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calculus

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Find the absolute maximum and absolute minimum values of f on the given interval.
f(t) = (36 − t^2)^ 1/t, [−1, 6]

  • calculus - ,

    nasty equation, none of the standard methods of derivatives fit.
    I am going to take logs of both sides

    ln y = ln (36- t^2)^(1/t)
    = (1/t)(ln (36-t^2)
    now product rule on right side

    (dy/dt) / y = (1/t)(-2t)/(36-t^2) + (-1/t^2)(ln(36-t^2) )
    dy/dt = y[(1/t)(-2t)/(36-t^2) + (-1/t^2)(ln(36-t^2) )]

    so y = 0
    (36-t^2)^(1/t) = 0
    there is an intuitive solution of t = 6

    or
    (1/t)(-2t)/(36-t^2) + (-1/t^2)(ln(36-t^2) ) = 0
    -2/(36-t^2) = ln(36-t^2) /t^2
    ln(36-t^2) = -2t^2/(36-t^2)

    what a messy equation, ran it through Wolfram
    and there are no real solutions, (4 complex)
    http://www.wolframalpha.com/input/?i=solve++log%2836-t%5E2%29+%3D+-2t%5E2%2F%2836-t%5E2%29

    unless I made an algebraic error or typo

    so evaluate f(6) which happens to be at the end of your interval.
    f(6) = (36-36)^(1/6 = 0
    Also evalute f(-1) to see which is the max or min.
    f(-1) = (36 - 1)^-1
    =1/35 = appr .02857

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