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anonymous

  • one year ago

Is the value of (1-exp(-x))/(1+exp(-x)) between (-1,1)? How?

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    are you asking if that is the range?

  3. anonymous
    • one year ago
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    Yes

  4. misty1212
    • one year ago
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    \[f(x)=\frac{1-\frac{1}{e^x}}{1+\frac{1}{e^x}}\] easier to see if you multiply by \(e^x\)

  5. misty1212
    • one year ago
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    you get \[f(x)=\frac{e^x-1}{e^x+1}\]

  6. misty1212
    • one year ago
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    now it is pretty clearly always less than 1, since \(e^x>0\) for all \(x\) and the numerator is one less than the denominator

  7. anonymous
    • one year ago
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    But the min value is not -1, right?

  8. misty1212
    • one year ago
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    it cannot be \(-1\) but that is the lower limit, which is why the range is the open interval \((-1,1)\) not the closed one \([-1,1]\)

  9. anonymous
    • one year ago
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    If xtends to -inf, the value is -1 but if x tends to inf it is not 1

  10. misty1212
    • one year ago
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    why not?

  11. anonymous
    • one year ago
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    The upper limit is not one, right?

  12. misty1212
    • one year ago
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    sure it is

  13. anonymous
    • one year ago
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    But e to the power inf is infinity, so if x tends to inf, we have inf divided by inf which is not defined

  14. misty1212
    • one year ago
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    ?

  15. misty1212
    • one year ago
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    you are talking about a limit right?

  16. anonymous
    • one year ago
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    Ya

  17. misty1212
    • one year ago
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    then lets go slow, because the statement "if x tends to inf, we have inf divided by inf which is not defined" makes no sense in terms of limits

  18. anonymous
    • one year ago
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    I m nr sure how the upper limit of the range is 1

  19. misty1212
    • one year ago
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    what do you make of \[\lim_{x\to \infty}\frac{x^2+1}{x^2-1}\]?

  20. anonymous
    • one year ago
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    The numerator goes to inf faster so the limit is inf

  21. anonymous
    • one year ago
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    Or Do we divide by x n see

  22. misty1212
    • one year ago
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    oh no i see the confusion \[x^2+1\] goes to infinity just as fast as \(x^2-1\) the number out at the end makes no difference at all

  23. anonymous
    • one year ago
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    So it tends to 1

  24. misty1212
    • one year ago
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    you remember in some class, maybe called "pre calculus" finding the horizontal asymptote of a rational function?

  25. anonymous
    • one year ago
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    Ya

  26. misty1212
    • one year ago
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    that is what you are doing exactly if the degrees are the same, it is the ratio of the leading coefficients you do not say "it is infinity over infinity therefore undefined"

  27. anonymous
    • one year ago
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    I m still not getting why the upper limit of my expression shd be 1, cn u pls explain that

  28. misty1212
    • one year ago
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    of the expression \[\frac{e^x-1}{e^x-1}\]?

  29. anonymous
    • one year ago
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    Ya

  30. misty1212
    • one year ago
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    ooops i made a typo there, should be \[\frac{e^x-1}{e^x+1}\]

  31. misty1212
    • one year ago
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    ok we agree that the numerator is one less then the denominator right? and therefore it can never be one, because a fraction is only one if the numerator and denominator are the same

  32. anonymous
    • one year ago
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    Ya

  33. misty1212
    • one year ago
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    and also since the numerator is less than the denominator, this is always smaller than 1, not larger right?

  34. anonymous
    • one year ago
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    Right

  35. misty1212
    • one year ago
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    so let me ask you a question it can never be one it is always smaller than one what do you think the limit might be? \(\frac{1}{2}\)?

  36. misty1212
    • one year ago
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    \(e^x\) grows very very quickly, much faster than \(x^2\) or in fact any power of \(x\)

  37. anonymous
    • one year ago
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    Ok

  38. misty1212
    • one year ago
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    already \(e^{10}\) is \(22026\) rounded so you would have \[\frac{22025}{22027}\] which is pretty close to 1, with \(x=10\) only

  39. anonymous
    • one year ago
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    I see

  40. misty1212
    • one year ago
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    imagine what it would look like if \(x=100\)

  41. anonymous
    • one year ago
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    Can u tell me what kind of function will have a range between 1 and infinity

  42. misty1212
    • one year ago
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    i can make up one sure

  43. misty1212
    • one year ago
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    how about \[f(x)=\sqrt{x}+1\]

  44. misty1212
    • one year ago
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    or \[f(x)=x^2+1\]

  45. anonymous
    • one year ago
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    I see that

  46. misty1212
    • one year ago
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    those would have range \([1,\infty)\) if you want \((1,\infty)\) open you could use for example \[f(x)=e^x+1\]

  47. anonymous
    • one year ago
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    I was looking for functions close to logit

  48. misty1212
    • one year ago
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    ok wise guy

  49. perl
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @misty1212 now it is pretty clearly always less than 1, since \(e^x>0\) for all \(x\) and the numerator is one less than the denominator \(\color{blue}{\text{End of Quote}}\) The numerator* is 2 less than the denominator

  50. misty1212
    • one year ago
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    you must be bored

  51. anonymous
    • one year ago
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    No way

  52. anonymous
    • one year ago
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    U explain so well

  53. anonymous
    • one year ago
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    R u prof of math?

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