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Weird function!

  1. Nov 6, 2006 #1
    Hey people,
    Consider the function:
    At 0,f(x)=0 and d/dx=1
    At 1,f(x)=1 and d/dx=1
    At 2,f(x)=16 and d/dx=107.11
    At 3,f(x)=7.625597485000*10^12 and d/dx=5.43324993100000*10^14

    Why this??? Check the slopes...

    And: Slopes between 0-1
    At 0.1, f(x)=0.16057 d/dx= 1.658
    At 0.2, f(x)=0.31146 d/dx=1.350
    At 0.3, f(x)=0.43215 d/dx=1.070
    At 0.4, f(x)=0.52987 d/dx=0.890
    At 0.5, f(x)=0.61255 d/dx=0.777
    At 0.6, f(x)=0.68662 d/dx=0.716
    At 0.7, f(x)=0.75740 d/dx=0.708
    At 0.8, f(x)=0.82972 d/dx=0.747
    At 0.9, f(x)=0.90862 d/dx=0.840
    At 1.0, f(x)=1.00000 d/dx=1.000
    The slopes between 0.2-0.7 are decreasing...why's that??? Can anyone explain?

    And one more function f(x)=x^x^x^x
    At 0, d/dx=-infinity
    At 0.1, d/dx=-1.528330000
    At 0.2, d/dx=-0.37292
    At 0.3 d/dx~0
    At 0.4 d/dx=0.31333
    At 0.5 d/dx=0.45030
    At 0.6 d/dx=0.54327
    At 0.7 d/dx=0.63323
    At 0.8 d/dx=0.72329
    At 0.9 d/dx=0.83695
    At 1.0 d/dx=1.000

    So can anyone say why does the slope decrease and then increase so rapidly??? Hope my questions are not falling on deaf ears!
  2. jcsd
  3. Nov 6, 2006 #2


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    Dearly Missed

    Well, you might try to differentiate the animal and verify your observations!
  4. Nov 6, 2006 #3
    :rofl: :rofl: Good one!
  5. Nov 6, 2006 #4


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    Is this ((x^x)^x)^x or (what I'm assuming) x^(x^(x^x))?

    EDIT: Or I can just be smart and plug in one of the values you've so kindly given us. Never mind
  6. Nov 6, 2006 #5


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    Dearly Missed

    It isn't that difficult to differentiate the creature.
    Let's tackle the first one:
    [tex]f(x)=x^{g(x)}=e^{\ln(x)g(x)}, g(x)=x^{x}=e^{x\ln(x)}[/tex]

    This might then be written out more fully, if that's important to you.
  7. Nov 6, 2006 #6
    Just plot the function, alongside x^x. The reason the slopes decrease is beacaues x^x is a decreasing function for a short time in and around that range.
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