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calculus

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(a) Find the critical points of f(x) = x^(1/3)-x^(1/5)

(b) Determine the maximum and minimum values of f on the interval [0; 2] and where they occur.

  • calculus - ,

    I will assume that you know that critical points would be points such as the x and y intercepts, the max and min points, and any points of inflection.

    I think the problem here lies in trying to solve different equations containing fractional exponents.

    the y-intercept is easy, let x = 0
    y = 0

    for the x-intercept let y = 0

    so x^(1/2 - x^(1/5) = 0
    x^(1/3) = x^(1/5)
    raise it to the 15th power, (15 is the LCM of 3 and 5)
    x^5 = x^3
    x^5 - x^3 - 0
    x^3(x^2 - 1) = 0
    x = 0 or x = ± 1

    After you get dy/dx = (1/3)x^(-2/3) - (1/5)x^(-4/5)
    you will have to set this equal to zero as well
    so the same sort of thing as above
    x^(-2/3) = x^(-4/5) , after dividing out the 1/3
    so 1/x^(2/3) = 1/x^(4/5)
    x^(4/5) = x^(2/3) after cross-multiplying
    raise to the 15th
    x^12 = x^10
    x^10(x^2 - 1) = 0
    x = 0 or x = ± 1 (well, isn't that interesting)

    etc.

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