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March 27, 2015

March 27, 2015

Posted by **moises** on Sunday, April 28, 2013 at 3:40am.

- calculus -
**Steve**, Sunday, April 28, 2013 at 6:10amcheck f(-2) and f(2).

f' = -4x^3 + 6x^2

= -2x^2(2x-3)

so there are local extrema at x = 3/2

Not at x=0, since that's a double root of f'=0.

Now pick the max/min values of f at -2,3/2,2

- calculus -
**Reiny**, Sunday, April 28, 2013 at 7:09amf ' (x) = - 4x^3 + 6x^2

= 0 for max/min

x^2(-4x + 6) = 0

x = 0 or x = 3/2

f(0) = 5

f(3/2) = -81/64 + 54/8 + 5 = 671/64 = appr 10.48

f(-2) = - 16 + 16 + 5 = 5

f(2) = -16 - 16 + 5 = -27

so the absolute max is 671/64 and the min is -27

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