Posted by Rachal on Wednesday, May 18, 2011 at 5:35am.
f '(x) = 16x^3 + 6x^2 - 16x - 5 , ( I assumed the second last term was -8x^2 )
16x^3 + 6x^2 - 16x - 5 = 0
zeros for this equation exist at
appr. -1,1 and -.3
f ''(x) = 48x^2+ 12x - 16
both f ''(1) and f ''(-1) > 0
but f ''(-.3) < 0
so when x = -.3 (appr), we have a local maximum
I will let you find f(-.3)
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