Compound Inequalities with Rationals - Math 3

1. The inequality $\frac{x^2-1}{(x-2)^2} \geq 0$ is part of a compound inequality system. At which value(s) of $x$ does this rational expression equal zero?
   Answer: Both $x = -1$ and $x = 1$, as both make the numerator zero The expression $\frac{x^2-1}{(x-2)^2} = \frac{(x-1)(x+1)}{(x-2)^2}$ equals zero when the numerator equals zero and the denominator is non-zero. The numerator $(x-1)(x+1) = 0$ when $x = 1$ or $x = -1$. The denominator $(x-2)^2$ is never zero at these points (only at $x = 2$), so both $x = -1$ and $x = 1$ make the expression equal to zero. The squared denominator doesn't affect where the expression equals zero, only where it's undefined.
2. A rational function $f(x) = \frac{ax+b}{cx+d}$ satisfies the compound inequality $f(x) < -1$ OR $f(x) \geq 3$ for all $x \in (-2, 1] \cup [4, \infty)$. What can be concluded about the vertical asymptote of $f(x)$?
   Answer: The vertical asymptote must be at $x = -2$ or between 1 and 4 Since the solution set has a gap at $x = -2$ (open interval) and between $x = 1$ and $x = 4$, 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 $x = -2$ creates a boundary and there's a gap from $(1, 4)$, the asymptote is most likely at $x = -2$ or somewhere in $(1, 4)$.
3. The inequality $\frac{|x-1|}{x+2} \leq \frac{3}{2}$ can be solved by considering cases. How many distinct intervals must be analyzed?
   Answer: Three intervals considering both critical points and sign changes The absolute value $|x-1|$ creates a case split at $x = 1$, and the denominator $x+2$ is undefined at $x = -2$. This creates three intervals to consider: $x < -2$, $-2 < x < 1$, and $x > 1$. In each interval, we need to determine the sign of $x-1$ for the absolute value and solve the resulting rational inequality. The critical points $x = -2$ and $x = 1$ divide the real line into three regions that must be analyzed separately.
4. A student claims that the solution to $\frac{x}{x-2} \geq 1$ AND $\frac{x+1}{x-3} < 2$ is $x \in (3, 5)$. Which statement best describes this claim?
   Answer: Incorrect, because the intersection of the two solution sets is actually empty For $\frac{x}{x-2} \geq 1$: Rearranging gives $\frac{x-(x-2)}{x-2} \geq 0$, which simplifies to $\frac{2}{x-2} \geq 0$. Since the numerator is always positive, we need $x-2 > 0$, so $x > 2$. For $\frac{x+1}{x-3} < 2$: Rearranging gives $\frac{x+1-2(x-3)}{x-3} < 0$, which simplifies to $\frac{x+1-2x+6}{x-3} < 0$, so $\frac{7-x}{x-3} < 0$. Critical points are $x = 7$ and $x = 3$. Using sign analysis: for $x < 3$, numerator positive, denominator negative, ratio negative ✓; for $3 < x < 7$, numerator positive, denominator positive, ratio positive ✗; for $x > 7$, numerator negative, denominator positive, ratio negative ✓. So the second inequality gives $x \in (-\infty, 3) \cup (7, \infty)$. The intersection of $(2, \infty)$ and $((-\infty, 3) \cup (7, \infty))$ is $(2, 3) \cup (7, \infty)$, not $(3, 5)$.

Start this quiz