# calculus

posted by on .

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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