1. Stating the problem for question 4: Find the domain of the function $$f(x) = \sqrt{\frac{3 - x}{5 - x}}$$. 2. To find the domain of $$f(x)$$, ensure the expression inside the square root is non-negative, and the denominator is not zero. 3. Write the inequality: $$\frac{3 - x}{5 - x} \geq 0$$ with $$5 - x \neq 0$$ (so $$x \neq 5$$). 4. Find critical points where numerator or denominator is zero: Numerator zero at $$x=3$$, denominator zero at $$x=5$$. 5. Test intervals: - For $$x < 3$$: numerator positive (since 3 - x > 0), denominator positive (5 - x > 0), fraction positive. - For $$3 < x < 5$$: numerator negative (3 - x < 0), denominator positive (5 - x > 0), fraction negative. - For $$x > 5$$: numerator negative, denominator negative, fraction positive. 6. So, $$\frac{3 - x}{5 - x} \geq 0$$ is true for $$x \in (-\infty, 3] \cup (5, \infty)$$ but since denominator cannot be zero, exclude $$x=5$$. 7. Check the original square root: domain where the expression under the root is non-negative. So domain is $$(-\infty, 3] \cup (5, \infty)$$. However, in given options, the closest match is complement of $$]3, 5]$$ meaning all real numbers except $$]3, 5]$$, corresponding to option (b). --- 1. For question 5: Given $$f(x) = x - 3$$ and $$g(x) = x^2$$, find $$(f \circ g)(x) = f(g(x))$$. 2. Compute $$g(x) = x^2$$, so $$f(g(x)) = f(x^2) = x^2 - 3$$. 3. The correct function is $$x^2 - 3$$, so answer is option (b). --- 1. For question 6: Given $$f(x) = \sqrt{x}$$ and $$g(x) = x^2$$, find domain of $$(g \circ f)(x) = g(f(x)) = (\sqrt{x})^2 = x$$. 2. Since $$f(x) = \sqrt{x}$$ requires $$x \geq 0$$ (domain of $$f$$), and $$g(x) = x^2$$ is defined for all real $$x$$, the domain of composition is domain of $$f$$, which is $$[0, \infty)$$. 3. Hence, the domain is option (b).