1. **Problem:** Find the domain and range of the function $f(x) = \sqrt{\frac{x - 1}{x^2 - 4}}$. 2. **Step 1: State the domain conditions.** - The expression inside the square root must be non-negative: $$\frac{x - 1}{x^2 - 4} \geq 0$$ - The denominator cannot be zero: $$x^2 - 4 \neq 0 \Rightarrow x \neq \pm 2$$ 3. **Step 2: Solve the inequality.** - Factor denominator: $$x^2 - 4 = (x - 2)(x + 2)$$ - Critical points: $x = 1, 2, -2$ 4. **Step 3: Test intervals around critical points:** - For $x < -2$: numerator $(x-1)<0$, denominator $(x-2)(x+2)>0$ (both negative times negative is positive), so fraction is negative over positive = negative, so fraction < 0 (not allowed). - For $-2 < x < 1$: numerator $(x-1)<0$, denominator $(x-2)(x+2)<0$ (one factor negative, one positive), so denominator negative, fraction negative over negative = positive, allowed. - For $1 < x < 2$: numerator $(x-1)>0$, denominator $(x-2)(x+2)<0$, fraction positive over negative = negative, not allowed. - For $x > 2$: numerator $(x-1)>0$, denominator $(x-2)(x+2)>0$, fraction positive over positive = positive, allowed. 5. **Step 4: Domain summary:** $$\boxed{(-2,1] \cup (2, \infty)}$$ 6. **Step 5: Range:** - Since $f(x)$ is a square root of a non-negative expression, range is $[0, \infty)$. **Final answer:** - Domain: $(-2,1] \cup (2, \infty)$ - Range: $[0, \infty)$