Posted by Julie on Friday, March 23, 2012 at 12:08pm.
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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