Inverse function of binomial where both are fractions (I need help)

There is a procedure for that. Have you seen it? Done it on a simpler one?
 
I know about the procedure but only if it's one fraction or a binomial where one is a fraction and the other is a whole number. The answer I ended up with is [-9±√(400x2+120x-79)] / 8x-4
Seems a bit of a stretch hahahaha
 
Couple of comments.
1) You did not write that y = what you wrote!
2) In the denominator I would not have multiplied out. It could have been possible that when you factored the numerator it could have cancelled out with one of the factors in the denominator but you would not have seen it since you multiplied out.


Most importantly did you notice that based on your result that the inverse does not exist. Why?
 
I've actually inputted the same question on Mathway, and it also told me that an inverse of it does not exist. Apparently though, in the exam, there is a correct answer: (5x-2)/(2x-1)
I'm confused hahahaha
 
You needn't be confused. Things are just wrong, sometimes. That "correct answer" is not correct at all - not all that related, frankly. Answer to a different problem, maybe? The inverse FUNCTION may not exist, but there is no need to throw out everything that may be called an "inverse".

You can prove the existence of the inverse mechanically.
1) Draw your function, y = f(x) on some thin paper.
2) On the same set of coordinate axes, add the line y = x and extend it until it hits the borders of the paper in both directions.
3) With the graph facing you, grab the line y = x at both points where it leaves the edge of the paper.
4) Flip the paper half way over, with the lower right corner ending up in the upper left corner and the graph now facing away from you..
5) Hold the paper up to the light and there it is..
No need to claim that it doesn't exist.
 
I tell my student to draw f(x) like they normally would. Just interchange where you label the x and y axes. Now turn the paper such that x and y are in there normal positions and there is your inverse--if it exists. It must be a function to exist, that is it must pass the vertical line test.
 
Jomo said:
f(x)=(x-3)/(2x+5) + 1/(2x-5)
and we claim f-1(x) = (5x-2)/(2x-1)

Now f(2) = -10/9 . Then f-1(-10/9) should be 2. Well f-1(-10/9) = 68/29
Click to expand...
It's close, for this value. One would not need to imagine that it would be this close for all other valuues.
 
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