Posted by **Anonymous** on Sunday, November 3, 2013 at 5:02pm.

Find the absolute maximum and absolute minimum values of f on the given interval.

f(t) = (36 − t^2)^ 1/t, [−1, 6]

- calculus -
**Reiny**, Sunday, November 3, 2013 at 5:54pm
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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