College Algebra Quiz: Composition Of Functions Domain Of Composition

What this quiz covers

This quiz focuses on Composition Of Functions Domain Of Composition, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $(-4, \infty)$
2. $[-4, \infty) \setminus \{5\}$ (correct answer)
3. $[-4, \infty) \setminus \{3\}$
4. $(-\infty, \infty) \setminus \{3\}$

Explanation: For $(f \circ g)(x) = f(g(x))$, we need both $g(x)$ to be defined and $f(g(x))$ to be defined. First, $g(x) = \sqrt{x + 4}$ requires $x + 4 \geq 0$, so $x \geq -4$. Next, $f(g(x)) = \frac{2\sqrt{x + 4} + 1}{\sqrt{x + 4} - 3}$ requires the denominator to be nonzero: $\sqrt{x + 4} - 3 \neq 0$, which means $\sqrt{x + 4} \neq 3$. Since $\sqrt{x + 4} = 3$ when $x + 4 = 9$, we get $x = 5$. Therefore, the domain is $[-4, \infty) \setminus \{5\}$.

Question 2

1. All real numbers (correct answer)
2. $x \geq 3$
3. $x \leq 3$
4. $x \neq 3$

Explanation: For $(t \circ s)(x) = t(s(x)) = t(x - 3)$, we need to determine when this composition is defined. The function $s(x) = x - 3$ is defined for all real numbers. The piecewise function $t(x)$ is also defined for all real numbers: $t(x) = x^2$ when $x \leq 0$ and $t(x) = 2x - 1$ when $x > 0$. For the composition, we evaluate $t$ at $s(x) = x - 3$. When $s(x) = x - 3 \leq 0$ (i.e., when $x \leq 3$), we use $t(s(x)) = (s(x))^2 = (x - 3)^2$. When $s(x) = x - 3 > 0$ (i.e., when $x > 3$), we use $t(s(x)) = 2(s(x)) - 1 = 2(x - 3) - 1 = 2x - 7$. Both pieces are defined for their respective domains, so the overall composition is defined for all real numbers.