calculus

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Find the value or values of c that satisfy the equation
f(b)-f(a)/b-a = f'(c)
in the conclusion of the mean value theorem for the given function and interval
f(x)= x^(2/3) , [0,1]

  • calculus -

    y = x^(2/3)
    at x = 1 which is b
    y = 1^(2/3) = 1
    at x = 0 which is a
    y = 1^0 = 1

    y(1) - y(0) = 1 - 0 = 1
    b - a = 1 - 0 = 1
    so
    1/1 = 1

    so according to the mean value theorem somewhere between x = 0 and x = 1 the derivative = 1

    dy/dx = (2/3) x^-1/3 = (2/3)/x^(1/3)
    when is that equal to one?

    x^(1/3) = 2/3
    (1/3) log x = log 2/3
    log x = -.52827
    x= .296

  • calculus typo -

    y = x^(2/3)
    at x = 1 which is b
    y = 1^(2/3) = 1
    at x = 0 which is a
    y = 0^0 = 0

    y(1) - y(0) = 1 - 0 = 1
    b - a = 1 - 0 = 1
    so
    1/1 = 1

    so according to the mean value theorem somewhere between x = 0 and x = 1 the derivative = 1

    dy/dx = (2/3) x^-1/3 = (2/3)/x^(1/3)
    when is that equal to one?

    x^(1/3) = 2/3
    (1/3) log x = log 2/3
    log x = -.52827
    x= .296

  • calculus -

    The answer in my textbook says 1 for the value of c. I don't understand how it arrived there.

  • calculus -

    Nor do I

  • calculus -

    Sounds like they wanted f'(c) rather than c.

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