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

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 (36t^2)
now product rule on right side
(dy/dt) / y = (1/t)(2t)/(36t^2) + (1/t^2)(ln(36t^2) )
dy/dt = y[(1/t)(2t)/(36t^2) + (1/t^2)(ln(36t^2) )]
so y = 0
(36t^2)^(1/t) = 0
there is an intuitive solution of t = 6
or
(1/t)(2t)/(36t^2) + (1/t^2)(ln(36t^2) ) = 0
2/(36t^2) = ln(36t^2) /t^2
ln(36t^2) = 2t^2/(36t^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%2836t%5E2%29+%3D+2t%5E2%2F%2836t%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) = (3636)^(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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