College Algebra Quiz: Composition Of Functions Domain Of Composition
2 questions · exam conditions
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Composition Of Functions Domain Of CompositionQuestion 1 of 2

Let f(x)=2x+1x−3f(x) = \frac{2x + 1}{x - 3} and g(x)=x+4g(x) = \sqrt{x + 4}. What is the domain of the composition (f∘g)(x)(f \circ g)(x)?

(−4,∞)(-4, \infty)
[−4,∞)∖{5}[-4, \infty) \setminus \{5\}
[−4,∞)∖{3}[-4, \infty) \setminus \{3\}
(−∞,∞)∖{3}(-\infty, \infty) \setminus \{3\}
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College Algebra Quiz: Composition Of Functions Domain Of Composition

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

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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.

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Question 1

Let f(x)=2x+1x−3f(x) = \frac{2x + 1}{x - 3} and g(x)=x+4g(x) = \sqrt{x + 4}. What is the domain of the composition (f∘g)(x)(f \circ g)(x)?

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

Question 2

Given the piecewise function t(x)={x2if x≤02x−1if x>0t(x) = \begin{cases} x^2 & \text{if } x \leq 0 \\ 2x - 1 & \text{if } x > 0 \end{cases} and s(x)=x−3s(x) = x - 3, what is the domain of (t∘s)(x)(t \circ s)(x)?

  1. All real numbers (correct answer)
  2. x≥3x \geq 3
  3. x≤3x \leq 3
  4. x≠3x \neq 3
Explanation: For (t∘s)(x)=t(s(x))=t(x−3)(t \circ s)(x) = t(s(x)) = t(x - 3), we need to determine when this composition is defined. The function s(x)=x−3s(x) = x - 3 is defined for all real numbers. The piecewise function t(x)t(x) is also defined for all real numbers: t(x)=x2t(x) = x^2 when x≤0x \leq 0 and t(x)=2x−1t(x) = 2x - 1 when x>0x > 0. For the composition, we evaluate tt at s(x)=x−3s(x) = x - 3. When s(x)=x−3≤0s(x) = x - 3 \leq 0 (i.e., when x≤3x \leq 3), we use t(s(x))=(s(x))2=(x−3)2t(s(x)) = (s(x))^2 = (x - 3)^2. When s(x)=x−3>0s(x) = x - 3 > 0 (i.e., when x>3x > 3), we use t(s(x))=2(s(x))−1=2(x−3)−1=2x−7t(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.