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, $f^{-1}(x)=◻$.
The domain of $\mathrm{f}$ is $◻$.
(Type your answer in interval notation.)
The domain of ${\mathrm{f}}^{-1}$ is $◻$.
(Type your answer in interval notation.)
The range of $f$ is $◻$.
(Type your answer in interval notation.)
The range of $f^{-1}$ is $◻$.
(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 $(-\mathrm{\infty },4)\cup (4,\mathrm{\infty })$

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 $(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$

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

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

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

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

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

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