College Algebra Quiz: Determine Domain And Range

What this quiz covers

This quiz focuses on Determine Domain And Range, 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. $[-4, 0) \cup [1, \infty)$ (correct answer)
3. $[-4, \infty)$
4. $(-\infty, 0) \cup [1, \infty)$

Explanation: For $x < 2$, we have $h(x) = x^2 - 4$. As $x$ approaches 2 from the left, $h(x)$ approaches $4-4 = 0$. As $x \to -\infty$, $h(x) \to \infty$. The minimum value on this piece occurs at $x = 0$, giving $h(0) = -4$. So the first piece contributes $[-4, 0)$ to the range. For $x \geq 2$, we have $h(x) = \sqrt{x-1}$. At $x = 2$, $h(2) = \sqrt{1} = 1$. As $x \to \infty$, $h(x) \to \infty$. So the second piece contributes $[1, \infty)$. The complete range is $[-4, 0) \cup [1, \infty)$. Choice A misses the gap, choice C incorrectly fills the gap, and choice D has the wrong left endpoint.

Question 2

1. Domain: $(-2, 2)$; Range: $(-\infty, 2]$ (correct answer)
2. Domain: $[-2, 2]$; Range: $(-\infty, 2]$
3. Domain: $(-2, 2)$; Range: $(-\infty, \infty)$
4. Domain: $(-\infty, -2) \cup (2, \infty)$; Range: $(-\infty, 2]$

Explanation: For the domain, we need $4-x^2 > 0$, which means $x^2 < 4$, so $-2 < x < 2$. The domain is $(-2, 2)$. For the range, note that as $x$ approaches $\pm 2$, $4-x^2$ approaches 0, so $\log_2(4-x^2) \to -\infty$. The maximum value occurs at $x = 0$, where $f(0) = \log_2(4) = 2$. Since the argument $4-x^2$ can take any value in $(0, 4]$ and $\log_2$ is increasing, the range is $(-\infty, 2]$. Choice B incorrectly includes endpoints in domain, choice C has wrong range, and choice D has incorrect domain.

Question 3

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

Explanation: When finding the domain of a function involving square roots and fractions, you need to ensure two conditions are met: the expression under the square root must be non-negative, and the denominator cannot equal zero. For $r(x) = \sqrt{\frac{x-1}{x+2}}$, you need $\frac{x-1}{x+2} \geq 0$ and $x + 2 \neq 0$. First, note that $x = -2$ makes the denominator zero, so it's excluded from the domain. To solve $\frac{x-1}{x+2} \geq 0$, find where the numerator and denominator change signs. The numerator $x-1 = 0$ when $x = 1$, and the denominator $x+2 = 0$ when $x = -2$. Test the sign of the fraction in each interval:

• For $x < -2$: both $x-1 < 0$ and $x+2 < 0$, so $\frac{x-1}{x+2} > 0$ ✓
• For $-2 < x < 1$: $x-1 < 0$ and $x+2 > 0$, so $\frac{x-1}{x+2} < 0$ ✗
• For $x > 1$: both $x-1 > 0$ and $x+2 > 0$, so $\frac{x-1}{x+2} > 0$ ✓
• At $x = 1$: $\frac{1-1}{1+2} = 0 \geq 0$ ✓

Question 4

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

Explanation: For the domain, we need two conditions: (1) $x^2 - 9 \geq 0$ for the square root, and (2) $x - 5 \neq 0$ for the denominator. From condition (1), $x^2 \geq 9$, so $x \leq -3$ or $x \geq 3$. This gives us $(-\infty, -3] \cup [3, \infty)$. From condition (2), $x \neq 5$. Combining both conditions, we get $(-\infty, -3] \cup [3, \infty) \setminus \{5\}$. Note that the endpoints $x = -3$ and $x = 3$ are included because $\sqrt{x^2-9} = 0$ is valid. Choice B incorrectly excludes the endpoints, choice C has the wrong interval, and choice D ignores the square root restriction.

Question 5

1. $[5, \infty)$
2. $[3, 5]$
3. $(3, \infty)$
4. $(3, 5]$ (correct answer)

Explanation: To find the range of a rational function like this, you need to analyze how the function behaves by examining its components and any restrictions on the output values. Start by examining the fraction $\frac{2}{x^2+1}$. Since $x^2 \geq 0$ for all real numbers, we know $x^2+1 \geq 1$. This means $\frac{2}{x^2+1} \leq \frac{2}{1} = 2$. The fraction approaches 2 as $x$ approaches 0, and approaches 0 as $|x|$ approaches infinity. However, since $x^2+1$ is never zero, the fraction is always positive and never actually reaches 0. Therefore, $\frac{2}{x^2+1}$ takes on all values in the interval $(0, 2]$. Adding 3 to this entire interval gives us $p(x) = 3 + \frac{2}{x^2+1}$, which takes on all values in $(3, 5]$. The function approaches but never reaches 3, and achieves its maximum value of 5 when $x = 0$. Choice A, $[5, \infty)$, incorrectly suggests the function can exceed 5 and includes values above the maximum. Choice B, $[3, 5]$, wrongly includes 3 as an achievable value, when the function only approaches it. Choice C, $(3, \infty)$, correctly excludes 3 but incorrectly suggests the function can take arbitrarily large values. Strategy tip: For rational functions, always check the behavior of denominators to find restrictions, then trace how those restrictions affect the overall function's output range.