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Math Help - Second derivative test to find local min and max

  1. #1
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    Second derivative test to find local min and max

    HI

    The goal is the find the local minimum and local maximum values of the following function using the second derivative test

    f(x) = x^3 + 1/x

    so I calculate

    f'(x) = 3x^2 + -1

    f"(x) = 6x

    so f'(x) = 0 when x = √(1/3)

    f"(√1/3) = 6 √1/3

    which I believe is the local minimum since f(√1/3) is >0

    now I am questioning my calculations since I am suppose to also have a maximum based on the question and I don't have one

    Please review my work and tell me where I have gone wrong

    Thanks

    calculus beginner
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  2. #2
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    Quote Originally Posted by calcbeg View Post
    HI

    The goal is the find the local minimum and local maximum values of the following function using the second derivative test

    f(x) = x^3 + 1/x

    so I calculate

    f'(x) = 3x^2 + -1 ... this is incorrect, should be f'(x) = 3x^2 - (1/x^2)

    f"(x) = 6x ... f''(x) = 6x + (2/x^3)

    so f'(x) = 0 when x = √(1/3)

    f"(√1/3) = 6 √1/3

    which I believe is the local minimum since f(√1/3) is >0

    now I am questioning my calculations since I am suppose to also have a maximum based on the question and I don't have one

    Please review my work and tell me where I have gone wrong

    Thanks

    calculus beginner
    .
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  3. #3
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    I have done some corrections - please check I am on the right track ...so

    f'(x) = 3x^2 - 1/x^2

    f"(x) = 6x + 2x^-3

    f'(x) is zero when 3x^2 - x^-2 = 0

    3x^2= x^-2
    x^4 = 1/3
    x= 4√(1/3)

    so x could be + 4√(1/3) or -4√(1/3)

    which means the local minimum is at 4√(1/3) since f'(4√1/3) is > 0

    and local maximum is at 4√(1/3) since f"(4√(1/3) is < 0

    Do I have it right now??
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  4. #4
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    correct, f'(x) = 0 at x = \pm \sqrt[4]{\frac{1}{3}}

    f''\left(\sqrt[4]{\frac{1}{3}}\right) > 0 ... f(x) has a minimum there

    f''\left(-\sqrt[4]{\frac{1}{3}}\right) < 0 ... f(x) has a maximum there
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