Honors Algebra 2: Domain Worksheet

For each of the problems below, find the domain.

  1. \( \displaystyle f(x) = \frac{9}{x^2} \)
    2. Solution \( \{ x \in \Re \mid x \neq 0 \} \)
  2. \( g(x) = -5\)
    3. Solution \( \{ x \in \Re \} \)
  3. \( h(x) = \sqrt{ 4 - x^2} \)
    4. Solution \(4 - x^2 \geq 0\). Factor this to get \((2 - x)(2 + x) \). This is negative to the left of -2, positive between -2 and 2, and negative again to the right of 2. So \( \{ x \in \Re \mid -2 \leq x \leq 2 \} \)
  4. \( f(x) = |x + 7| \)
    5. Solution \( \{ x \in \Re \} \)
  5. \( f(x) = |x| + 7\)
    6. Solution \( \{ x \in \Re \} \)
  6. \( g(x) = \sqrt{ x^2 - 81} \)
    7. Solution \( x^2 - 81 \geq 0\). Factor this to get \((x + 9)(x - 9)\). This is positive to the left of -9, negative between -9 and 9, and positive again to the right of 9. So \( \{ x \in \Re \mid x \leq -9 \bigvee x \geq 9 \} \)
  7. \( \displaystyle h(x) = \frac{ x^2 - 36}{ x + 6}\)
    8. Solution \( \{ x \in \Re \mid x \neq -6 \} \)
  8. \( \displaystyle m(x) = \frac{ x + 6}{ x^2 - 36}\)
    9. Solution \( \{ x \in \Re \mid x \neq -6, 6 \} \)
  9. \( \displaystyle f(x) = \frac{-2}{ x + 3}\)
    10. Solution \( \{ x \in \Re \mid x \neq -3 \} \)
    11. Solution \( \{ x \in \Re \mid x \leq 3 \bigvee x \ge 4 \)