1. **Problem statement:** Given functions $f(x)$ and $g(x)$, find the compositions $f \circ g$ and $g \circ f$, and state their domains. 2. **Recall:** - The composition $f \circ g$ means $f(g(x))$. - The composition $g \circ f$ means $g(f(x))$. - The domain of $f \circ g$ is all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$. - The domain of $g \circ f$ is all $x$ in the domain of $f$ such that $f(x)$ is in the domain of $g$. --- ### (a) $f(x) = 2 - \frac{1}{x + |x|}$, $g(x) = x^2 - 4$ 3. **Find $f \circ g$: $f(g(x))$** $$f(g(x)) = 2 - \frac{1}{g(x) + |g(x)|} = 2 - \frac{1}{x^2 - 4 + |x^2 - 4|}$$ 4. **Domain of $f$:** - Denominator $x + |x| \neq 0$. - Note $x + |x| = 0$ when $x = 0$ or $x < 0$ (since for $x<0$, $x + |x| = x - x = 0$). - Actually, for $x>0$, $x + |x| = x + x = 2x \neq 0$. - For $x=0$, denominator is $0$. - For $x<0$, $x + |x| = 0$. - So domain of $f$ is $x > 0$. 5. **Domain of $g$:** all real numbers. 6. **Domain of $f \circ g$:** all $x$ such that $g(x) > 0$ (since $f$ requires input $>0$). Solve $g(x) = x^2 - 4 > 0$: $$x^2 - 4 > 0 \implies (x-2)(x+2) > 0$$ This holds when $x < -2$ or $x > 2$. So domain of $f \circ g$ is $(-\infty, -2) \cup (2, \infty)$. --- 7. **Find $g \circ f$: $g(f(x))$** $$g(f(x)) = (f(x))^2 - 4 = \left(2 - \frac{1}{x + |x|}\right)^2 - 4$$ 8. **Domain of $g$:** all real numbers. 9. **Domain of $f$:** $x > 0$ (from step 4). 10. **Domain of $g \circ f$:** all $x$ in domain of $f$, so $x > 0$. --- ### (b) $f(x) = \sqrt{x}$, $g(x) = |x|$ 11. **Find $f \circ g$: $f(g(x))$** $$f(g(x)) = \sqrt{|x|}$$ 12. **Domain of $g$:** all real numbers. 13. **Domain of $f$:** $x \geq 0$ (since square root requires nonnegative input). 14. **Domain of $f \circ g$:** all $x$ such that $g(x) \geq 0$. Since $|x| \geq 0$ for all $x$, domain is all real numbers. --- 15. **Find $g \circ f$: $g(f(x))$** $$g(f(x)) = |\sqrt{x}| = \sqrt{x}$$ 16. **Domain of $f$:** $x \geq 0$. 17. **Domain of $g$:** all real numbers. 18. **Domain of $g \circ f$:** domain of $f$, so $x \geq 0$. --- **Final answers:** - (a) $f \circ g = 2 - \frac{1}{x^2 - 4 + |x^2 - 4|}$ with domain $(-\infty, -2) \cup (2, \infty)$. - (a) $g \circ f = \left(2 - \frac{1}{x + |x|}\right)^2 - 4$ with domain $(0, \infty)$. - (b) $f \circ g = \sqrt{|x|}$ with domain $(-\infty, \infty)$. - (b) $g \circ f = \sqrt{x}$ with domain $[0, \infty)$.