College Algebra Quiz: Function Notation Evaluate And Interpret
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Function Notation Evaluate And InterpretQuestion 1 of 1

Consider the function f(x)=∣x−3∣+2f(x) = |x - 3| + 2. If f(a)=f(b)f(a) = f(b) where a≠ba \neq b, and both aa and bb are in the domain [−1,7][-1, 7], which of the following must be true?

a+b=6a + b = 6 and one of a,ba, b is less than 3 while the other is greater than 3
a+b=6a + b = 6 and both aa and bb are greater than or equal to 3
a+b=3a + b = 3 and one of a,ba, b is less than 3 while the other is greater than 3
∣a−3∣=∣b−3∣|a - 3| = |b - 3| but a+ba + b could be any value in the given domain
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College Algebra Quiz

College Algebra Quiz: Function Notation Evaluate And Interpret

Practice Function Notation Evaluate And Interpret 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 Function Notation Evaluate And Interpret, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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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

Consider the function f(x)=∣x−3∣+2f(x) = |x - 3| + 2. If f(a)=f(b)f(a) = f(b) where a≠ba \neq b, and both aa and bb are in the domain [−1,7][-1, 7], which of the following must be true?

  1. a+b=6a + b = 6 and one of a,ba, b is less than 3 while the other is greater than 3 (correct answer)
  2. a+b=6a + b = 6 and both aa and bb are greater than or equal to 3
  3. a+b=3a + b = 3 and one of a,ba, b is less than 3 while the other is greater than 3
  4. ∣a−3∣=∣b−3∣|a - 3| = |b - 3| but a+ba + b could be any value in the given domain
Explanation: The function f(x)=∣x−3∣+2f(x) = |x - 3| + 2 has its vertex at x=3x = 3, where it achieves its minimum value of f(3)=2f(3) = 2. For x<3x < 3, we have f(x)=−(x−3)+2=−x+5f(x) = -(x - 3) + 2 = -x + 5. For x≥3x \geq 3, we have f(x)=(x−3)+2=x−1f(x) = (x - 3) + 2 = x - 1. If f(a)=f(b)f(a) = f(b) with a≠ba \neq b, then we need ∣a−3∣+2=∣b−3∣+2|a - 3| + 2 = |b - 3| + 2, which simplifies to ∣a−3∣=∣b−3∣|a - 3| = |b - 3|. This occurs when either a−3=b−3a - 3 = b - 3 (giving a=ba = b, which contradicts a≠ba \neq b) or a−3=−(b−3)a - 3 = -(b - 3), which gives a−3=−b+3a - 3 = -b + 3, so a+b=6a + b = 6. For this to happen with a≠ba \neq b, one value must be on the left side of the vertex (<3< 3) and the other on the right side (>3> 3). For example, if a=1a = 1, then f(1)=2+2=4f(1) = 2 + 2 = 4, and we need f(b)=4f(b) = 4. From b−1=4b - 1 = 4, we get b=5b = 5. Indeed, a+b=1+5=6a + b = 1 + 5 = 6. Choice D is incorrect because while ∣a−3∣=∣b−3∣|a - 3| = |b - 3| is necessary, it doesn't capture the full constraint that a+b=6a + b = 6.