College Algebra Quiz: Determine Domain And Range
5 questions · exam conditions
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Determine Domain And RangeQuestion 1 of 5

The function f(x)=log⁡2(4−x2)f(x) = \log_2(4-x^2) has domain and range, respectively:

Domain: (−2,2)(-2, 2); Range: (−∞,2](-\infty, 2]
Domain: [−2,2][-2, 2]; Range: (−∞,2](-\infty, 2]
Domain: (−2,2)(-2, 2); Range: (−∞,∞)(-\infty, \infty)
Domain: (−∞,−2)∪(2,∞)(-\infty, -2) \cup (2, \infty); Range: (−∞,2](-\infty, 2]
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College Algebra Quiz

College Algebra Quiz: Determine Domain And Range

Practice Determine Domain And Range in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

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Question 1

The function f(x)=log⁡2(4−x2)f(x) = \log_2(4-x^2) has domain and range, respectively:

  1. Domain: (−2,2)(-2, 2); Range: (−∞,2](-\infty, 2] (correct answer)
  2. Domain: [−2,2][-2, 2]; Range: (−∞,2](-\infty, 2]
  3. Domain: (−2,2)(-2, 2); Range: (−∞,∞)(-\infty, \infty)
  4. Domain: (−∞,−2)∪(2,∞)(-\infty, -2) \cup (2, \infty); Range: (−∞,2](-\infty, 2]
Explanation: For the domain, we need 4−x2>04-x^2 > 0, which means x2<4x^2 < 4, so −2<x<2-2 < x < 2. The domain is (−2,2)(-2, 2). For the range, note that as xx approaches ±2\pm 2, 4−x24-x^2 approaches 0, so log⁡2(4−x2)→−∞\log_2(4-x^2) \to -\infty. The maximum value occurs at x=0x = 0, where f(0)=log⁡2(4)=2f(0) = \log_2(4) = 2. Since the argument 4−x24-x^2 can take any value in (0,4](0, 4] and log⁡2\log_2 is increasing, the range is (−∞,2](-\infty, 2]. Choice B incorrectly includes endpoints in domain, choice C has wrong range, and choice D has incorrect domain.

Question 2

For the function p(x)=3+2x2+1p(x) = 3 + \frac{2}{x^2+1}, what is the range?

  1. [5,∞)[5, \infty)
  2. [3,5][3, 5]
  3. (3,∞)(3, \infty)
  4. (3,5](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 2x2+1\frac{2}{x^2+1}. Since x2≥0x^2 \geq 0 for all real numbers, we know x2+1≥1x^2+1 \geq 1. This means 2x2+1≤21=2\frac{2}{x^2+1} \leq \frac{2}{1} = 2. The fraction approaches 2 as xx approaches 0, and approaches 0 as ∣x∣|x| approaches infinity. However, since x2+1x^2+1 is never zero, the fraction is always positive and never actually reaches 0. Therefore, 2x2+1\frac{2}{x^2+1} takes on all values in the interval (0,2](0, 2]. Adding 3 to this entire interval gives us p(x)=3+2x2+1p(x) = 3 + \frac{2}{x^2+1}, which takes on all values in (3,5](3, 5]. The function approaches but never reaches 3, and achieves its maximum value of 5 when x=0x = 0. Choice A, [5,∞)[5, \infty), incorrectly suggests the function can exceed 5 and includes values above the maximum. Choice B, [3,5][3, 5], wrongly includes 3 as an achievable value, when the function only approaches it. Choice C, (3,∞)(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.

Question 3

For the piecewise function $$h(x) = \begin{cases} x^2-4 & \text{if } x < 2 \ \sqrt{x-1} & \text{if } x \geq 2 \end{cases}

  1. (−4,∞)(-4, \infty)
  2. [−4,0)∪[1,∞)[-4, 0) \cup [1, \infty) (correct answer)
  3. [−4,∞)[-4, \infty)
  4. (−∞,0)∪[1,∞)(-\infty, 0) \cup [1, \infty)
Explanation: For x<2x < 2, we have h(x)=x2−4h(x) = x^2 - 4. As xx approaches 2 from the left, h(x)h(x) approaches 4−4=04-4 = 0. As x→−∞x \to -\infty, h(x)→∞h(x) \to \infty. The minimum value on this piece occurs at x=0x = 0, giving h(0)=−4h(0) = -4. So the first piece contributes [−4,0)[-4, 0) to the range. For x≥2x \geq 2, we have h(x)=x−1h(x) = \sqrt{x-1}. At x=2x = 2, h(2)=1=1h(2) = \sqrt{1} = 1. As x→∞x \to \infty, h(x)→∞h(x) \to \infty. So the second piece contributes [1,∞)[1, \infty). The complete range is [−4,0)∪[1,∞)[-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 4

Consider the function r(x)=x−1x+2r(x) = \sqrt{\frac{x-1}{x+2}}. What is its domain?

  1. [1,∞)∖{−2}[1, \infty) \setminus \{-2\}
  2. (−2,1](-2, 1]
  3. (−∞,−2)∪[1,∞)(-\infty, -2) \cup [1, \infty) (correct answer)
  4. (−∞,−2)∪(1,∞)(-\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)=x−1x+2r(x) = \sqrt{\frac{x-1}{x+2}}, you need x−1x+2≥0\frac{x-1}{x+2} \geq 0 and x+2≠0x + 2 \neq 0. First, note that x=−2x = -2 makes the denominator zero, so it's excluded from the domain. To solve x−1x+2≥0\frac{x-1}{x+2} \geq 0, find where the numerator and denominator change signs. The numerator x−1=0x-1 = 0 when x=1x = 1, and the denominator x+2=0x+2 = 0 when x=−2x = -2. Test the sign of the fraction in each interval:
  • For x<−2x < -2: both x−1<0x-1 < 0 and x+2<0x+2 < 0, so x−1x+2>0\frac{x-1}{x+2} > 0 ✓
  • For −2<x<1-2 < x < 1: x−1<0x-1 < 0 and x+2>0x+2 > 0, so x−1x+2<0\frac{x-1}{x+2} < 0 ✗
  • For x>1x > 1: both x−1>0x-1 > 0 and x+2>0x+2 > 0, so x−1x+2>0\frac{x-1}{x+2} > 0 ✓
  • At x=1x = 1: 1−11+2=0≥0\frac{1-1}{1+2} = 0 \geq 0 ✓
Therefore, the domain is (−∞,−2)∪[1,∞)(-\infty, -2) \cup [1, \infty), which is answer C. Choice A incorrectly includes values between -2 and 1. Choice B only includes the interval between the critical points. Choice D excludes x=1x = 1, but this value makes the expression equal zero, which is acceptable under a square root. Study tip: For rational expressions under square roots, always check both where the expression equals zero (include these points) and where it's undefined (exclude these points).

Question 5

The function m(x)=x2−9x−5m(x) = \frac{\sqrt{x^2-9}}{x-5} has which domain?

  1. (−∞,−3]∪[3,∞)∖{5}(-\infty, -3] \cup [3, \infty) \setminus \{5\} (correct answer)
  2. (−∞,−3)∪(3,∞)∖{5}(-\infty, -3) \cup (3, \infty) \setminus \{5\}
  3. [−3,3]∖{5}[-3, 3] \setminus \{5\}
  4. (−∞,∞)∖{−3,3,5}(-\infty, \infty) \setminus \{-3, 3, 5\}
Explanation: For the domain, we need two conditions: (1) x2−9≥0x^2 - 9 \geq 0 for the square root, and (2) x−5≠0x - 5 \neq 0 for the denominator. From condition (1), x2≥9x^2 \geq 9, so x≤−3x \leq -3 or x≥3x \geq 3. This gives us (−∞,−3]∪[3,∞)(-\infty, -3] \cup [3, \infty). From condition (2), x≠5x \neq 5. Combining both conditions, we get (−∞,−3]∪[3,∞)∖{5}(-\infty, -3] \cup [3, \infty) \setminus \{5\}. Note that the endpoints x=−3x = -3 and x=3x = 3 are included because x2−9=0\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.