Rational Expressions and Their Domain

What Is a Rational Expression?

A rational expression is the quotient of two polynomials — a fraction whose numerator and denominator are both polynomials. Because a rational expression is a ratio, it is undefined whenever the denominator equals zero. The domain of a rational expression is therefore the set of all real numbers except the values that make the denominator zero.

How to Find the Domain

For a rational expression \(\dfrac{N(x)}{D(x)}\):

  1. Set the denominator equal to zero and solve \(D(x) = 0\).
  2. The domain is the set of all real numbers excluding those solutions.

The domain can be written in three equivalent ways:

Worked Examples: Finding the Domain

Find the domain of each rational expression below.

Example 1: \( \dfrac{x-1}{x+2} \)

View Step-by-Step Solution

Step 1. Solve \(x + 2 = 0 \Rightarrow x = -2\).

Step 2. Exclude \(x = -2\) from the real numbers.

  • Words: all real numbers except \(x = -2\).
  • Interval notation: \((-\infty, -2) \cup (-2, +\infty)\).
  • Inequalities: \(x < -2\) or \(x > -2\).

Example 2: \( \dfrac{2x+1}{(x-2)(x+3)} \)

View Step-by-Step Solution

Step 1. Solve \((x - 2)(x + 3) = 0 \Rightarrow x = 2\) or \(x = -3\).

Step 2. Exclude \(x = 2\) and \(x = -3\).

  • Words: all real numbers except \(x = -3\) and \(x = 2\).
  • Interval notation: \((-\infty, -3) \cup (-3, 2) \cup (2, +\infty)\).
  • Inequalities: \(x < -3\), \(-3 < x < 2\), or \(x > 2\).

Example 3: \( \dfrac{-x+3}{x^2+1} \)

View Step-by-Step Solution

Step 1. Solve \(x^2 + 1 = 0\). Since \(x^2 \ge 0\) for all real \(x\), \(x^2 + 1 \ge 1 > 0\), so this equation has no real solutions.

Step 2. The denominator never equals zero, so no values are excluded.

  • Words: all real numbers.
  • Interval notation: \((-\infty, +\infty)\).

Example 4: \( \dfrac{x^2+4}{x^3+2x^2-3x} \)

View Step-by-Step Solution

Step 1. Solve \(x^3 + 2x^2 - 3x = 0\). Factoring:

\[ x(x^2 + 2x - 3) = 0 \] \[ x(x - 1)(x + 3) = 0 \]

The solutions are \(x = -3,\; x = 0,\; x = 1\).

  • Words: all real numbers except \(x = -3\), \(x = 0\), and \(x = 1\).
  • Interval notation: \((-\infty, -3) \cup (-3, 0) \cup (0, 1) \cup (1, +\infty)\).
  • Inequalities: \(x < -3\), \(-3 < x < 0\), \(0 < x < 1\), or \(x > 1\).

Example 5: \( \dfrac{x^2+2x+1}{x^2+x-1} \)

View Step-by-Step Solution

Step 1. Solve \(x^2 + x - 1 = 0\) using the quadratic formula:

\[ \Delta = (1)^2 - 4(1)(-1) = 5 \] \[ x_1 = \dfrac{-1 - \sqrt{5}}{2}, \qquad x_2 = \dfrac{-1 + \sqrt{5}}{2} \]

Step 2. Exclude both roots.

  • Words: all real numbers except \(x = \dfrac{-1-\sqrt{5}}{2}\) and \(x = \dfrac{-1+\sqrt{5}}{2}\).
  • Interval notation: \( \left(-\infty, \dfrac{-1-\sqrt{5}}{2}\right) \cup \left(\dfrac{-1-\sqrt{5}}{2}, \dfrac{-1+\sqrt{5}}{2}\right) \cup \left(\dfrac{-1+\sqrt{5}}{2}, +\infty\right) \).
  • Inequalities: \(x < \dfrac{-1-\sqrt{5}}{2}\), or \(\dfrac{-1-\sqrt{5}}{2} < x < \dfrac{-1+\sqrt{5}}{2}\), or \(x > \dfrac{-1+\sqrt{5}}{2}\).

Practice Exercises

Find the domain of each rational expression below.

Exercise 1: \( \dfrac{x+9}{x-10} \)

View Solution

Solve \(x - 10 = 0 \Rightarrow x = 10\).

  • Words: all real numbers except \(x = 10\).
  • Interval notation: \((-\infty, 10) \cup (10, +\infty)\).
  • Inequalities: \(x < 10\) or \(x > 10\).

Exercise 2: \( \dfrac{1}{(x-9)(x+1)} \)

View Solution

Solve \((x - 9)(x + 1) = 0 \Rightarrow x = -1\) or \(x = 9\).

  • Words: all real numbers except \(x = -1\) and \(x = 9\).
  • Interval notation: \((-\infty, -1) \cup (-1, 9) \cup (9, +\infty)\).
  • Inequalities: \(x < -1\), \(-1 < x < 9\), or \(x > 9\).

Exercise 3: \( \dfrac{1}{x^3+1} \)

View Solution

Factor \(x^3 + 1 = (x + 1)(x^2 - x + 1)\). The quadratic factor has discriminant \(\Delta = 1 - 4 = -3 < 0\), so it has no real roots. The only real solution is \(x = -1\).

  • Words: all real numbers except \(x = -1\).
  • Interval notation: \((-\infty, -1) \cup (-1, +\infty)\).
  • Inequalities: \(x < -1\) or \(x > -1\).

Exercise 4: \( \dfrac{-x+7}{x^4 - 16} \)

View Solution

Factor the denominator:

\[ x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x - 2)(x + 2)(x^2 + 4) \]

The quadratic factor \(x^2 + 4\) has no real zeros. The real solutions are \(x = -2\) and \(x = 2\).

  • Words: all real numbers except \(x = -2\) and \(x = 2\).
  • Interval notation: \((-\infty, -2) \cup (-2, 2) \cup (2, +\infty)\).
  • Inequalities: \(x < -2\), \(-2 < x < 2\), or \(x > 2\).

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