check 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
find the absolute minimum and maximum valies of f(x)=-x^4+2x^3+5 on the interval [-2,2]
2 answers
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
= 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