\(\frac{x+2}{x-3} > 0\)
Rational inequality: An inequality involving a rational expression (fraction with polynomials).
- Find critical points (zeros of numerator and denominator)
- Create sign chart with intervals
- Determine sign of rational expression in each interval
- Select intervals where inequality is satisfied
Numerator: \(x+2=0 \Rightarrow x=-2\)
Denominator: \(x-3=0 \Rightarrow x=3\)
We want \(\frac{x+2}{x-3} > 0\), so select intervals where expression is positive
\(x \in (-∞,-2) \cup (3,∞)\)
• Critical points: Zeros of numerator and undefined points of denominator
• Sign analysis: Test values in each interval
• Domain restriction: Exclude values where denominator is zero
\(\frac{x-1}{x+4} \leq 0\)
Less than or equal: Include zeros of numerator (where expression equals zero)
Numerator: \(x-1=0 \Rightarrow x=1\)
Denominator: \(x+4=0 \Rightarrow x=-4\)
We want \(\frac{x-1}{x+4} \leq 0\), so select intervals where expression is negative or zero
Include \(x=1\) (numerator zero) but exclude \(x=-4\) (denominator zero)
\(x \in (-4,1]\)
• Inequality with ≤: Include zeros of numerator
• Domain restriction: Exclude values where denominator is zero
• Sign analysis: Test values in each interval
\(\frac{2x-1}{x+3} \geq 1\)
Standard form: Move all terms to one side: \(\frac{2x-1}{x+3} - 1 \geq 0\)
\(\frac{2x-1}{x+3} - 1 \geq 0\)
\(\frac{2x-1}{x+3} - \frac{x+3}{x+3} \geq 0\)
\(\frac{(2x-1)-(x+3)}{x+3} \geq 0\)
\(\frac{x-4}{x+3} \geq 0\)
Numerator: \(x-4=0 \Rightarrow x=4\)
Denominator: \(x+3=0 \Rightarrow x=-3\)
We want \(\frac{x-4}{x+3} \geq 0\), so select intervals where expression is positive or zero
Include \(x=4\) (numerator zero) but exclude \(x=-3\) (denominator zero)
\(x \in (-∞,-3) \cup [4,∞)\)
• Standard form: Move all terms to one side before solving
• Combine fractions: Find common denominator to create single rational expression
• Sign analysis: Test values in each interval
\(\frac{x^2-4}{x-1} < 0\)
Factor numerator: \(x^2-4 = (x-2)(x+2)\)
\(x^2-4 = (x-2)(x+2)\)
So \(\frac{x^2-4}{x-1} = \frac{(x-2)(x+2)}{x-1}\)
Numerator zeros: \(x-2=0 \Rightarrow x=2\), \(x+2=0 \Rightarrow x=-2\)
Denominator zero: \(x-1=0 \Rightarrow x=1\)
We want \(\frac{(x-2)(x+2)}{x-1} < 0\), so select intervals where expression is negative
\(x \in (-∞,-2) \cup (1,2)\)
• Factor polynomials: Factor numerator and denominator completely
• Sign analysis: Consider sign of each factor in each interval
• Domain restriction: Exclude values where denominator is zero
\(\frac{x^2+x-6}{x^2-4} \geq 0\)
Factor both: Numerator: \(x^2+x-6 = (x+3)(x-2)\), Denominator: \(x^2-4 = (x-2)(x+2)\)
Numerator: \(x^2+x-6 = (x+3)(x-2)\)
Denominator: \(x^2-4 = (x-2)(x+2)\)
\(\frac{(x+3)(x-2)}{(x-2)(x+2)} = \frac{x+3}{x+2}\) for \(x \neq 2\)
Note: \(x=2\) makes both numerator and denominator zero, but still creates a restriction
Numerator zero: \(x+3=0 \Rightarrow x=-3\)
Denominator zero: \(x+2=0 \Rightarrow x=-2\)
Also consider \(x=2\) as a restriction
We want \(\frac{x^2+x-6}{x^2-4} \geq 0\), so select intervals where expression is positive or zero
Include \(x=-3\) (numerator zero) but exclude \(x=-2\) and \(x=2\) (denominator zero)
\(x \in (-∞,-3] \cup (-2,∞)\)
• Factor completely: Factor both numerator and denominator
• Cancel common factors: Note domain restrictions
• Domain restrictions: Exclude all values that make denominator zero
\(-1 < \frac{3x+1}{x-2} < 2\)
Compound inequality: Solve as two separate inequalities: \(\frac{3x+1}{x-2} > -1\) AND \(\frac{3x+1}{x-2} < 2\)
\(-1 < \frac{3x+1}{x-2} < 2\) becomes:
1) \(\frac{3x+1}{x-2} > -1\) AND
2) \(\frac{3x+1}{x-2} < 2\)
\(\frac{3x+1}{x-2} > -1\)
\(\frac{3x+1}{x-2} + 1 > 0\)
\(\frac{3x+1 + x-2}{x-2} > 0\)
\(\frac{4x-1}{x-2} > 0\)
Critical points: \(x=\frac{1}{4}, x=2\)
Solution: \(x \in (-∞,\frac{1}{4}) \cup (2,∞)\)
\(\frac{3x+1}{x-2} < 2\)
\(\frac{3x+1}{x-2} - 2 < 0\)
\(\frac{3x+1 - 2(x-2)}{x-2} < 0\)
\(\frac{3x+1 - 2x+4}{x-2} < 0\)
\(\frac{x+5}{x-2} < 0\)
Critical points: \(x=-5, x=2\)
Solution: \(x \in (-5,2)\)
Combine: \([(-∞,\frac{1}{4}) \cup (2,∞)] \cap (-5,2)\)
Result: \(x \in (-5,\frac{1}{4})\)
\(x \in (-5,\frac{1}{4})\)
• Compound inequalities: Solve each part separately
• Intersection: Take intersection of individual solutions
• Domain restrictions: Apply to all parts of the compound inequality
\(\frac{x^3-x}{x^2+1} \leq 0\)
Factor numerator: \(x^3-x = x(x^2-1) = x(x-1)(x+1)\)
\(x^3-x = x(x^2-1) = x(x-1)(x+1)\)
Denominator: \(x^2+1 > 0\) for all real \(x\)
Numerator zeros: \(x=0, x=1, x=-1\)
Since \(x^2+1 > 0\) always, denominator never zero
We want \(\frac{x^3-x}{x^2+1} \leq 0\), so select intervals where expression is negative or zero
Include zeros: \(x=-1, 0, 1\)
\(x \in (-\infty,-1] \cup [0,1]\)
• Always positive denominators: Don't affect sign of expression
• Factor completely: Factor numerator completely
• Include zeros: When inequality includes equality
\(\frac{x-a}{x-b} > 0\) where \(a < b\)
Parameterized inequality: Critical points are \(x=a\) and \(x=b\)
Numerator zero: \(x-a=0 \Rightarrow x=a\)
Denominator zero: \(x-b=0 \Rightarrow x=b\)
For \(x < a\): Both \(x-a < 0\) and \(x-b < 0\), so \(\frac{x-a}{x-b} > 0\)
For \(a < x < b\): \(x-a > 0\) and \(x-b < 0\), so \(\frac{x-a}{x-b} < 0\)
For \(x > b\): Both \(x-a > 0\) and \(x-b > 0\), so \(\frac{x-a}{x-b} > 0\)
We want \(\frac{x-a}{x-b} > 0\), so select intervals where expression is positive
\(x \in (-∞,a) \cup (b,∞)\)
• Parameterized problems: Treat parameters as constants
• Order matters: Consider relative positions of critical points
• Sign analysis: Test sign of each factor in each interval
\(\frac{(x-1)^2}{x+2} \geq 0\)
Squared terms: \((x-1)^2 \geq 0\) for all real \(x\), equals 0 only at \(x=1\)
\((x-1)^2 \geq 0\) always, and equals 0 only when \(x=1\)
This means the numerator is always non-negative
Numerator zero: \(x=1\) (but doesn't change sign since it's squared)
Denominator zero: \(x=-2\)
Since \((x-1)^2 \geq 0\) always, the sign depends on the denominator:
For \(x > -2\): denominator positive, so expression ≥ 0
For \(x < -2\): denominator negative, so expression ≤ 0
At \(x=1\): expression = 0, which satisfies ≥ 0
\(x \in (-2,∞)\)
• Squared terms: Always non-negative, don't change sign
• Zero numerators: Make entire expression zero
• Sign determination: Only denominator affects sign when numerator is always positive
\(\frac{x}{x-1} + \frac{2}{x+1} \leq 1\)
Combine fractions: Move all terms to one side and find common denominator
\(\frac{x}{x-1} + \frac{2}{x+1} - 1 \leq 0\)
\(\frac{x(x+1) + 2(x-1) - (x-1)(x+1)}{(x-1)(x+1)} \leq 0\)
\(\frac{x^2+x + 2x-2 - (x^2-1)}{(x-1)(x+1)} \leq 0\)
\(\frac{x^2+x + 2x-2 - x^2+1}{(x-1)(x+1)} \leq 0\)
\(\frac{3x-1}{(x-1)(x+1)} \leq 0\)
Wait, let me recalculate:
\(\frac{x^2+x + 2x-2 - x^2+1}{(x-1)(x+1)} = \frac{3x-1}{(x-1)(x+1)}\)
Actually, let's recalculate carefully:
\(\frac{x(x+1) + 2(x-1) - (x-1)(x+1)}{(x-1)(x+1)}\)
\(= \frac{x^2+x + 2x-2 - (x^2-1)}{(x-1)(x+1)}\)
\(= \frac{x^2+x + 2x-2 - x^2+1}{(x-1)(x+1)}\)
\(= \frac{3x-1}{(x-1)(x+1)}\)
This is incorrect. Let's recalculate properly:
\(\frac{x}{x-1} + \frac{2}{x+1} - 1 = \frac{x(x+1) + 2(x-1) - (x-1)(x+1)}{(x-1)(x+1)}\)
\(= \frac{x^2+x + 2x-2 - (x^2-1)}{(x-1)(x+1)}\)
\(= \frac{x^2+x + 2x-2 - x^2+1}{(x-1)(x+1)}\)
\(= \frac{3x-1}{(x-1)(x+1)}\)
Let me restart: \(\frac{x}{x-1} + \frac{2}{x+1} \leq 1\)
\(\frac{x(x+1) + 2(x-1)}{(x-1)(x+1)} \leq 1\)
\(\frac{x^2+x + 2x-2}{(x-1)(x+1)} \leq 1\)
\(\frac{x^2+3x-2}{(x-1)(x+1)} - 1 \leq 0\)
\(\frac{x^2+3x-2 - (x-1)(x+1)}{(x-1)(x+1)} \leq 0\)
\(\frac{x^2+3x-2 - (x^2-1)}{(x-1)(x+1)} \leq 0\)
\(\frac{x^2+3x-2 - x^2+1}{(x-1)(x+1)} \leq 0\)
\(\frac{3x-1}{(x-1)(x+1)} \leq 0\)
Numerator: \(3x-1=0 \Rightarrow x=\frac{1}{3}\)
Denominator: \((x-1)(x+1)=0 \Rightarrow x=1, x=-1\)
We want \(\frac{3x-1}{(x-1)(x+1)} \leq 0\), so select intervals where expression is negative or zero
Include \(x=\frac{1}{3}\) (numerator zero) but exclude \(x=-1,1\) (denominator zero)
\(x \in (-1,\frac{1}{3}] \cup (1,∞)\)
• Combine fractions: Find common denominator to create single rational expression
• Sign analysis: Test values in each interval defined by critical points
• Domain restrictions: Exclude values where denominator is zero
A rational inequality compares a rational expression to 0 or another value:
Values that make the denominator zero are excluded from the solution set.
Zeros of the numerator and undefined points of the denominator:
Test values in each interval between critical points to determine the sign of the rational expression.
Express solutions using proper interval notation, including or excluding endpoints based on inequality symbol.
Questions & Answers
Question: When solving rational inequalities, why do we need to consider the zeros of the denominator separately from the zeros of the numerator?
Answer: This is a crucial distinction that affects both the solution process and the final answer:
- Zeros of the numerator: These make the rational expression equal to zero. If your inequality includes equality (≥ or ≤), these points are included in the solution.
- Zeros of the denominator: These make the rational expression undefined. These points are never included in the solution, regardless of the inequality symbol.
For example, in \(\frac{x-2}{x+3} \geq 0\):
- \(x = 2\) (numerator zero) is included because the inequality allows equality
- \(x = -3\) (denominator zero) is excluded because the expression is undefined there
These points also serve as boundary points that divide the number line into intervals for sign analysis.
Question: How do I handle rational inequalities that involve squared terms in the numerator or denominator?
Answer: Squared terms behave differently in sign analysis:
- Squared terms in numerator: \((x-a)^2 \geq 0\) for all real \(x\), and equals zero only at \(x = a\). This means they don't change the sign of the rational expression.
- Squared terms in denominator: \((x-a)^2 > 0\) for all \(x \neq a\), and undefined at \(x = a\). Again, they don't change the sign of the expression.
For example, in \(\frac{(x-1)^2}{x+2} \geq 0\):
- \((x-1)^2\) is always non-negative (positive except at \(x=1\) where it's zero)
- So the sign of the expression depends only on the denominator \(x+2\)
- The expression is positive when \(x > -2\) and negative when \(x < -2\)
Key insight: Squared factors don't change sign, so focus on non-squared factors for sign determination.
Question: What's the best approach when solving compound rational inequalities like \(-2 < \frac{x+1}{x-3} < 4\)?
Answer: The most effective approach is to split the compound inequality into two separate inequalities:
For \(-2 < \frac{x+1}{x-3} < 4\), solve:
- \(-2 < \frac{x+1}{x-3}\) which becomes \(\frac{x+1}{x-3} + 2 > 0\) or \(\frac{3x-5}{x-3} > 0\)
- \(\frac{x+1}{x-3} < 4\) which becomes \(\frac{x+1}{x-3} - 4 < 0\) or \(\frac{-3x+13}{x-3} < 0\)
Then solve each rational inequality separately using standard techniques (find critical points, create sign charts). Finally, take the intersection of the two solution sets since both conditions must be satisfied simultaneously.
Important: Maintain the same domain restrictions throughout both parts of the compound inequality. The intersection gives you the final solution that satisfies both parts of the original compound inequality.