What is the extreme f(x)=2x-(800)/(x^2) ?

The extreme f(x)=2x-(800)/(x^2) is Maximum(-2\sqrt[3]{100},-4\sqrt[3]{100}-(200)/(100^{2/3)})

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extreme f(x)=2x-(800)/(x^2)

extreme points

f (x) = 2 x - \frac{800}{x ^{2}}

Solution steps

f^{'} (x) = 2 + \frac{1600}{x ^{3}}

Find intervals:

Increasing

Decreasing

Increasing

: 0 < x < \infty

Plug

into

f (x) = ∣ - 4 x - 5 ∣, - 5 < x < 1

f (x, y) = 3 x^{6} y - 5 x y^{2}

f (x) = x^{3} e^{- 4 x}, 0 \leq x \leq 2

f (x) = 3 x^{2} + 12 x y + 9 x^{2} + y^{3}

V (u, w) = \frac{1}{2} (3 u + w)

What is the extreme f(x)=2x-(800)/(x^2) ?
The extreme f(x)=2x-(800)/(x^2) is Maximum(-2\sqrt[3]{100},-4\sqrt[3]{100}-(200)/(100^{2/3)})