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Math Help - Absolute minimum and maximum of a function involving e

  1. #1
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    Absolute minimum and maximum of a function involving e

    I have no problem finding maximum and minimums of functions, except when there's an e involved. For example finding the max and min values of e^((x^3)-x) on the closed interval -1 and 0. First off I don't know how to find the derivative of a power raised to another power, and even when I cheat and use a derivative calculator to find that, I don't know how to set the remaining derivative with e equal to 0. Would someone mind helping me out here?
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  2. #2
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    Hi

    The derivative of e^{u(x)} is u'(x)e^{u(x)}
    Here u(x) = x^3 - x
    u'(x) = 3x^2 - 1

    Therefore the derivative of e^{x^3-x} is (3x^2 - 1)e^{x^3-x}

    To find the sign you have to remember that e^{anything} is always strictly positive

    Therefore the sign of the derivative of e^{x^3-x} is the same as the sign of 3x^2 - 1

    Now I think that you can finish the job !
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  3. #3
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    But I can't...I still don't understand how to set that derivative equal to 0 since there's an e in the term.
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  4. #4
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    The derivative of e^{x^3-x} is (3x^2 - 1)e^{x^3-x}
    This is a product therefore to set the derivative equal to 0 means solving 3x-1 = 0 (since e^anything can never be equal to 0)
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  5. #5
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    But that makes no sense because and absolute min I already found to be 1. For that to be equal to 0, x would have to be much smaller than one...
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  6. #6
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    You are studying the function e^{x^3-x} on interval [-1,0]

    The derivative of e^{x^3-x} is (3x^2 - 1)e^{x^3-x}

    To find the max and min of the function you have to solve (3x^2 - 1)e^{x^3-x} = 0 on [-1,0]
    This is equivalent to solve 3x^2 - 1 = 0 on [-1,0]

    On [-1,0] you can find only one value for which 3x^2 - 1 = 0
    This value is - \frac{\sqrt{3}}{3}

    Now you have to find the sign of the derivative on [-1,0] in order to know where the function is increasing and decreasing
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  7. #7
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    Quote Originally Posted by running-gag View Post
    You are studying the function e^{x^3-x} on interval [-1,0]

    The derivative of e^{x^3-x} is (3x^2 - 1)e^{x^3-x}

    To find the max and min of the function you have to solve (3x^2 - 1)e^{x^3-x} = 0 on [-1,0]
    This is equivalent to solve 3x^2 - 1 = 0 on [-1,0]

    On [-1,0] you can find only one value for which 3x^2 - 1 = 0
    This value is - \frac{\sqrt{3}}{3}

    Now you have to find the sign of the derivative on [-1,0] in order to know where the function is increasing and decreasing

    You seem to be misunderstanding. I don't need to find where the function is increasing or decreasing, I just have to find the absolute min and absolute max values of the function. That's it. Because it's absolute min/max and not just relative min/max, I plugged in the roots -1 and 0, which gave me 1, which I tried for both the min and max answer, and it turned out it is indeed the min. Now when I solve 3x^2-1=0, the resulting fraction that you provided, and I already tried, gives a number LOWER than the min when plugged into the function so it CANNOT be the max. That's precisely why I'm confused about this particular problem.
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  8. #8
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    I can confirm you that the absolute min are given for x=-1 and x=0 and this value is 1
    I can also confirm you that the absolute max is given for x = - \frac{\sqrt{3}}{3} and this value is approximately 1.469
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  9. #9
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    Here is the curve

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  10. #10
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    Oh jeez. I kept plugging in only POSITIVE sqrt 3/3, rather than negative. My apologies. This leads me to another similar question, however. What would you do if you have a similar function with e involved, and are asked to find the same thing (absolute min/max values) but the exponent's derivative no longer contains an x.

    F(x) = e^-x - e^-2x

    When you pull the exponents out in front there are no longer x's attached so how would you set it equal to 0 if e to any power can never equal 0.
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  11. #11
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    In the case F(x) = e^{-x} - e^{-2x}
    the derivative is F'(x) = -e^{-x} +2 e^{-2x}

    F'(x) = 0 leads to -e^{-x} +2 e^{-2x} = 0

    2 e^{-2x} = e^{-x}

    2 = e^x

    x = ln(2) which is an absolute max

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