SOLUTION: determine the local minimum and maximum values of f(x)=(3x^2-6x)(3x^2+6x), to the nearest hundredth.

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Question 427186: determine the local minimum and maximum values of f(x)=(3x^2-6x)(3x^2+6x), to the nearest hundredth.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+%283x%5E2-6x%29%283x%5E2%2B6x%29+=+9x%5E4+-+36x%5E2
==> dy%2Fdx+=+36x%5E3+-+72x
Let this equal 0:
36x%5E3+-+72x+=+36x%28x%5E2+-+2%29+=+0==> x = 0, sqrt%282%29,+-sqrt%282%29.
Also, d%5E2y%2Fdx%5E2+=+108x%5E2+-+72+=+36%283x%5E2+-+2%29
When x = 0, d%5E2y%2Fdx%5E2+%3C+0 ==> local max there, and so local max value is y = 0.
When x = sqrt%282%29 or -sqrt%282%29, d%5E2y%2Fdx%5E2+%3E+0, ==> local mins there, and so local (absolute) min value is y = -36.
(There is no absolute max value.)