Inverse Functions - Help, Please

[OpixMK]

I'm not an expert at math, and I'm taking advanced analysis II. Could someone please help me?
How do you find the inverse function of f(x) = x/(x-2)
I really appreciate any help. Thanks.

[TTitan44]

I'm a senior in high school and I'm taking Algebra II Academic.

[kingedward]

Im a sophomore and Im taking advanced algebra II and trig. :bigsmile:

[zzzaacckk]

to find the inverse, just remember that f(x) is another way of saying y, so switch your ys with your x's so the eq become: x = y/(y-2) and resolve for y.

[bludhoundz]

I'm not an expert at math, and I'm taking advanced analysis II. Could someone please help me?
How do you find the inverse function of f(x) = x/(x-2)
I really appreciate any help. Thanks.

To invert a function, switch y with x, and then rearrange.

In this case, since you have multiple x's, it won't work (or at least I haven't done it in long enough to make that work). I don't remember another way to work it out algebraically.

Graphically, you can just flip the graph over the y = x line and try to figure out the equation for that relation.

zack - you can't simplify that. It normally works, but when you have 2 y's in the equation, it's not possible to simplify it down to one (or at least my brief try didn't work out very well).

[kingedward]

You could get rid of the fraction by multiplying...you would get x = y squared - 2y.

Dunno if that helps, Ive never done inverse functions.

[bludhoundz]

Nah, you have to completely isolate y.

I looked up how to do it quickly, and tried their method, but it doesn't work for this function.

[HorNeT]

Normally you don't have two of the same variables in a problem like this.

[Maverik07]

Graph it in the x direction, then just flip it around into the y direction.

I forget how to really do it, but that'll get you the answer.

[TeXaS LoNgHoRnS]

To invert a function, switch y with x, and then rearrange.

In this case, since you have multiple x's, it won't work (or at least I haven't done it in long enough to make that work). I don't remember another way to work it out algebraically.

Graphically, you can just flip the graph over the y = x line and try to figure out the equation for that relation.

zack - you can't simplify that. It normally works, but when you have 2 y's in the equation, it's not possible to simplify it down to one (or at least my brief try didn't work out very well).

f(x) = ax + b
f(x) = x2 , f '(x) = 2 x
f(x) = x2
f(x) = √x
f(x) = xn
[f(x) + g(x)]' = f '(x) + g'(x)
[f(x) · g(x)]' = f '(x) · g(x) + f(x) · g'(x)

These are some i have done, tho i live in Denmark, so i don't know how to explain. Therefor i thought maybe i just show you mine.


:huh::confused:

What the hell...

Im still in algebra 2...damn smart people lol

[Maverik07]

In Europe, the learning curve for math is much higher than here. Hes doing derivatives, thats calc AB stuff

[Azer]

You could make it into a vector function and reflect that over <1,1>, but that's probably making things unnecessarily complicated.

Lets see here..
f(x) = x/(x-2)
x = f(x)'/(f(x)' - 2) [like substituting y for x, except I'm using f(x)']
x*f(x)' - 2x = f(x)' [multiply both sides by f(x)']
x*f(x)' -f(x)' = 2x [subtract f(x)' from both sides; add 2x to both sides]
f(x)'*(x-1) = 2x [factored out a f(x)' term from the left side]
f(x)' = 2x/(x-1) [divided both sides by (x-1)]

Thus the inverse of f(x) = x/(x-2) is f(x)' = 2x/(x-1).

[Geoffrey]

I did derivatives in year 10 here...

[Maverik07]

Exactly, US has different math standards for some reason, i did derivatives as a junior, but im in higher math levels than most kids. Most kids dont do derivatives until senior year or college, in US.

[Azer]

/me is learning non-Euclidean math.

And it is confusing as hell.

[bludhoundz]

You could make it into a vector function and reflect that over <1,1>, but that's probably making things unnecessarily complicated.

Lets see here..
f(x) = x/(x-2)
x = f(x)'/(f(x)' - 2) [like substituting y for x, except I'm using f(x)']
x*f(x)' - 2x = f(x)' [multiply both sides by f(x)']
x*f(x)' -f(x)' = 2x [subtract f(x)' from both sides; add 2x to both sides]
f(x)'*(x-1) = 2x [factored out a f(x)' term from the left side]
f(x)' = 2x/(x-1) [divided both sides by (x-1)]

Thus the inverse of f(x) = x/(x-2) is f(x)' = 2x/(x-1).
Yeah, that works. I was trying to do something similar last night, but I made the mistake of solving everything for x and then interchanging x with y instead of the other way around.

But you really shouldn't label the inverse function with a ' because that can be confused with derivatives, though it's in a different place. If possible, I'd try to find something on the forum which allows you to write superscript, but I haven't found it in the past 5 minutes which I've been looking.

[steve12]

to find the inverse, just remember that f(x) is another way of saying y, so switch your ys with your x's so the eq become: x = y/(y-2) and resolve for y.

To invert a function, switch y with x, and then rearrange.

In this case, since you have multiple x's, it won't work (or at least I haven't done it in long enough to make that work). I don't remember another way to work it out algebraically.

Graphically, you can just flip the graph over the y = x line and try to figure out the equation for that relation.

zack - you can't simplify that. It normally works, but when you have 2 y's in the equation, it's not possible to simplify it down to one (or at least my brief try didn't work out very well).


Read those for the simplest way to solve.

[abyaly]

Yeah, that works. I was trying to do something similar last night, but I made the mistake of solving everything for x and then interchanging x with y instead of the other way around.That should give you the same answer :p

[bludhoundz]

That should give you the same answer :p

Well then I was just tired and did it horribly horribly wrong :confused: