1. **Problem Statement:** Find the domains and ranges of the functions $f(x) = 2$, $g(x) = x^2 + 1$, $\frac{f}{g}$, and $\frac{g}{f}$.\n\n2. **Recall definitions:**\n- The **domain** of a function is the set of all input values $x$ for which the function is defined.\n- The **range** is the set of all possible output values of the function.\n- For quotients like $\frac{f}{g}$ and $\frac{g}{f}$, the denominator cannot be zero.\n\n3. **Analyze $f(x) = 2$:**\n- Domain: Since $f$ is a constant function, it is defined for all real numbers, so domain is $\mathbb{R}$.\n- Range: The output is always 2, so range is $\{2\}$.\n\n4. **Analyze $g(x) = x^2 + 1$:**\n- Domain: $g$ is a polynomial, defined for all real numbers, so domain is $\mathbb{R}$.\n- Range: Since $x^2 \geq 0$ for all $x$, $x^2 + 1 \geq 1$. The minimum value is 1 at $x=0$, so range is $[1, \infty)$.\n\n5. **Analyze $\frac{f}{g} = \frac{2}{x^2 + 1}$:**\n- Domain: Denominator $x^2 + 1$ is never zero (since $x^2 \geq 0$, $x^2 + 1 \geq 1$), so domain is all real numbers $\mathbb{R}$.\n- Range: Since $x^2 + 1 \geq 1$, the fraction $\frac{2}{x^2 + 1} \leq 2$ and is always positive. The maximum value is $\frac{2}{1} = 2$ at $x=0$, and as $|x| \to \infty$, $\frac{2}{x^2 + 1} \to 0$. So range is $(0, 2]$.\n\n6. **Analyze $\frac{g}{f} = \frac{x^2 + 1}{2}$:**\n- Domain: $f(x) = 2$ is never zero, so domain is all real numbers $\mathbb{R}$.\n- Range: Since $x^2 + 1 \geq 1$, dividing by 2 gives $\frac{x^2 + 1}{2} \geq \frac{1}{2}$. So range is $[\frac{1}{2}, \infty)$.\n\n**Final answers:**\n- $f$: domain $\mathbb{R}$, range $\{2\}$\n- $g$: domain $\mathbb{R}$, range $[1, \infty)$\n- $\frac{f}{g}$: domain $\mathbb{R}$, range $(0, 2]$\n- $\frac{g}{f}$: domain $\mathbb{R}$, range $[\frac{1}{2}, \infty)$