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Math Help - Trying again

  1. #1
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    Post Trying again

    Still unsure of how to find the inflection points with an e^x problem...

    Take this problem:
    f(t) = 1 - e^-0.03t
    f'(t) = -e^-.03t + (-.03t)
    f'(t) = -0.03te^-0.03t

    It seems logical that if t=0 then f'(t) =0, but is there a way to solve it without guessing?

    f"(t) = -.03t(e^-.03t)(-.03)
    f"(t) = .0009te^-.03t

    In other words, can you set .0009t = e^-.03t and solve? or do you just pick some points near 0 and plot?
    Last edited by confusedagain; June 3rd 2007 at 11:03 AM. Reason: forgot to add sentence
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  2. #2
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    Hello, confusedagain!

    Still unsure of how to find the inflection points with an e^x problem...

    Take this problem:
    f(t) \:= \:1 - e^{-0.03t}
    f'(t) \:= \:-e^{-0.03t} + (-0.03t) ?

    The derivative of: . y \:=\:e^u is: . y' \:=\:e^u\cdot u'

    In baby-talk: the exponential function itself times the derivative of the exponent.


    If f(t)\:=\:1 - e^{-0.03t}

    . . then: . f'(t)\:=\:-e^{-0.03t}(-0.03) \:=\:0.03e^{-0.03t}

    . . then: . f''(t)\:=\:0.03e^{-0.03t}(-0.03) \:=\:-0.0009e^{-0.03t}


    To find inflection points, solve f''(t) = 0

    So we have: . -0.0009e^{-0.03t} \:=\:0

    . . Divide by -0.0009: . e^{-0.003t} \:=\:0

    . . Multiply both sides by e^{0.03t}\!:\;\;1 \:=\:0 . . . . What?

    Therefore, there are no inflection points.

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  3. #3
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    Quote Originally Posted by confusedagain View Post
    Still unsure of how to find the inflection points with an e^x problem...

    Take this problem:
    f(t) = 1 - e^-0.03t
    f'(t) = -e^-.03t + (-.03t)
    f'(t) = -0.03te^-0.03t

    It seems logical that if t=0 then f'(t) =0, but is there a way to solve it without guessing?

    f"(t) = -.03t(e^-.03t)(-.03)
    f"(t) = .0009te^-.03t

    In other words, can you set .0009t = e^-.03t and solve? or do you just pick some points near 0 and plot?
    Hello,

    I assume that your function is:
    f(t)=1-e^{-0.03t}
    To calculate the first derivative use chain rule:

    f'(t) = -e^{-0.03t} \cdot (-0.03) = 0.03 \cdot e^{-0.03t}

    Both factors are positiv thus the product is positiv too, that means f'(t) \neq 0

    To calculate the second drivative use chain rule again:
    f''(t) = 0.03 \cdot e^{-0.03t} \cdot (-0.03) = -0.0009 \cdot e^{-0.03t}
    Both factors have always different signs thus the product is negative for all t.

    The function don't have any extreme points or inflection points if the domain is \mathbb{R}
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