1. Problem Q1: Find the domain and range of $f(x)=x^2$. 2. Formula and rules: For polynomial functions there are no denominators or even roots to restrict domain, so domain is all real numbers. 3. Work for domain: There is no denominator and no even root, so any real $x$ is allowed. 4. Express domain: Domain $=\mathbb{R}$. 5. Work for range: Squaring yields $y=x^2\ge 0$ for all real $x$; minimum occurs at $x=0$ with $y=0$ and $y$ increases without bound as $|x|\to\infty$. 6. Express range: Range $=[0,\infty)$. 7. Final Answer Q1: Domain $=\mathbb{R}$, Range $=[0,\infty)$. 8. Problem Q2: Find the domain of $f(x)=\frac{1}{x-2}$. 9. Rules: A rational function has domain all real numbers except where the denominator is zero. 10. Work: Set denominator $x-2\ne 0$ which gives $x\ne 2$. 11. Final Answer Q2: Domain $=\mathbb{R}\setminus\{2\}$. 12. Problem Q3: Find the domain and range of $f(x)=\sqrt{x+3}$. 13. Rules: For the principal square root the radicand must satisfy $x+3\ge 0$ and the output is $\ge 0$. 14. Work for domain: Solve $x+3\ge 0$ gives $x\ge -3$ so Domain $=[-3,\infty)$. 15. Work for range: The square root yields $y=\sqrt{x+3}\ge 0$ with minimum $0$ at $x=-3$ and grows without bound, so Range $=[0,\infty)$. 16. Final Answer Q3: Domain $=[-3,\infty)$, Range $=[0,\infty)$. 17. Problem Q4: Let $f(x)=x+1$ and $g(x)=x^2$. Find $(f\circ g)(2)$ and $(g\circ f)(2)$. 18. Formula and rules: Composition is defined by $(f\circ g)(x)=f(g(x))$ and we evaluate at the given input. 19. Work for $(f\circ g)(2)$: Compute $g(2)=2^2=4$ then $f(g(2))=f(4)=4+1=5$. 20. Work for $(g\circ f)(2)$: Compute $f(2)=2+1=3$ then $g(f(2))=g(3)=3^2=9$. 21. Final Answer Q4: $(f\circ g)(2)=5$, $(g\circ f)(2)=9$. 22. Problem Q5: Find the inverse of $f(x)=3x+4$ and verify by composition. 23. Formula and rules: To find an inverse swap $x$ and $y$ and solve for $y$, then check $f(f^{-1}(x))=x$. 24. Work: Let $y=3x+4$ then swap to $x=3y+4$ and solve $3y=x-4$ so $y=\frac{x-4}{3}$, hence $f^{-1}(x)=\frac{x-4}{3}$. 25. Verification: Compute $f(f^{-1}(x))=3\left(\frac{x-4}{3}\right)+4=x-4+4=x$ which confirms the inverse. 26. Final Answer Q5: $f^{-1}(x)=\frac{x-4}{3}$. 27. Problem Q6: Find the inverse of $f(x)=\frac{x-2}{5}$ and verify by composition. 28. Rules: Use the swap-and-solve method for linear functions. 29. Work: Let $y=\frac{x-2}{5}$ then swap to $x=\frac{y-2}{5}$ and multiply to get $5x=y-2$ so $y=5x+2$, thus $f^{-1}(x)=5x+2$. 30. Verification: $f(f^{-1}(x))=\frac{(5x+2)-2}{5}=\frac{5x}{5}=x$ so the inverse is correct. 31. Final Answer Q6: $f^{-1}(x)=5x+2$.