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Math Help - differentiate and find maximum and minimum turning point

  1. #1
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    differentiate and find maximum and minimum turning point

    2(x^2/3) + (x^4/5) +11

    i differnetiated this to get
    4/3(x^-1/3) + 4/5(x^-1/5)

    i know then, that i have to set the differentiated formula = 0. Please help with what i have to do next.

    Thanks
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: differentiate and find maximum and minimum turning point

    The first derivative is indeed:
    f'(x)=\frac{4}{3}x^{\frac{-1}{3}}+\frac{4}{5}x^{\frac{-1}{5}}
    To find the minumum/maximum then you have to solve f'(x)=0. What do you get? Afterwards make a sign table to determine where the function increases/decreases.
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  3. #3
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    Re: differentiate and find maximum and minimum turning point

    I can't remember how to solve it. i know im missing something simple. so can anyone help me solve this please. thank you
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  4. #4
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    Re: differentiate and find maximum and minimum turning point

    Quote Originally Posted by salma2011 View Post
    I can't remember how to solve it. i know im missing something simple. so can anyone help me solve this please. thank you
    I suggest finding the lowest common denominators of 3 and 5 and change your exponents into that form:

    f'(x) = \dfrac{4}{3}x^{-5/15} + \dfrac{4}{5}x^{-3/15} = 0

    You can then factor out 4x^{-1/15} as a common factor

    4x^{-3/15}\left(\dfrac{x^2}{3} + \dfrac{1}{5}\right) = 0

    Can you solve now?


    To test whether you have a maximum or minimum check the sign of f''(x).

    • If f''(x) < 0 you have a maximum
    • If f''(x) > 0 you have a minimum
    • If f''(x) = 0 you have a point of inflection
    Last edited by e^(i*pi); October 1st 2011 at 12:04 PM. Reason: sign error
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: differentiate and find maximum and minimum turning point

    There're probably different ways to solve the equation, but if you have to solve:
    0=\frac{4}{3}x^{\frac{-1}{3}}+\frac{4}{5}x^{\frac{-1}{5}}
    \Leftrightarrow 4x^{\frac{-1}{3}}\left(\frac{1}{3}+\frac{1}{5}x^{\frac{2}{15}  }\right)=0
    \Leftrightarrox x^{\frac{-1}{3}}=0 \ \mbox{or} \ x^{\frac{2}{15}}=\frac{-5}{3}

    This means there're no solutions, because there's no x value wherefore:
    x^{\frac{-1}{3}}=0 or x^{\frac{2}{15}}=\frac{-5}{3}.

    So what's your conclusion?
    Last edited by Siron; October 1st 2011 at 11:25 AM.
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  6. #6
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    Re: differentiate and find maximum and minimum turning point

    i solved for x and got x = + or - square root of 3/5

    (sorry don't know how to put mathematical functions onto the computer)

    and then from the derivative i work out whether it's a maximum or a minimum.

    thank you for all your help
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