1. **State the problem:** Given functions $f(x) = 3x - 2$ and $g(x) = x - 3$, find the inverse functions $f^{-1}(x)$ and $g^{-1}(x)$. 2. **Find $f^{-1}(x)$:** Start with $y = 3x - 2$. Swap $x$ and $y$ to find the inverse: $x = 3y - 2$. Solve for $y$: $$x + 2 = 3y$$ $$y = \frac{x + 2}{3}$$ So, $$f^{-1}(x) = \frac{x + 2}{3}$$ 3. **Find $g^{-1}(x)$:** Start with $y = x - 3$. Swap $x$ and $y$: $$x = y - 3$$ Solve for $y$: $$y = x + 3$$ So, $$g^{-1}(x) = x + 3$$ 4. **Relation between $(fg)^{-1}(x)$ and $g^{-1}f^{-1}(x)$:** Recall the property of inverse functions: $$(fg)^{-1} = g^{-1}f^{-1}$$ This means the inverse of the composition $fg$ is the composition of the inverses in reverse order. 5. **Sketch graphs of $y = f(x)$ and $y = f^{-1}(x)$:** - $f(x) = 3x - 2$ is a straight line with slope 3 and y-intercept -2. - $f^{-1}(x) = \frac{x + 2}{3}$ is also a straight line with slope $\frac{1}{3}$ and y-intercept $\frac{2}{3}$. 6. **Relationship between the graphs:** The graph of $f^{-1}(x)$ is the reflection of the graph of $f(x)$ across the line $y = x$. **Final answers:** $$f^{-1}(x) = \frac{x + 2}{3}$$ $$g^{-1}(x) = x + 3$$ $$(fg)^{-1} = g^{-1}f^{-1}$$ The graphs of $f$ and $f^{-1}$ are symmetric about the line $y = x$.