1. **Problem Statement:** Find the inverse function $f^{-1}$ of $f(x) = 3x$, check the answer, and find the domain and range of both $f$ and $f^{-1}$. Also, graph $f$, $f^{-1}$, and $y=x$ on the same axes. 2. **Formula and Rules:** To find the inverse function, swap $x$ and $y$ in the equation $y = f(x)$ and solve for $y$. The inverse function $f^{-1}(x)$ satisfies $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. 3. **Find the inverse:** Start with $y = 3x$. Swap $x$ and $y$: $x = 3y$. Solve for $y$: $$y = \frac{x}{3}$$ This is the inverse function: $$f^{-1}(x) = \frac{x}{3}$$ 4. **Check the inverse:** Check $f(f^{-1}(x))$: $$f\left(f^{-1}(x)\right) = f\left(\frac{x}{3}\right) = 3 \times \frac{x}{3} = \cancel{3} \times \frac{x}{\cancel{3}} = x$$ Check $f^{-1}(f(x))$: $$f^{-1}(f(x)) = f^{-1}(3x) = \frac{3x}{3} = \cancel{3}x / \cancel{3} = x$$ Both checks confirm the inverse is correct. 5. **Domain and Range:** - For $f(x) = 3x$, domain is all real numbers $(-\infty, \infty)$. - Range of $f$ is also all real numbers $(-\infty, \infty)$. - For $f^{-1}(x) = \frac{x}{3}$, domain and range are also all real numbers $(-\infty, \infty)$. 6. **Graphing:** - $f(x) = 3x$ is a straight line through the origin with slope 3. - $f^{-1}(x) = \frac{x}{3}$ is a straight line through the origin with slope $\frac{1}{3}$. - $y = x$ is the identity line. - The graph of $f$ and $f^{-1}$ are reflections of each other across the line $y = x$. **Final answer:** $$f^{-1}(x) = \frac{x}{3}$$ Domain and range of both $f$ and $f^{-1}$ are $(-\infty, \infty)$.