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F (f inverse (5)) = undefined?

  1. Feb 5, 2005 #1
    let f(x) = 3+x^2+tan(pi x/2), where -1 < x < 1

    -> find f(f_inverse(5))

    The teacher says f(f_inverse(5)) = 5

    but I think f(f_inverse(5)) = undefined

    I don't think the teacher considered the domain of f. You can't get 5 from -1 < x < 1, so who's wrong?
     
  2. jcsd
  3. Feb 5, 2005 #2

    Hurkyl

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    What's the domain of f^-1?
     
  4. Feb 5, 2005 #3
    the domain of f^-1 is the range of f, which doesn't include 5 (when I graphed it)

    Is there a way i can prove that f doesn't include 5 or can i just say something like 5 is not a member of the range of f?
     
  5. Feb 5, 2005 #4

    learningphysics

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    Your teacher is right.

    What is the domain and range of f?

    What is the domain and range of f_inverse?

    If f has an inverse and a is in the range of f... then f(f_inverse(a))=a by the very definition of the inverse.
     
    Last edited: Feb 5, 2005
  6. Feb 5, 2005 #5
    That's my point; a = 5, and with a domain of -1<x<1 for f, a is NOT in the range of f, so f(f_inverse(a)) must be undefined, right?
     
  7. Feb 5, 2005 #6

    learningphysics

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    You made a mistake somewhere. 5 should be in the range of f.
     
  8. Feb 5, 2005 #7

    Hurkyl

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    What's the range of tan (pi x/2) for -1 < x < 1?
     
  9. Feb 5, 2005 #8

    dextercioby

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    Yes,trust me,5 is definitely in the range of "f".I would go as far as telling you that the range of "f" is the entire R.

    Can u see why??

    Daniel.
     
  10. Feb 5, 2005 #9
    EDIT- remember the domain was restricted to (-1 < x < 1); if you stick the equation in a TI-8X you'll see that 5 is not in the range of f.

    Okay I think can "prove" this. Be gentle cause I'm just taking calculus:

    For all one to one functions we have,

    a) f^-1(y) = x <-> f(x) = y;
    and
    b) f(x1) cannot equal f(x2), whenever x1 does not equal x2.

    so,
    f(x) = 5 when x = +/- 1.40056553...

    But x is not a member of (-1 < x < 1)
    (I can stop here)
    and
    f(-x) = f(x), so when f(x) = 5, f(x) is not a one to one function by (b). Thus, f(f^-1(5)) is undefined.
     
  11. Feb 5, 2005 #10

    Hurkyl

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    (Without using your calculator)
     
  12. Feb 5, 2005 #11

    dextercioby

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    Take the calculator & compute the alue for x=0.999.

    Daniel.
     
  13. Feb 5, 2005 #12
    I guess what I'm trying to say is that given the domain of -1 < x < 1, no one can give me a number that will make 3+x^2+tan(pi x/2) = 5. If I'm wrong, then please share that number.

    I understand that tan has R for its range, but there are other components to the function, and with the restriction of the domain, you can't get the function to equal 5.
     
  14. Feb 5, 2005 #13

    Hurkyl

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    So tell me, if the range of tan is all of R... how can adding a small value (how big can 3 + x^2 be?) keep f from reaching 5?
     
  15. Feb 5, 2005 #14

    dextercioby

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    Okay,Hurkyl,i'll leave it up to you... :wink:

    Daniel.
     
  16. Feb 5, 2005 #15
    This is funny because I'm seeing both sides of the picture here. But...

    While the range of tan may be R, the range of tan under a -1<x<1 is limited; it sure isn't R. If you look at this pretty picture here http://www.ping.be/~ping1339/tan.gif you'll see that tan doesn't hit 5 until after 1 (or -1).

    If I'm wrong then please give me a number that's w/in the domain that will make the function equal five. That's all I ask, a simple request.
     
  17. Feb 5, 2005 #16

    Hurkyl

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    Yes, there is something funny going on...


    You drew the graph of tan x, and saw that it reached 5 sometime after 1... but what about tan (pi x/2)?
     
  18. Feb 5, 2005 #17
    The number is 0.6421... I must have graphed it wrong. My apologies.

    Wow I see what went wrong; I graphed the function in degree mode, not radians, which is why i got a "bad" graph
     
    Last edited: Feb 5, 2005
  19. Feb 5, 2005 #18

    Hurkyl

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    That's the answer I got. I hypothesized your calculator was in degree mode.

    Anyways, I hope you see the importance of not relying entirely on your calculator! You should usually try doing some analysis of a problem without it.
     
  20. Feb 5, 2005 #19
    yeah im supposed to do math with my head, not my calculator. Live and learn I guess; the trig func should've tipped me off. Thanks all who contributed
     
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