Rational Functions
Complete Notes & Formulae for Twelfth Grade (Precalculus)
1. Rational Function Definition
A rational function is the ratio of two polynomials:
\[ f(x) = \frac{P(x)}{Q(x)} \]
where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) \neq 0 \)
2. Asymptotes and Excluded Values
Excluded Values (Domain Restrictions):
Values that make the denominator zero
To Find:
1. Set denominator \( Q(x) = 0 \)
2. Solve for x
3. These values are excluded from the domain
Vertical Asymptotes:
Vertical lines where the function approaches ±∞
To Find:
1. Factor both numerator and denominator completely
2. Cancel any common factors
3. Set remaining denominator factors equal to zero
4. These x-values give vertical asymptotes: \( x = a \)
Example:
\( f(x) = \frac{x + 1}{(x - 2)(x + 3)} \)
Vertical asymptotes: \( x = 2 \) and \( x = -3 \)
Horizontal Asymptotes:
Horizontal lines that the function approaches as \( x \to \pm\infty \)
Rules (compare degrees of numerator and denominator):
| Condition | Horizontal Asymptote |
|---|---|
| Degree of numerator < Degree of denominator | \( y = 0 \) |
| Degree of numerator = Degree of denominator | \( y = \frac{a}{b} \) (ratio of leading coefficients) |
| Degree of numerator > Degree of denominator | No horizontal asymptote (may have slant asymptote) |
Examples:
• \( f(x) = \frac{3x + 1}{x^2 - 4} \): \( y = 0 \) (degree 1 < degree 2)
• \( f(x) = \frac{2x^2 + 5}{3x^2 - 1} \): \( y = \frac{2}{3} \) (same degrees)
• \( f(x) = \frac{x^3 + 1}{x^2 - 4} \): No horizontal asymptote (degree 3 > degree 2)
Holes (Removable Discontinuities):
Points where factors cancel from numerator and denominator
To Find:
1. Factor completely
2. Identify common factors that cancel
3. Set canceled factor equal to zero to find x-coordinate
4. Substitute x-value into simplified function to find y-coordinate
Example:
\( f(x) = \frac{(x - 2)(x + 1)}{(x - 2)(x - 3)} = \frac{x + 1}{x - 3} \) (after canceling)
Hole at \( x = 2 \): \( y = \frac{2 + 1}{2 - 3} = \frac{3}{-1} = -3 \)
Hole at point: (2, -3)
Vertical asymptote: \( x = 3 \)
3. Solve Rational Equations
LCD Method:
Steps:
1. Identify restrictions: Find values that make any denominator zero (these cannot be solutions)
2. Find the LCD: Least Common Denominator of all fractions
3. Multiply both sides by LCD: This eliminates all denominators
4. Solve the resulting equation: Usually polynomial or linear
5. Check solutions: Verify answers don't make any denominator zero
6. Reject extraneous solutions: Discard any that violate restrictions
Example 1 (Simple):
Solve: \( \frac{3}{x} + \frac{2}{x - 1} = 1 \)
Step 1: Restrictions: \( x \neq 0, x \neq 1 \)
Step 2: LCD = \( x(x - 1) \)
Step 3: Multiply by LCD:
\( x(x-1) \cdot \frac{3}{x} + x(x-1) \cdot \frac{2}{x-1} = x(x-1) \cdot 1 \)
\( 3(x-1) + 2x = x(x-1) \)
Step 4: Solve:
\( 3x - 3 + 2x = x^2 - x \)
\( 5x - 3 = x^2 - x \)
\( 0 = x^2 - 6x + 3 \)
Use quadratic formula: \( x = \frac{6 \pm \sqrt{36 - 12}}{2} = \frac{6 \pm \sqrt{24}}{2} = 3 \pm \sqrt{6} \)
Step 5: Check: Neither solution equals 0 or 1 ✓
Solutions: \( x = 3 + \sqrt{6} \) and \( x = 3 - \sqrt{6} \)
Example 2 (Extraneous Solution):
Solve: \( \frac{x}{x - 2} = \frac{2}{x - 2} + 1 \)
Step 1: Restriction: \( x \neq 2 \)
Step 2: LCD = \( x - 2 \)
Step 3: Multiply by LCD:
\( x = 2 + (x - 2) \)
\( x = 2 + x - 2 \)
\( x = x \)
This is true for all x, BUT x ≠ 2 is restricted!
Solution: All real numbers except x = 2
4. Check Whether Two Rational Functions Are Inverses
Inverse Function Verification:
Two functions \( f \) and \( g \) are inverses if and only if:
\[ f(g(x)) = x \quad \text{AND} \quad g(f(x)) = x \]
Steps:
1. Compute \( f(g(x)) \) by substituting g(x) into f
2. Simplify completely
3. Check if result equals x
4. Compute \( g(f(x)) \) by substituting f(x) into g
5. Simplify completely
6. Check if result equals x
7. If BOTH equal x, they are inverses; otherwise, they are NOT
Example 1 (ARE Inverses):
Verify: \( f(x) = \frac{2x}{x + 1} \) and \( g(x) = \frac{x}{2 - x} \)
Check 1: \( f(g(x)) \)
\( f(g(x)) = f\left(\frac{x}{2-x}\right) = \frac{2 \cdot \frac{x}{2-x}}{\frac{x}{2-x} + 1} \)
\( = \frac{\frac{2x}{2-x}}{\frac{x + 2 - x}{2-x}} = \frac{\frac{2x}{2-x}}{\frac{2}{2-x}} \)
\( = \frac{2x}{2-x} \cdot \frac{2-x}{2} = \frac{2x}{2} = x \) ✓
Check 2: \( g(f(x)) \)
\( g(f(x)) = g\left(\frac{2x}{x+1}\right) = \frac{\frac{2x}{x+1}}{2 - \frac{2x}{x+1}} \)
\( = \frac{\frac{2x}{x+1}}{\frac{2(x+1) - 2x}{x+1}} = \frac{\frac{2x}{x+1}}{\frac{2}{x+1}} \)
\( = \frac{2x}{x+1} \cdot \frac{x+1}{2} = \frac{2x}{2} = x \) ✓
Both compositions equal x → They ARE inverses!
Example 2 (NOT Inverses):
Verify: \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{x + 1} \)
Check 1: \( f(g(x)) \)
\( f(g(x)) = f\left(\frac{1}{x+1}\right) = \frac{1}{\frac{1}{x+1}} = x + 1 \)
This equals \( x + 1 \), NOT x ✗
Since \( f(g(x)) \neq x \), they are NOT inverses!
(No need to check the second composition)
5. Finding the Inverse of a Rational Function
Steps:
1. Replace \( f(x) \) with y
2. Switch x and y
3. Solve for y (this may require cross-multiplication)
4. Replace y with \( f^{-1}(x) \)
Example:
Find inverse of: \( f(x) = \frac{3x - 1}{x + 2} \)
Step 1: \( y = \frac{3x - 1}{x + 2} \)
Step 2: \( x = \frac{3y - 1}{y + 2} \)
Step 3: Solve for y:
\( x(y + 2) = 3y - 1 \)
\( xy + 2x = 3y - 1 \)
\( xy - 3y = -2x - 1 \)
\( y(x - 3) = -2x - 1 \)
\( y = \frac{-2x - 1}{x - 3} = \frac{-(2x + 1)}{x - 3} \)
Step 4: \( f^{-1}(x) = \frac{-2x - 1}{x - 3} \)
6. Quick Reference Summary
Key Concepts:
Rational Function: \( f(x) = \frac{P(x)}{Q(x)} \)
Vertical Asymptote: Set simplified denominator = 0
Horizontal Asymptote:
• deg(num) < deg(den): y = 0
• deg(num) = deg(den): y = ratio of leading coefficients
• deg(num) > deg(den): No horizontal asymptote
Holes: Common factors that cancel
Solving: Multiply by LCD, check for extraneous solutions
Inverse Verification: \( f(g(x)) = x \) AND \( g(f(x)) = x \)
📚 Study Tips
✓ Always factor completely before finding asymptotes and holes
✓ Holes occur where factors cancel; vertical asymptotes where they don't
✓ Check for extraneous solutions when solving rational equations
✓ To verify inverses, BOTH compositions must equal x
✓ Horizontal asymptote depends on degree comparison