1. **Problem:** Find the inverse function for each given function and state the domain of the inverse. --- ### (a) $f(x) = x + 2$ 1. Write the function: $y = x + 2$ 2. Swap $x$ and $y$ to find the inverse: $x = y + 2$ 3. Solve for $y$: $$y = x - 2$$ 4. So, the inverse function is $$f^{-1}(x) = x - 2$$ 5. Since $f$ is defined for all real $x$, the domain of $f^{-1}$ is also all real numbers. --- ### (b) $f(x) = 4x - 2$ 1. Write the function: $y = 4x - 2$ 2. Swap $x$ and $y$: $x = 4y - 2$ 3. Solve for $y$: $$4y = x + 2$$ $$y = \frac{x + 2}{4}$$ 4. The inverse function is $$f^{-1}(x) = \frac{x + 2}{4}$$ 5. Domain of $f^{-1}$ is all real numbers. --- ### (c) $f(x) = x$ 1. Write the function: $y = x$ 2. Swap $x$ and $y$: $x = y$ 3. Solve for $y$: $$y = x$$ 4. The inverse function is $$f^{-1}(x) = x$$ 5. Domain of $f^{-1}$ is all real numbers. --- ### (d) $f(x) = \frac{3}{x}$, $x \neq 0$ 1. Write the function: $y = \frac{3}{x}$ 2. Swap $x$ and $y$: $$x = \frac{3}{y}$$ 3. Solve for $y$: $$y = \frac{3}{x}$$ 4. The inverse function is $$f^{-1}(x) = \frac{3}{x}$$ 5. Domain of $f$ is $x \neq 0$, so domain of $f^{-1}$ is also $x \neq 0$. --- ### (e) $f(x) = \frac{1}{x + 2}$, $x \neq -2$ 1. Write the function: $y = \frac{1}{x + 2}$ 2. Swap $x$ and $y$: $$x = \frac{1}{y + 2}$$ 3. Solve for $y$: $$x(y + 2) = 1$$ $$xy + 2x = 1$$ $$xy = 1 - 2x$$ $$y = \frac{1 - 2x}{x}$$ 4. So, the inverse function is $$f^{-1}(x) = \frac{1 - 2x}{x}$$ 5. Domain of $f$ is $x \neq -2$, so range excludes 0 because $f(-2)$ is undefined. Domain of $f^{-1}$ is $x \neq 0$ because division by zero is undefined. --- **Summary:** - (a) $f^{-1}(x) = x - 2$, domain $\mathbb{R}$ - (b) $f^{-1}(x) = \frac{x + 2}{4}$, domain $\mathbb{R}$ - (c) $f^{-1}(x) = x$, domain $\mathbb{R}$ - (d) $f^{-1}(x) = \frac{3}{x}$, domain $x \neq 0$ - (e) $f^{-1}(x) = \frac{1 - 2x}{x}$, domain $x \neq 0$