-3<2\]

The test intervals are \(( -\infty,-3)\), \((-3,2)\), and \((2,\infty)\).

Interval Test value Numerator Denominator Fraction
(−∞, −3) −4 − − Positive
(−3, 2) 0 − + Negative
(2, ∞) 3 + + Positive

Choose positive regions. Both critical values are excluded: −3 makes the denominator zero; 2 makes the fraction zero and the inequality is strict.

Final answer

\[(-\infty,-3)\cup(2,\infty)\]

01.6.5 Solving Non-Strict Rational Inequalities

The symbols \(\geq\) and \(\leq\) permit equality. A fraction equals zero only if the numerator is zero and the denominator is nonzero.

Example 6 · Less than or equal to zero

\[\frac{x-4}{x+1}\leq0\]

Numerator zero: \(x=4\). Denominator zero: \(x=-1\) (excluded).

Test the intervals separated by −1 and 4. The fraction is positive on the two outer intervals and negative between −1 and 4.

The zero at 4 is allowed because equality is permitted; −1 remains excluded.

Final answer

\[(-1,4]\]

Strict versus non-strict

Inequality Solution
\(\frac{x-4}{x+1}>0\) \(( -\infty,-1)\cup(4,\infty)\)
\(\frac{x-4}{x+1}\geq0\) \(( -\infty,-1)\cup[4,\infty)\)
\(\frac{x-4}{x+1}<0\) \((-1,4)\)
\(\frac{x-4}{x+1}\leq0\) \((-1,4]\)

Remember

The excluded denominator value \(x=-1\) is never included, even in non-strict inequalities.

01.6.6 Quadratic Numerators

Factorization methods from Lesson 01.5 also apply when the numerator is quadratic.

Example 7 · Factor a quadratic numerator

\[\frac{x^2-x-2}{x-3}\geq0\]

Factor:

\[\frac{(x-2)(x+1)}{x-3}\geq0\]

Critical values are −1, 2, and 3. Only 3 is a denominator zero.

Interval Test x Numerator sign Denominator sign Result
(−∞, −1) −2 + − −
(−1, 2) 0 − − +
(2, 3) 2.5 + − −
(3, ∞) 4 + + +

Select the positive intervals and include numerator zeros −1 and 2. Exclude denominator zero 3.

Final answer

\[[-1,2]\cup(3,\infty)\]

01.6.7 Why You Cannot Multiply by an Unknown Denominator

Common high-stakes error

Multiplication by a positive factor preserves an inequality; multiplication by a negative factor reverses it. The sign of a variable expression may be unknown.
\[\frac{1}{x-2}>1\]

Multiplying immediately by \(x-2\) while keeping the sign is not justified because \(x-2\) might be negative.

Correct method · Combine first

\[\frac{1}{x-2}-1>0\]
\[\frac{1-(x-2)}{x-2}>0\]
\[\frac{3-x}{x-2}>0\]

Critical values: 2 (excluded denominator zero) and 3 (numerator zero).

Interval Numerator Denominator Fraction
x < 2 + − Negative
2 < x < 3 + + Positive
x > 3 − + Negative

Final answer

\[(2,3)\]

You may multiply by a variable expression only after establishing its sign or separating all sign cases.

01.6.8 Multiple Rational Expressions

Combine fractions on both sides into one rational expression before sign analysis.

Example 8 · Fractions on both sides

\[\frac1{x-1}\geq\frac2{x+2}\]

Original restrictions: \(x\neq1\) and \(x\neq-2\).

\[\frac1{x-1}-\frac2{x+2}\geq0\]
\[\frac{(x+2)-2(x-1)}{(x-1)(x+2)}\geq0\]
\[\frac{4-x}{(x-1)(x+2)}\geq0\]

Critical values: −2, 1 (excluded) and 4 (possible zero).

Interval Test x Sign
(−∞, −2) −3 Positive
(−2, 1) 0 Negative
(1, 4) 2 Positive
(4, ∞) 5 Negative

Include 4, exclude −2 and 1.

Final answer

\[(-\infty,-2)\cup(1,4]\]

01.6.9 Quadratic Denominators

Factor quadratic denominators to find all excluded values and build a complete sign chart.

Example 9 · Factored quadratic denominator

\[\frac{x-2}{x^2+2x-3}<0\]
\[x^2+2x-3=(x+3)(x-1)\]
\[\frac{x-2}{(x+3)(x-1)}<0\]

Critical values: −3, 1 (both denominator zeros), 2 (numerator zero).

Interval x−2 x+3 x−1 Overall
x < −3 − − − Negative
−3 < x < 1 − + − Positive
1 < x < 2 − + + Negative
x > 2 + + + Positive

Final answer

\[(-\infty,-3)\cup(1,2)\]

All three critical values are excluded here: two are undefined and the third gives zero while the inequality is strict.

01.6.10 Canceling Common Factors and Removable Discontinuities

Canceling a common factor simplifies the formula only on its original domain. The denominator restriction remains.

Example 10 · A canceled factor

\[\frac{x^2-1}{x-1}\geq0\]

Original restriction: \(x\neq1\).

\[\frac{(x-1)(x+1)}{x-1}\geq0\]

For allowed x, this simplifies to:

\[x+1\geq0\]
\[x\geq-1\]

But x=1 is still undefined in the original expression: \(0/0\).

Final answer

\[[-1,1)\cup(1,\infty)\]
Situation Interpretation
Factor cancels Simplifies only on the original domain
Original denominator = 0 Value remains excluded
Simplified expression is defined there Original expression still has a hole

01.6.11 Repeated Factors and Sign Changes

An even-power factor does not reverse sign when crossing its zero. An odd-power factor does.

Example 11 · Repeated numerator factor

\[\frac{(x-2)^2}{x+1}>0\]

Numerator zero: 2. Denominator zero: −1.

The squared numerator is positive for every x except 2; at 2 it is zero. The denominator is positive for x > −1.

Region Numerator Denominator Fraction
x < −1 Positive Negative Negative
−1 < x < 2 Positive Positive Positive
x = 2 Zero Positive Zero
x > 2 Positive Positive Positive

Final answer

\[(-1,2)\cup(2,\infty)\]
Power of factor Sign on either side
Odd Factor changes sign
Even Factor does not change sign

At an even-power denominator zero, the rational expression is undefined even if its sign is unchanged across that point.

01.6.12 Special Cases

Case 1 · No solution

\[\frac{x^2+4}{x^2+1}<0\]

Both numerator and denominator are strictly positive for every real x. Thus the quotient cannot be negative.

Solution set: ∅

Case 2 · Exactly one solution

\[\frac{(x-1)^2}{x^2+1}\leq0\]

The denominator is always positive, and the numerator is nonnegative. Equality holds only at x=1.

\[\{1\}\]

Case 3 · All real numbers except an excluded point

\[\frac{x^2+1}{(x-2)^2}\geq0\]

The quotient is positive everywhere it is defined; x=2 is excluded.

\[(-\infty,2)\cup(2,\infty)\]
Case Solution type
Never satisfies inequality Empty set
Equality at one permitted point Singleton set
Always true on domain Entire domain
Changes sign One or more intervals

01.6.13 Alternative: Squared-Denominator Method

For \(Q(x)
eq0\), the square \([Q(x)]^2\) is positive. Multiplying by it does not reverse the inequality.

\[\frac{P(x)}{Q(x)}>0\quad\Longleftrightarrow\quad P(x)Q(x)>0,\quad Q(x)\neq0\]

Example 12 · Multiply by a positive square

\[\frac{x-1}{x+2}\geq0\]

Restriction: \(x\neq-2\). Multiply both sides by \((x+2)^2>0\) for allowed x:

\[(x-1)(x+2)\geq0\]
\[x\leq-2\quad\text{or}\quad x\geq1\]

Exclude the forbidden point −2.

Final answer

\[(-\infty,-2)\cup[1,\infty)\]

Exam strategy

Use sign charts as your default method. The squared-denominator method can shorten calculations, but never discard the original denominator restrictions.

01.6.14 Graphical Interpretation

\[y=\frac{x-2}{x+3}\]

The numerator zero at \(x=2\) gives x-intercept \((2,0)\). The denominator zero at \(x=-3\) gives a vertical asymptote in this example.

The inequality \(R(x)>0\) corresponds to portions of the graph above the x-axis; \(R(x)<0\) corresponds to portions below it.

Inequality Graph interpretation
\(R(x)>0\) Above the x-axis
\(R(x)<0\) Below the x-axis
\(R(x)\geq0\) Above or on the x-axis
\(R(x)\leq0\) Below or on the x-axis

Graph caution

Undefined points are excluded in every inequality. A canceled denominator factor may produce a hole rather than a vertical asymptote.

01.6.15 Word Problems

Rational inequalities arise in time, speed, cost, efficiency, and other ratio-based situations.

Example 13 · Travel time

A student must travel 120 kilometers in less than 3 hours. Let \(v\) be the average speed in km/h.

Physical restriction: \(v>0\).

\[t=\frac{120}{v}\]
\[\frac{120}{v}<3\]

Since v is positive, multiplying by v keeps the sign unchanged.

\[120<3v\]
\[v>40\]

The speed must be strictly greater than 40 km/h; at 40 km/h the trip takes exactly 3 hours.

01.6.16 Common Mistakes to Avoid

1 · Including denominator zeros

For \(\frac{x-1}{x+2}\geq0\), the forbidden point \(x=-2\) must not be included. Correct solution: \((-\infty,-2)\cup[1,\infty)\).

2 · Multiplying by an unknown-sign expression

For \(\frac1{x-3}>2\), multiplying by \(x-3\) requires knowing its sign. Combine into \(\frac{1-2(x-3)}{x-3}>0\) instead.

3 · Canceling and forgetting a hole

\(\frac{(x-1)(x+3)}{x-1}>0\) simplifies to \(x+3>0\) only for \(x\neq1\). Solution: \((-3,1)\cup(1,\infty)\).

4 · Automatically alternating signs

An even-power factor such as \((x-2)^2\) has the same sign on either side of x=2.

5 · Finding numerator zeros only

Always identify denominator zeros and determine signs on each interval.

6 · Forgetting original domain after simplifying

Reduced formulas do not restore values excluded by the original denominator.

7 · Confusing AND with OR

Use union for values satisfying at least one condition; use intersection when both conditions must hold.

01.6.17 General Strategy for Solving Rational Inequalities

  1. Move all terms to one side.
  2. Record all ORIGINAL denominator restrictions.
  3. Combine into one rational expression and factor.
  4. Find zeros of numerator and denominator (including canceled factors).
  5. Sort the critical values.
  6. Determine the sign on each interval.
  7. Select the intervals meeting the original inequality.
  8. Include permitted zeros where equality is allowed; exclude every forbidden point; write and verify the result.
Situation Efficient method
One factor in each position Basic sign chart
Many factors Factor sign table
Quadratic numerator/denominator Factor and test signs
Fractions on both sides Combine into one fraction
Canceling factors Simplify but retain excluded values
Repeated factors Account for even/odd powers
Predictable signs Use square properties
Known denominator sign Valid multiplication rules

01.6.18 Lesson Summary

  1. Every denominator zero is excluded from the original domain.
  2. A numerator zero can be included only for non-strict inequalities and when the denominator is nonzero.
  3. Same signs give a positive quotient; opposite signs give a negative quotient.
  4. Critical values split the number line into intervals for sign analysis.
  5. Test one point per interval and account for repeated factors.
  6. Never multiply by an expression with unknown sign without case analysis.
  7. Canceled factors do not remove original domain restrictions.
  8. State the complete solution set in interval notation.

For \(B
eq0\), the essential sign rules are:

\[\frac{A}{B}>0\quad\text{for same nonzero signs}\]
\[\frac{A}{B}<0\quad\text{for opposite signs}\]
\[\frac{A}{B}=0\quad\text{if }A=0\text{ and }B\neq0\]

Final reminder

Finding zeros, determining signs, and checking domain restrictions are separate requirements. Complete all three before finalizing your answer.
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