1. Problem: Evaluate and find the domain of the composite functions given: a. $f(x) = \sqrt{x - 2}$ and $g(x) = 5x$, find $(f \circ g)(x)$ and its domain. b. $f(x) = \sqrt{x - 2}$ and $g(x) = 5x$, find $(g \circ f)(x)$ and its domain. 2. Formula and rules: - Composite function $(f \circ g)(x) = f(g(x))$ means substitute $g(x)$ into $f$. - Domain of composite function is all $x$ in domain of $g$ such that $g(x)$ is in domain of $f$. - For $f(x) = \sqrt{x - 2}$, domain is $x - 2 \geq 0 \Rightarrow x \geq 2$. 3. Evaluate $(f \circ g)(x)$: $$ (f \circ g)(x) = f(g(x)) = f(5x) = \sqrt{5x - 2} $$ Domain condition: $5x - 2 \geq 2$ because inside the square root must be $\geq 0$. Simplify domain: $$ 5x - 2 \geq 2 $$ $$ 5x \geq 4 $$ $$ x \geq \frac{4}{5} $$ So domain of $(f \circ g)(x)$ is $[\frac{4}{5}, \infty)$. 4. Evaluate $(g \circ f)(x)$: $$ (g \circ f)(x) = g(f(x)) = g(\sqrt{x - 2}) = 5 \cdot \sqrt{x - 2} $$ Domain condition: $x - 2 \geq 0 \Rightarrow x \geq 2$. So domain of $(g \circ f)(x)$ is $[2, \infty)$. 5. Problem: Use $f(x) = x - 5$ and $g(x) = \frac{2}{x}$ to find $(f \circ g)(x)$ and its domain. 6. Evaluate $(f \circ g)(x)$: $$ (f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{x} - 5 $$ 7. Domain considerations: - $g(x) = \frac{2}{x}$ is defined for $x \neq 0$. - $f(x) = x - 5$ is defined for all real numbers. So domain of $(f \circ g)(x)$ is all real $x$ except $x \neq 0$. Final answers: - $(f \circ g)(x) = \sqrt{5x - 2}$ with domain $x \geq \frac{4}{5}$. - $(g \circ f)(x) = 5 \sqrt{x - 2}$ with domain $x \geq 2$. - $(f \circ g)(x) = \frac{2}{x} - 5$ with domain $x \neq 0$.