What is the extreme f(x)=2x-(2000)/(x^2) ?

The extreme f(x)=2x-(2000)/(x^2) is Maximum(-10\sqrt[3]{2},-30\sqrt[3]{2})

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extreme f(x)=2x-(2000)/(x^2)

extreme points

f (x) = 2 x - \frac{2000}{x ^{2}}

Solution steps

f^{'} (x) = 2 + \frac{4000}{x ^{3}}

Find intervals:

Increasing

Decreasing

Increasing

: 0 < x < \infty

Plug

into

f (x, y) = 3 x^{2} + 2 y^{2} - 18 x + 8 y

e^{4 x^{2} + 2 y^{2} - 24 x}

f (x) = x^{3} - 3 x^{2} + 9 x + 1

f (x) = - \frac{1}{3} x^{3} + 2 x^{2} + 12 x + 28

f (x) = - 0.015 x^{2} + 1.22 x - 7.2

What is the extreme f(x)=2x-(2000)/(x^2) ?
The extreme f(x)=2x-(2000)/(x^2) is Maximum(-10\sqrt[3]{2},-30\sqrt[3]{2})