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Finding the Range?

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the range of f

    2. Relevant equations

    F(x)= 1/(x-1)(x+1) , x>1

    3. The attempt at a solution

    I started by substituting x=2 and x=3 this gives me 1/3 and 1/8 respectively so i get that as x increases y decreases to infinity but how do i represent that as


    its just the notation and stuff i am struggling on.
  2. jcsd
  3. Dec 19, 2008 #2


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    Hi matt_crouch! :smile:
    Be systematic …

    wirte it out as an equation in x …

    you should get a quadratic equation …

    for what values of f does it have real solutions?
  4. Dec 19, 2008 #3


    Staff: Mentor

    Do you mean this as [tex]F(x) = \frac{1}{(x -1)(x + 1)}[/tex]? If so, you should have put another pair of parentheses around the two factors involving x, like so: F(x) = 1/((x - 1)(x + 1)).
    How can y decrease to infinity?

    You're given that x > 1, so both factors in the denominator (I'm assuming that's what you meant) are going to be positive. The numerator is 1, which is positive. The closer x gets to 1, the larger F(x) will be. As x gets large, F(x) will get closer to zero, but will remain positive.
  5. Dec 19, 2008 #4
    Because its a 1/x relationship isnt the graph asymptotic?
    Isnt the range a part of the graph where a horizontal line cuts the graph at two points?
  6. Dec 19, 2008 #5


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    No, the range is those parts of the vertical axis where a horizontal line cuts the graph at at least one point.

    Anyway, what quadratic equation did you get? :smile:
  7. Dec 19, 2008 #6
    Yes, the graph of 1/x is asymptotic by definition. However, when a horizontal line (that is not part of the original graph) intersects the graph at more than one point, then the function in not one-to-one.
  8. Dec 20, 2008 #7
    Im not quite sure how you can write it out as an quatratic. should i find the inverse of f(x)
  9. Dec 20, 2008 #8


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    x2 - 1 - 1/f = 0 … for what values of f are there solutions?
  10. Dec 20, 2008 #9
    ahh i think i got it.. is the range


  11. Dec 20, 2008 #10


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    how did you get that? :confused:

    hmm … i've just looked again at your original post …
    which didn't seem to me to make any sense originally, but now it's dawned on me :redface: that you probably meant "zero" not "infinity" …

    in which case the answer to your original question "how do i represent that as A<y<B" is that if you know that y decreases to zero (but never quite makes it), then you write that simply as y > 0. :smile:
  12. Dec 20, 2008 #11
    ye i was getting confused because as the graph was asymtopic i was thinking that it would go to infinity but the domain would go on for infinity whereas the range which is the "output" would never quite reach it.

    cheers though :D
  13. Dec 21, 2008 #12
    The "easy" way to find the range is to find the domain of a functions inverse. :wink:
  14. Dec 22, 2008 #13
    As x>1 then isnt the largest value that f(x) can be a 1/3 ? so isnt the range 0 < y < 1/3 ?
  15. Dec 22, 2008 #14


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    No … as you said …
    … f(2) = 1/3. :wink:

    f(1+) = … ? :smile:
  16. Dec 22, 2008 #15
    1/0 which is undefined?
  17. Dec 22, 2008 #16


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    Sort-of …

    f(1+) = ∞ (and f(1-) = -∞) …

    but we're not really supposed to regard ∞ as a number :wink:

    so instead of 0 < f < ∞, we just say 0 < f. :smile:
  18. Dec 22, 2008 #17
    ahhhh is seeeee aaawesome..
    massive help cheers Tiny Tim.
  19. Dec 22, 2008 #18


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    The substance (I don't know what jawbreaking terminology you are required to use but once you get the substance you can adapt it to that :biggrin:)

    Either do or don't write it as 1/(x2-1) according to what seems to you easiest.
    You were asked for x>1.
    So where do you start? At x=1 What is f(1)?
    Where do you end? At x=∞. What is f(∞)?
    f is continuous between these two values of x. So f has to include all values between f(1) and f(∞)?

    For x between those 1 and ∞, for f to go out of the range between f(1) and
    f(∞) f would have to have an extremum in the range of x. First does it look like it?

    Then you could formally demonstrate it hasn't.

    Actually I think the simplest argument is f would have to be increasing with x somewhere in the range. You don't need calculus to show that it can't.
    Last edited: Dec 22, 2008
  20. Dec 22, 2008 #19
    Ye cheers epenguin and tiny tim u have really helped me out =]
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