Posted by **kchik** on Thursday, November 22, 2012 at 5:58am.

Given that f(x)=2/3x^3 + 3/2x^2 -2x + 5. Find if exist, the maximum and minimum point using Second Derivative Test.

- calculus -
**Steve**, Thursday, November 22, 2012 at 7:38pm
f = 2/3 x^3 + 3/2 x^2 - 2x + 5

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

f'' = 4x + 3

f has min/max at f'=0, or x= -2 or 1/2

f''(-2) < 0 so f has a maximum there

f''(1/2) > 0 so f has a minimum there

From what you know about the general shape of cubics, this should make sense.

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