1. **Problem Statement:** Determine the domains of the functions $f(x) = \frac{x+1}{2x-3}$ and $g(x) = \sqrt{5-3x}$, and the domain of the composite function $(h \circ g)(x)$ where $h(x) = \frac{2}{x}$. 2. **Domain of $f(x)$:** The function $f(x)$ is a rational function. Its domain excludes values that make the denominator zero. Set denominator to zero: $$2x - 3 = 0 \implies x = \frac{3}{2}$$ So, domain of $f$ is all real numbers except $x = \frac{3}{2}$. 3. **Domain of $g(x)$:** Since $g(x) = \sqrt{5 - 3x}$, the expression inside the square root must be non-negative: $$5 - 3x \geq 0 \implies 3x \leq 5 \implies x \leq \frac{5}{3}$$ So, domain of $g$ is $(-\infty, \frac{5}{3}]$. 4. **Domain of $(h \circ g)(x)$:** The composite function is: $$(h \circ g)(x) = h(g(x)) = \frac{2}{g(x)} = \frac{2}{\sqrt{5 - 3x}}$$ For this to be defined: - $g(x)$ must be in the domain of $h$, so $g(x) \neq 0$ (denominator cannot be zero). - Also, $g(x)$ must be defined, so $5 - 3x \geq 0$. Set $g(x) \neq 0$: $$\sqrt{5 - 3x} \neq 0 \implies 5 - 3x \neq 0 \implies x \neq \frac{5}{3}$$ Combine with domain of $g$: $$x \leq \frac{5}{3} \text{ and } x \neq \frac{5}{3} \implies x < \frac{5}{3}$$ So, domain of $(h \circ g)(x)$ is $(-\infty, \frac{5}{3})$. --- **Final answers:** - Domain of $f(x)$: $\boxed{\mathbb{R} \setminus \left\{ \frac{3}{2} \right\}}$ - Domain of $g(x)$: $\boxed{(-\infty, \frac{5}{3}]}$ - Domain of $(h \circ g)(x)$: $\boxed{(-\infty, \frac{5}{3})}$