Find the range of function $f(x) = \frac{4-x}{x-4}$ We have to find the range of the function,

$y = \dfrac{4-x}{x-4}$

My approach:-
I know the first method to find the range of a function by finding  the domain of inverse function.
$y = \dfrac{4-x}{x-4}$
$\implies y(x-4) = 4-x$
$\implies xy - 4y = 4-x$
$\implies xy +x = 4+4y$
$\implies x(y+1) = 4(y+1)$
$\implies x = \dfrac{4(y+1)}{(y+1)}$
Now, here $y$ cannot be equal to $-1$.
Therefore the range of the function is $\mathbb{R} - \{-1\}$
The second method is :-
$y= \dfrac{4-x}{x-4}$
$y = \dfrac{-(x-4)}{(x-4)}$
$y  = -1$
Therefore the range of the function is $\{-1\}$
I checked my answer on wolfram alpha, $\{-1\}$ is the correct answer. But what's the mistake in the first method? Which step is incorrect?
Further more, I tried to check my solution step by step on desmos by plotting the graph. What I found is that the graphs from 1st to 6th step are same but the graph changed at the step when I got $x = \dfrac{4(1+y)}{(1+y)}$.
I need help here. Thanks in advance!

 A: The method of finding the domain of the inverse cannot work when the function has no inverse, which is the case here. Actually, your function is constant (and, in particular, not injective) and only takes the value $-1$. Therefore, the range is $\{-1\}$.
A: When finding the inverse of a function algebraically, you switch the X's and the Y's of the function to represent switching the x and y axis on a graph. Therefore,the inverse of Y = (4-X)/(X-4) is X = (4-Y)/(Y-4). Take a -1 out of the numerator and you get X = (-1)(Y-4) / (Y-4). Simplify by taking out Y-4 from both the numerator and denominator and you are left with X = -1. Which means that the range of the inital function is -1.
You went wrong at the very beggining, since you didn't initally switch X and Y.Graphing it on desmos, it becomes clear that the original function is simply a straight line on Y = -1, with it's "Inverse" (there is actually not an inverse function for this function, via the horizontal line test) being X = -1 (Not a function, via the vertical line test)
