Question

Set the denominator to zero and solve for those values since any number that makes the denominator of a rational expression equal zero renders it undefined. Remember that even if a factor cancels when the expression is simplified, the original expression remains undefined at the roots of the original denominator.

Answer 1

Good rule: a rational expression is undefined exactly where its denominator = 0. Steps and important notes:

Steps
1. Factor the denominator (and numerator if helpful).
2. Solve denominator = 0 to find the forbidden x-values.
3. Those x-values are excluded from the domain even if they cancel with a numerator factor.
4. After you find the excluded values you may simplify the expression algebraically — but do not re-include excluded points.

Examples

1) (x^2 − 1)/(x^2 − 1)
- Factor: numerator = (x − 1)(x + 1), denominator = (x − 1)(x + 1).
- Solve denominator = 0 → x = 1, −1.
- Domain: all real x except x = ±1.
- Although the expression simplifies algebraically to 1, the original expression is undefined at x = 1 and x = −1 (removable holes).

2) (x + 2)/(x^2 − 4)
- Factor denominator: x^2 − 4 = (x − 2)(x + 2).
- Solve: x = 2, −2 are excluded.
- Simplified form cancels (x + 2), giving 1/(x − 2), but both x = 2 and x = −2 remain excluded. At x = −2 the graph has a hole (removable), at x = 2 a vertical asymptote (nonremovable).

3) (x + 1)/(x^2 + 1)
- Denominator x^2 + 1 = 0 has solutions x = i, −i (nonreal).
- For real-domain problems: no real zeros → domain is all real numbers.

Extra notes
- If denominator is a polynomial of degree ≥ 2 and doesn’t factor over the reals, use the quadratic formula to find real roots.
- Always determine excluded values from the original (unsimplified) denominator. Even canceled roots are excluded.
- If you need to know the y-value of a removable hole, compute the limit of the simplified expression at that x.

If you want, give me a specific rational expression and I’ll find its excluded values and domain.