1. Problem 15a: Find the inverse function $f^{-1}(x)$ of $f(x)=3x^{-2}$. The user proposes: $x+2=3y\ y=\frac{x+2}{3}$ but this does not relate correctly to $f$. Let's find $f^{-1}$ correctly. Given $f(x) = 3x^{-2} = \frac{3}{x^2}$. To find $f^{-1}$, set $y=f(x)=\frac{3}{x^2}$ and solve for $x$ in terms of $y$: $$y = \frac{3}{x^2} \implies x^2 = \frac{3}{y} \implies x = \pm \sqrt{\frac{3}{y}}$$ So, the inverse function is: $$f^{-1}(x) = \pm \sqrt{\frac{3}{x}}$$ (We specify branch depending on domain). --- 2. Problem 15b: Find $g(f(x))$ where $g(x) = \frac{x+2}{x+1}$ and $f(x)=3x^{-2} = \frac{3}{x^2}$. Compute: $$g(f(x)) = g\left(\frac{3}{x^2}\right) = \frac{\frac{3}{x^2} + 2}{\frac{3}{x^2} + 1} = \frac{\frac{3+2x^2}{x^2}}{\frac{3+x^2}{x^2}} = \frac{3+2x^2}{3+x^2}$$ --- 3. Problem 15c: Simplify $g(f(x))$: $$g(f(x)) = \frac{3+2x^2}{3+x^2}$$ This expression is already simplified. --- 4. Problem 16a: Find $f\left(\frac{1}{2}\right)$ for $f(x)=\frac{1}{x^2+1}$: $$f\left(\frac{1}{2}\right) = \frac{1}{(\frac{1}{2})^2 + 1} = \frac{1}{\frac{1}{4} + 1} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$$ --- 5. Problem 16b: Find $f(g(x))$ where $g(x)=\sqrt{x-1}, x \geq 1$ and $f(x)=\frac{1}{x^2+1}$: First compute: $$f(g(x)) = f(\sqrt{x-1}) = \frac{1}{(\sqrt{x-1})^2 + 1} = \frac{1}{x - 1 + 1} = \frac{1}{x}$$ --- Final answers: 1. $$f^{-1}(x) = \pm \sqrt{\frac{3}{x}}$$ 2. $$g(f(x)) = \frac{3+2x^2}{3+x^2}$$ 3. $$g(f(x))$$ simplified is the same: $\frac{3+2x^2}{3+x^2}$ 4. $$f\left(\frac{1}{2}\right) = \frac{4}{5}$$ 5. $$f(g(x)) = \frac{1}{x}$$