Math 3 Quiz: Compound Inequalities With Rationals
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Compound Inequalities With RationalsQuestion 1 of 4

A rational function f(x)=ax+bcx+df(x) = \frac{ax+b}{cx+d} satisfies the compound inequality f(x)<−1f(x) < -1 OR f(x)≥3f(x) \geq 3 for all x∈(−2,1]∪[4,∞)x \in (-2, 1] \cup [4, \infty). What can be concluded about the vertical asymptote of f(x)f(x)?

The vertical asymptote must be at x=1x = 1 or x=4x = 4
The vertical asymptote must be at x=−2x = -2 or between 1 and 4
The vertical asymptote cannot exist within the interval (−2,4)(-2, 4)
The vertical asymptote must be at x=2x = 2 or x=3x = 3
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Math 3 Quiz

Math 3 Quiz: Compound Inequalities With Rationals

Practice Compound Inequalities With Rationals in Math 3 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 Compound Inequalities With Rationals, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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Question 1

A rational function f(x)=ax+bcx+df(x) = \frac{ax+b}{cx+d} satisfies the compound inequality f(x)<−1f(x) < -1 OR f(x)≥3f(x) \geq 3 for all x∈(−2,1]∪[4,∞)x \in (-2, 1] \cup [4, \infty). What can be concluded about the vertical asymptote of f(x)f(x)?

  1. The vertical asymptote must be at x=1x = 1 or x=4x = 4
  2. The vertical asymptote must be at x=−2x = -2 or between 1 and 4 (correct answer)
  3. The vertical asymptote cannot exist within the interval (−2,4)(-2, 4)
  4. The vertical asymptote must be at x=2x = 2 or x=3x = 3
Explanation: Since the solution set has a gap at x=−2x = -2 (open interval) and between x=1x = 1 and x=4x = 4, these gaps likely correspond to where the rational function changes behavior dramatically. The vertical asymptote (where the denominator equals zero) would cause the function to be undefined and create natural boundaries in the solution set. Since x=−2x = -2 creates a boundary and there's a gap from (1,4)(1, 4), the asymptote is most likely at x=−2x = -2 or somewhere in (1,4)(1, 4).

Question 2

The inequality ∣x−1∣x+2≤32\frac{|x-1|}{x+2} \leq \frac{3}{2} can be solved by considering cases. How many distinct intervals must be analyzed?

  1. Two intervals separated by the zero of the absolute value
  2. Three intervals considering both critical points and sign changes (correct answer)
  3. Four intervals accounting for absolute value and asymptote behavior
  4. Two intervals determined solely by the vertical asymptote location
Explanation: The absolute value ∣x−1∣|x-1| creates a case split at x=1x = 1, and the denominator x+2x+2 is undefined at x=−2x = -2. This creates three intervals to consider: x<−2x < -2, −2<x<1-2 < x < 1, and x>1x > 1. In each interval, we need to determine the sign of x−1x-1 for the absolute value and solve the resulting rational inequality. The critical points x=−2x = -2 and x=1x = 1 divide the real line into three regions that must be analyzed separately.

Question 3

The inequality x2−1(x−2)2≥0\frac{x^2-1}{(x-2)^2} \geq 0 is part of a compound inequality system. At which value(s) of xx does this rational expression equal zero?

  1. x=1x = 1 only, since x=−1x = -1 creates a sign change issue
  2. x=−1x = -1 only, since x=1x = 1 is excluded by domain restrictions
  3. Both x=−1x = -1 and x=1x = 1, as both make the numerator zero (correct answer)
  4. Neither x=−1x = -1 nor x=1x = 1, due to the squared denominator
Explanation: The expression x2−1(x−2)2=(x−1)(x+1)(x−2)2\frac{x^2-1}{(x-2)^2} = \frac{(x-1)(x+1)}{(x-2)^2} equals zero when the numerator equals zero and the denominator is non-zero. The numerator (x−1)(x+1)=0(x-1)(x+1) = 0 when x=1x = 1 or x=−1x = -1. The denominator (x−2)2(x-2)^2 is never zero at these points (only at x=2x = 2), so both x=−1x = -1 and x=1x = 1 make the expression equal to zero. The squared denominator doesn't affect where the expression equals zero, only where it's undefined.

Question 4

A student claims that the solution to xx−2≥1\frac{x}{x-2} \geq 1 AND x+1x−3<2\frac{x+1}{x-3} < 2 is x∈(3,5)x \in (3, 5). Which statement best describes this claim?

  1. Correct, because both individual inequalities are solved properly and intersected correctly
  2. Incorrect, because the student forgot to exclude x=2x = 2 from the final answer
  3. Incorrect, because the solution to the first inequality should include x=2x = 2
  4. Incorrect, because the intersection of the two solution sets is actually empty (correct answer)
Explanation: For xx−2≥1\frac{x}{x-2} \geq 1: Rearranging gives x−(x−2)x−2≥0\frac{x-(x-2)}{x-2} \geq 0, which simplifies to 2x−2≥0\frac{2}{x-2} \geq 0. Since the numerator is always positive, we need x−2>0x-2 > 0, so x>2x > 2. For x+1x−3<2\frac{x+1}{x-3} < 2: Rearranging gives x+1−2(x−3)x−3<0\frac{x+1-2(x-3)}{x-3} < 0, which simplifies to x+1−2x+6x−3<0\frac{x+1-2x+6}{x-3} < 0, so 7−xx−3<0\frac{7-x}{x-3} < 0. Critical points are x=7x = 7 and x=3x = 3. Using sign analysis: for x<3x < 3, numerator positive, denominator negative, ratio negative ✓; for 3<x<73 < x < 7, numerator positive, denominator positive, ratio positive ✗; for x>7x > 7, numerator negative, denominator positive, ratio negative ✓. So the second inequality gives x∈(−∞,3)∪(7,∞)x \in (-\infty, 3) \cup (7, \infty). The intersection of (2,∞)(2, \infty) and ((−∞,3)∪(7,∞))((-\infty, 3) \cup (7, \infty)) is (2,3)∪(7,∞)(2, 3) \cup (7, \infty), not (3,5)(3, 5).