1. **Problem Statement:** Find the inverse functions for the given functions: - $f(x) = 3x + 5$ - $y = -2x + 7$ - $f(x) = \frac{2}{3}x - 1$ 2. **Formula and Important Rules:** To find the inverse function $f^{-1}(x)$ of a function $f(x)$, we swap $x$ and $y$ and then solve for $y$. If $y = f(x)$, then the inverse satisfies $x = f(y)$. 3. **Step-by-step Solutions:** **For $f(x) = 3x + 5$:** - Start with $y = 3x + 5$ - Swap $x$ and $y$: $x = 3y + 5$ - Solve for $y$: $$x - 5 = 3y$$ $$\Rightarrow y = \frac{x - 5}{3}$$ - Using cancellation notation: $$y = \frac{\cancel{3} \cdot \frac{x - 5}{\cancel{3}}}{1} = \frac{x - 5}{3}$$ - So, the inverse function is: $$f^{-1}(x) = \frac{x - 5}{3}$$ **For $y = -2x + 7$:** - Start with $y = -2x + 7$ - Swap $x$ and $y$: $x = -2y + 7$ - Solve for $y$: $$x - 7 = -2y$$ $$\Rightarrow y = \frac{x - 7}{-2} = -\frac{x - 7}{2}$$ - Using cancellation notation: $$y = -\frac{\cancel{2} \cdot \frac{x - 7}{\cancel{2}}}{1} = -\frac{x - 7}{2}$$ - So, the inverse function is: $$f^{-1}(x) = -\frac{x - 7}{2}$$ **For $f(x) = \frac{2}{3}x - 1$:** - Start with $y = \frac{2}{3}x - 1$ - Swap $x$ and $y$: $x = \frac{2}{3}y - 1$ - Solve for $y$: $$x + 1 = \frac{2}{3}y$$ $$\Rightarrow y = \frac{3}{2}(x + 1)$$ - Using cancellation notation: $$y = \frac{\cancel{3}}{\cancel{2}}(x + 1) = \frac{3}{2}(x + 1)$$ - So, the inverse function is: $$f^{-1}(x) = \frac{3}{2}(x + 1)$$ 4. **Summary:** - $f^{-1}(x) = \frac{x - 5}{3}$ - $f^{-1}(x) = -\frac{x - 7}{2}$ - $f^{-1}(x) = \frac{3}{2}(x + 1)$ These are the inverse functions for the bottom three questions.