Calculus
posted by Cam .
Find relative maxima and minima of
g(x)= 2x^2+ 4000/x +10

g ' (x) = 4x  4000/x^2 + 0
= 0 for a max/min
4x = 4000/x^2
4x^3 = 4000
x^3 = 1000
x = 10
when x = 10
f(10) = 200 + 400 + 10 = 610
there is a relative min 610 at x=10
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