1. **State the problem:** We want to analyze the function $$y = \sqrt{\frac{x+1}{x-1}} + \sqrt{2x+3}$$ and find its domain. 2. **Identify domain restrictions:** - For the first square root $$\sqrt{\frac{x+1}{x-1}}$$, the expression inside must be non-negative: $$\frac{x+1}{x-1} \geq 0$$ Also, the denominator cannot be zero: $$x - 1 \neq 0$$ - For the second square root $$\sqrt{2x+3}$$, the radicand must be non-negative: $$2x + 3 \geq 0$$ 3. **Solve the inequality $$\frac{x+1}{x-1} \geq 0$$:** - Critical points are where numerator or denominator is zero: $$x = -1$$ and $$x = 1$$. - Test intervals: - For $$x < -1$$, pick $$x = -2$$: $$\frac{-2+1}{-2-1} = \frac{-1}{-3} = \frac{1}{3} > 0$$ (true) - For $$-1 < x < 1$$, pick $$x = 0$$: $$\frac{0+1}{0-1} = \frac{1}{-1} = -1 < 0$$ (false) - For $$x > 1$$, pick $$x = 2$$: $$\frac{2+1}{2-1} = \frac{3}{1} = 3 > 0$$ (true) - Since denominator cannot be zero, exclude $$x=1$$. - So solution for this inequality is $$(-\infty, -1] \cup (1, \infty)$$. 4. **Solve the inequality $$2x + 3 \geq 0$$:** - $$2x \geq -3$$ - $$x \geq -\frac{3}{2}$$ 5. **Combine domain restrictions:** - From step 3: $$x \in (-\infty, -1] \cup (1, \infty)$$ - From step 4: $$x \in [-\frac{3}{2}, \infty)$$ - Intersection is: - For $$(-\infty, -1]$$ and $$[-\frac{3}{2}, \infty)$$, intersection is $$[-\frac{3}{2}, -1]$$ - For $$(1, \infty)$$ and $$[-\frac{3}{2}, \infty)$$, intersection is $$(1, \infty)$$ 6. **Final domain:** $$D = \left[-\frac{3}{2}, -1\right] \cup (1, \infty)$$ 7. **Summary:** The function is defined for $$x$$ in $$\left[-\frac{3}{2}, -1\right]$$ and for $$x > 1$$.