Posted by kchik on .
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,
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.