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Math Help - The critical numbers of the function

  1. #1
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    The critical numbers of the function

    f(t)=6^(2/3)+t^(5/3)

    I understand I take the derivative of the function and set it equal to zero to find the critical numbers, however I get stuck after I take the derivative

    f'(t)=4t^(-1/3)+5/3t^(2)=0
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  2. #2
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    Quote Originally Posted by Asuhuman18 View Post
    f(t)=6^(2/3)+t^(5/3)

    I understand I take the derivative of the function and set it equal to zero to find the critical numbers, however I get stuck after I take the derivative

    f'(t)=4t^(-1/3)+5/3t^(2)=0
    I guess you meant f(t) to be ...

    f(t) = 6t^{2/3} + t^{5/3}

    f'(t) = 4t^{-1/3} + \frac{5}{3}t^{2/3} = t^{-1/3}\left(4 + \frac{5}{3}t \right) = 0

    can you finish?
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  3. #3
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    Quote Originally Posted by skeeter View Post
    I guess you meant f(t) to be ...

    f(t) = 6t^{2/3} + t^{5/3}

    f'(t) = 4t^{-1/3} + \frac{5}{3}t^{2/3} = t^{-1/3}\left(4 + \frac{5}{3}t \right) = 0

    can you finish?
    yeah sorry about that...Thanks for all the help
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