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Quick Algebra question

  1. Apr 25, 2008 #1
    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
     
  2. jcsd
  3. Apr 25, 2008 #2

    symbolipoint

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    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.
     
  4. Apr 25, 2008 #3
    umm my book saids y= x+1/x
     
  5. Apr 25, 2008 #4

    symbolipoint

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    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.
     
  6. Apr 25, 2008 #5
    yea thats i did but what i dont know is where i went wrong in my algebra
     
  7. Apr 25, 2008 #6

    symbolipoint

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    ... just to continue, your next step would be to multiply both sides by 'y'.
     
  8. Apr 25, 2008 #7
    k this is what i got so far x(y)= y-1
    should i move the y to the other side or the '-1'?
     
  9. Apr 25, 2008 #8

    symbolipoint

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    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?
     
  10. Apr 25, 2008 #9

    HallsofIvy

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    If this were y= (x+1)/x, becoming x= (y+1)/y, then you would have xy= y+ 1.
    First factor out "y": y(x- 1)= 1 and then divide by x-1.

    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.
     
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