1. **Problem statement:** Find $(f \circ g)(x)$ and its domain, then find $(g \circ f)(x)$ and its domain for the functions given. 2. **Part a:** Given $f(x) = \sqrt{x}$ and $g(x) = \frac{1}{x - 4}$. 3. **Find $(f \circ g)(x)$:** $$(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x - 4}\right) = \sqrt{\frac{1}{x - 4}}$$ 4. **Domain of $(f \circ g)(x)$:** - The expression inside the square root must be $\geq 0$: $$\frac{1}{x - 4} \geq 0$$ - The denominator cannot be zero: $$x - 4 \neq 0 \Rightarrow x \neq 4$$ - Since $1/(x-4)$ is positive when $x-4 > 0$, domain is: $$x - 4 > 0 \Rightarrow x > 4$$ 5. **Find $(g \circ f)(x)$:** $$(g \circ f)(x) = g(f(x)) = g(\sqrt{x}) = \frac{1}{\sqrt{x} - 4}$$ 6. **Domain of $(g \circ f)(x)$:** - $f(x) = \sqrt{x}$ requires $x \geq 0$. - Denominator $\sqrt{x} - 4 \neq 0 \Rightarrow \sqrt{x} \neq 4 \Rightarrow x \neq 16$. - So domain is: $$[0,16) \cup (16, \infty)$$ 7. **Observation:** - Domains of $(f \circ g)(x)$ and $(g \circ f)(x)$ are different. - Composition order affects both the function and its domain. 8. **Part b:** Given $f(x) = x^2$ and $g(x) = \sqrt{16 - x^2}$. 9. **Find $(f \circ g)(x)$:** $$(f \circ g)(x) = f(g(x)) = f\left(\sqrt{16 - x^2}\right) = \left(\sqrt{16 - x^2}\right)^2 = 16 - x^2$$ 10. **Domain of $(f \circ g)(x)$:** - $g(x) = \sqrt{16 - x^2}$ requires: $$16 - x^2 \geq 0 \Rightarrow -4 \leq x \leq 4$$ 11. **Find $(g \circ f)(x)$:** $$(g \circ f)(x) = g(f(x)) = g(x^2) = \sqrt{16 - (x^2)^2} = \sqrt{16 - x^4}$$ 12. **Domain of $(g \circ f)(x)$:** - Inside the square root must be $\geq 0$: $$16 - x^4 \geq 0 \Rightarrow x^4 \leq 16 \Rightarrow |x| \leq 2$$ 13. **Observation:** - Domains differ: $(f \circ g)(x)$ domain is $[-4,4]$, $(g \circ f)(x)$ domain is $[-2,2]$. - Composition order changes the domain and the function. **Final answers:** - Part a: $$(f \circ g)(x) = \sqrt{\frac{1}{x - 4}}, \quad \text{domain: } (4, \infty)$$ $$(g \circ f)(x) = \frac{1}{\sqrt{x} - 4}, \quad \text{domain: } [0,16) \cup (16, \infty)$$ - Part b: $$(f \circ g)(x) = 16 - x^2, \quad \text{domain: } [-4,4]$$ $$(g \circ f)(x) = \sqrt{16 - x^4}, \quad \text{domain: } [-2,2]$$