Use the second derivative test to find the local minimum and maximum values of each function, wherever possible.
A) f(x) = 16/x + x^2
B) s(t) = t^4 - 8t^3
A.
f ' = -16/x^2 + 2x
f '' = 32/x^2 + 2
max/min where f'=0 and f''≠0
f'(2) = 0
f''(2) > 0, so f is concave up,
so f(2) = 12 is a minimum
similarly for (B)