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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

1 answer

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)

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