📊 Domain of a Rational Function
When a rational function is defined and when it is not
🎯 Why Does This Matter?
A rational function (quotient) is one of the most common function types in advanced mathematics!
The first step in any investigation of a rational function is to find the domain – because there are values of \(x\) for which the function simply does not exist!
🔑 Basic rule: you cannot divide by zero!
📚 What Is a Rational Function?
Rational function = quotient of two polynomials
\(f(x) = \frac{P(x)}{Q(x)}\)
where \(P(x)\) and \(Q(x)\) are polynomials
Examples:
⛔ Domain – Core Rule
A rational function is defined for every \(x\) for which the denominator is not zero
📋 Steps for finding the domain:
| Step | What to do |
|---|---|
| 1 | Identify the denominator of the function |
| 2 | Solve the equation: denominator = 0 |
| 3 | The domain is all real numbers except the solutions |
✏️ Detailed Examples
Example 1: Linear denominator
Find the domain of \(f(x) = \frac{x+3}{x-5}\)
Solution:
Denominator: \(x - 5\)
Solve: \(x - 5 = 0\) → \(x = 5\)
Domain: \(x \neq 5\)
Or in other notation: \(\mathbb{R} \setminus \{5\}\) or \((-\infty, 5) \cup (5, \infty)\)
Example 2: Quadratic denominator (two solutions)
Find the domain of \(f(x) = \frac{2x}{x^2-9}\)
Solution:
Denominator: \(x^2 - 9\)
Solve: \(x^2 - 9 = 0\)
\(x^2 = 9\)
\(x = 3\) or \(x = -3\)
Domain: \(x \neq 3\) and \(x \neq -3\)
Example 3: Denominator with a common factor
Find the domain of \(f(x) = \frac{x}{x^2+3x}\)
Solution:
Denominator: \(x^2 + 3x\)
Factor: \(x^2 + 3x = x(x + 3)\)
Solve: \(x(x + 3) = 0\)
\(x = 0\) or \(x = -3\)
Domain: \(x \neq 0\) and \(x \neq -3\)
Example 4: Quadratic denominator with no real roots
Find the domain of \(f(x) = \frac{x-1}{x^2+4}\)
Solution:
Denominator: \(x^2 + 4\)
Solve: \(x^2 + 4 = 0\) → \(x^2 = -4\) → no real solution!
\(x^2 + 4 > 0\) for all \(x\) – the denominator never equals zero.
Domain: all real numbers \(\mathbb{R}\) ✓
📝 Domain Notation Types
If the domain is "all real numbers except \(x = 2\) and \(x = 5\)", you can write:
| Notation type | Example |
|---|---|
| Condition | \(x \neq 2\) and \(x \neq 5\) |
| Set notation | \(\mathbb{R} \setminus \{2, 5\}\) |
| Interval notation | \((-\infty, 2) \cup (2, 5) \cup (5, \infty)\) |
💡 In advanced exams: the most common and simplest notation is \(x \neq ...\)
⚠️ Common Traps
❌ Trap 1: Looking at the numerator
The domain depends only on the denominator!
The numerator can equal zero – that is fine.
❌ Trap 2: Missing a solution
\(x^2 - 9 = 0\) has two solutions!
\(x = 3\) and also \(x = -3\)
❌ Trap 3: Forgetting to factor
\(x^2 + 3x = 0\)
Must factor: \(x(x+3) = 0\)
There is a solution at \(x = 0\)!
📊 Summary Table – Denominator Types
| Denominator type | Example | Solutions | Domain |
|---|---|---|---|
| Linear | \(x - 3\) | One solution | \(x \neq 3\) |
| Quadratic with roots | \(x^2 - 4\) | Two solutions | \(x \neq \pm 2\) |
| Quadratic without roots | \(x^2 + 4\) | No solutions | All \(\mathbb{R}\) |
| With common factor | \(x^2 + 3x\) | Factor first! | \(x \neq 0, -3\) |
📝 Summary
Domain of a rational function: denominator ≠ 0
Solve "denominator = 0" and exclude the solutions from the domain
Now you are ready to continue to the next topic: Vertical Asymptote and Removable Discontinuity!