1. **Problem statement:** Given the function $f(x) = \frac{1}{x+5}$, find the domain and range of its inverse $f^{-1}$, and determine if it is possible to do this without finding the expression for $f^{-1}$. 2. **Recall the domain and range of $f(x)$:** - The domain of $f$ is all real numbers except where the denominator is zero, so $x \neq -5$. Thus, domain of $f$ is $(-5, \infty)$ or $(-\infty, -5) \cup (-5, \infty)$. - The range of $f$ is all real numbers except zero, since $\frac{1}{x+5}$ never equals zero. So range of $f$ is $\mathbb{R} \setminus \{0\}$. 3. **Domain and range of the inverse function:** - The domain of $f^{-1}$ is the range of $f$, so domain of $f^{-1}$ is $\mathbb{R} \setminus \{0\}$. - The range of $f^{-1}$ is the domain of $f$, so range of $f^{-1}$ is $\mathbb{R} \setminus \{-5\}$. 4. **Finding $f^{-1}(x)$ explicitly:** Start with $y = \frac{1}{x+5}$. Swap $x$ and $y$ to find inverse: $$x = \frac{1}{y+5}$$ Multiply both sides by $y+5$: $$x(y+5) = 1$$ $$xy + 5x = 1$$ Solve for $y$: $$xy = 1 - 5x$$ $$y = \frac{1 - 5x}{x}$$ Simplify by splitting the fraction: $$y = \frac{1}{x} - 5$$ So, $$f^{-1}(x) = \frac{1}{x} - 5$$ 5. **Domain and range from the inverse expression:** - Domain of $f^{-1}$ is all real numbers except $x \neq 0$ (since division by zero is undefined). - Range of $f^{-1}$ is all real numbers except $y \neq -5$ (since $\frac{1}{x}$ never equals zero). 6. **Is it possible to find domain and range of $f^{-1}$ without finding $f^{-1}$?** Yes, by using the fact that the domain of $f^{-1}$ is the range of $f$, and the range of $f^{-1}$ is the domain of $f$. However, to understand asymptotes and behavior fully, finding the explicit expression helps. --- 7. **Problem statement:** Given a one-to-one function $g$ with a vertical asymptote at $x = -5$ and the table of values: \begin{tabular}{c|c} x & g(x) \\ \hline -5 & \text{undefined} \\ -2 & -14 \\ 0 & -11 \\ 3 & -1 \\ 6 & 0 \end{tabular} (a) Find the $y$-intercept of $g^{-1}$. (b) Write the equation for the horizontal asymptote of $g^{-1}$. 8. **Solution for (a):** The $y$-intercept of $g^{-1}$ is the value of $g^{-1}(0)$. Since $g^{-1}$ swaps $x$ and $y$, the $y$-intercept of $g^{-1}$ corresponds to the $x$ value where $g(x) = 0$. From the table, $g(6) = 0$, so $$g^{-1}(0) = 6$$ Thus, the $y$-intercept of $g^{-1}$ is 6. 9. **Solution for (b):** The vertical asymptote of $g$ at $x = -5$ becomes a horizontal asymptote of $g^{-1}$ at $y = -5$. Therefore, the equation of the horizontal asymptote of $g^{-1}$ is $$y = -5$$