1. **State the problem:** Find the domain of the function $$h(x) = \sqrt{\frac{1 - x}{x + 1}}$$ which means we need all values of $$x$$ for which the expression under the square root is defined and non-negative. 2. **Set the radicand non-negative:** The expression inside the square root must be $$\geq 0$$. $$\frac{1 - x}{x + 1} \geq 0$$. 3. **Identify restrictions:** The denominator cannot be zero. $$x + 1 \neq 0 \Rightarrow x \neq -1$$. 4. **Solve the inequality:** To solve $$\frac{1 - x}{x + 1} \geq 0$$, consider critical points where the numerator or denominator is zero: - Numerator zero at $$x = 1$$ - Denominator zero at $$x = -1$$ 5. **Test intervals:** Split the real line into intervals determined by $$-1$$ and $$1$$: - For $$x < -1$$: Choose $$x = -2$$, numerator $$1 - (-2) = 3 > 0$$, denominator $$-2 + 1 = -1 < 0$$, so fraction is negative. - For $$-1 < x < 1$$: Choose $$x = 0$$, numerator $$1 - 0 = 1 > 0$$, denominator $$0 + 1 = 1 > 0$$, fraction is positive. - For $$x > 1$$: Choose $$x = 2$$, numerator $$1 - 2 = -1 < 0$$, denominator $$2 + 1 = 3 > 0$$, fraction is negative. 6. **Include points if inequality allows:** - At $$x = 1$$, numerator zero, fraction $$0$$, valid. - At $$x = -1$$, denominator zero, undefined, exclude. 7. **Conclusion:** The domain is $$-1 < x \leq 1$$. **Answer:** Option c. $$-1 < x \leq 1$$