# Honors Algebra 2: Domain Worksheet

• Class: Honors Algebra 2
• Author: Peter Atlas
• Text: Algebra and Trigonometry: Structure and Method, Brown

For each of the problems below, find the domain.

1. $$\displaystyle f(x) = \frac{9}{x^2}$$
2. Solution $$\{ x \in \Re \mid x \neq 0 \}$$
3. $$g(x) = -5$$
4. Solution $$\{ x \in \Re \}$$
5. $$h(x) = \sqrt{ 4 - x^2}$$
6. 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 $$\{ x \in \Re \mid -2 \leq x \leq 2 \}$$
7. $$f(x) = |x + 7|$$
8. Solution $$\{ x \in \Re \}$$
9. $$f(x) = |x| + 7$$
10. Solution $$\{ x \in \Re \}$$
11. $$g(x) = \sqrt{ x^2 - 81}$$
12. 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 $$\{ x \in \Re \mid x \leq -9 \bigvee x \geq 9 \}$$
13. $$\displaystyle h(x) = \frac{ x^2 - 36}{ x + 6}$$
14. Solution $$\{ x \in \Re \mid x \neq -6 \}$$
15. $$\displaystyle m(x) = \frac{ x + 6}{ x^2 - 36}$$
16. Solution $$\{ x \in \Re \mid x \neq -6, 6 \}$$
17. $$\displaystyle f(x) = \frac{-2}{ x + 3}$$
18. Solution $$\{ x \in \Re \mid x \neq -3 \}$$
19. Solution $$\{ x \in \Re \mid x \leq 3 \bigvee x \ge 4$$