Find the inverse of the given one-to-one function $f$. Give the domain and the range of $f$ and of $f^{-1}$, and graph both $f$ and $f{ }^{-1}$ on the same set of axes. \[ f(x)=\frac{x+1}{x-4} \] The inverse function, $f^{-1}(x)=\square$. The domain of $\mathrm{f}$ is $\square$. (Type your answer in interval notation.) The domain of $\mathrm{f}^{-1}$ is $\square$. (Type your answer in interval notation.) The range of $f$ is $\square$. (Type your answer in interval notation.) The range of $f^{-1}$ is $\square$. (Type your answer in interval notation.)

Step 1 ：Replace $f(x)$ with $y$ to get $y = \frac{x+1}{x-4}$

Step 2 ：Swap $x$ and $y$ to get $x = \frac{y+1}{y-4}$

Step 3 ：Solve for $y$ to find the inverse function: $f^{-1}(x) = \frac{4x + 1}{x - 1}$

Step 4 ：The domain of $f$ is all real numbers except for the value that makes the denominator zero, so the domain of $f$ is $(-\infty, 4) \cup (4, \infty)$

Step 5 ：The range of $f$ is all real numbers except for the value that $f(x)$ approaches as $x$ approaches the value excluded from the domain, so the range of $f$ is $(-\infty, 0) \cup (0, \infty)$

Step 6 ：The domain and range of $f^{-1}$ are the range and domain of $f$, respectively, so the domain of $f^{-1}$ is $(-\infty, 0) \cup (0, \infty)$ and the range of $f^{-1}$ is $(-\infty, 4) \cup (4, \infty)$

Step 7 ：\(\boxed{\text{The inverse function, } f^{-1}(x) \text{, is } \frac{4x + 1}{x - 1}}\)

Step 8 ：\(\boxed{\text{The domain of } f \text{ is } (-\infty, 4) \cup (4, \infty)}\)

Step 9 ：\(\boxed{\text{The domain of } f^{-1} \text{ is } (-\infty, 0) \cup (0, \infty)}\)

Step 10 ：\(\boxed{\text{The range of } f \text{ is } (-\infty, 0) \cup (0, \infty)}\)

Step 11 ：\(\boxed{\text{The range of } f^{-1} \text{ is } (-\infty, 4) \cup (4, \infty)}\)