Solving for Inverse Function of f(x)= x+1 / x

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k this is a function and i have to find the inversed

f(x)= x+1 / x

attempt: x= y + 1 / y
multiply both sides by 'y'

x(y)= y + 1
subtract both sides by y

x(y) -y=1

divide 'x'

im lost from here help please:D
 
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symbolipoint said:
You must improve your notation. You seem to mean, f(x) = (x +1)/y from which you may like to
find an inverse such as x = (y + 1)/x and the rest is simple algebraic steps.

umm my book saids y= x+1/x
 
symbolipoint said:
If that is what your book said, then the exchange will be from y = x + 1/x
to x = y + 1/y

I made a variable writing error in my first response.

yea that's i did but what i don't know is where i went wrong in my algebra
 
symbolipoint said:
... just to continue, your next step would be to multiply both sides by 'y'.

k this is what i got so far x(y)= y-1
should i move the y to the other side or the '-1'?
 
Larrytsai said:
k this is what i got so far x(y)= y-1
should i move the y to the other side or the '-1'?

No. That step is wrong. I obtained an xy term like you did, but I see no way to obtain a clear y as a function of x. Either I have become deficient in some of my inverse function skills, or your original function cannot be converted to an inverse according to "Intermediate Algebra" methods. Certainly someone will advise us. Maybe a different coordinate system? Polar?
 
Larrytsai said:
k this is a function and i have to find the inversed

f(x)= x+1 / x

attempt: x= y + 1 / y
multiply both sides by 'y'

x(y)= y + 1

If this were y= (x+1)/x, becoming x= (y+1)/y, then you would have xy= y+ 1.
subtract both sides by y

x(y) -y=1

divide 'x'
First factor out "y": y(x- 1)= 1 and then divide by x-1.

im lost from here help please:D
However, if, as you appear to be saying, the problem really is y= x+ 1/x, becoming x= y+ 1/y, then multiplying both sides by y gives you xy= y2+ 1. That you would write as the quadratic equation y2- xy+ 1= 0 and solve using the quadratic formula with a= 1, b= -x, c=1. Notice that you will have a "[itex]\pm[/itex]" which means that the original function was not "one to one" and so does not have a true inverse.