Still looking for the inverse function of y=x/(x^2-x-2) ... Thanks =]

Question

anonymous · Answer

Hmm, it doesn't have inverse actually unless you restrict the domain.

jamesj · Answer

(For the record, you really shouldn't spam this question to @Physics.)

Yes, you do need to think carefully about the domains and ranges.  But as a first step, you can find the naive inverse by solving your equation for x.  You can do that, can't you?  Just work through the algebra.

I.e., if  y=x/(x^2-x-2) then

yx^2 - yx - 2y -x = 0
i.e.,
yx^2 -(y+1)x - 2y = 0

Now use the quadratic formula.

anonymous · Answer

@jamesj i think we got exactly that before.  quadratic formula may have been the issue from there on

jamesj · Answer

(@sat73, ah, I hadn't seen the earlier version.)

anonymous · Answer

maybe that was someone else. exact same question though.

anonymous · Answer

thanks for the reply, but i sill can't get it though. I've noticed that the range cannot be for x=2 and x=-1, but how to use the quadratic formula from this step exactly ?

jamesj · Answer

If yx^2 -(y+1)x - 2y = 0
then in standard form --- ax^2 + bx + c = 0 --- 
a = y, b = -(y+1), c = -2y.

anonymous · Answer

put 
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ with 
$$a=y,b=-(y+1), c=2y$$

anonymous · Answer

lol 
$$c=-2y$$ what jamesj said

anonymous · Answer

yes i did that but then i got $$\Delta$$ which i don't know how to make sqrt out of it

jamesj · Answer

That's fine, it is what it is.

anonymous · Answer

and the plus, minus business comes from that fact that your original function is not one to one