calculus
posted by moises .
find the absolute minimum and maximum valies of f(x)=x^4+2x^3+5 on the interval [2,2]

check f(2) and f(2).
f' = 4x^3 + 6x^2
= 2x^2(2x3)
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 
f ' (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