What is the extreme f(x)=x+1/(x^2) ?

The extreme f(x)=x+1/(x^2) is Minimum(\sqrt[3]{2},\sqrt[3]{2}+1/(2^{2/3)})

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extreme f(x)=x+1/(x^2)

extreme points

f (x) = x + \frac{1}{x ^{2}}

Solution steps

f^{'} (x) = 1 - \frac{2}{x ^{3}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

f (x) = \frac{x ^{2}}{( x ^{2} - 9 )}

f (x) = \frac{x ^{2} + 9}{x}

g (x, y) = x - c + y

f (x, y) = \frac{x ^{3}}{3} + y^{2} - 2 x + 2 y - 2 x y

f (x) = 1 + \frac{7}{x} - \frac{9}{x ^{2}}

What is the extreme f(x)=x+1/(x^2) ?
The extreme f(x)=x+1/(x^2) is Minimum(\sqrt[3]{2},\sqrt[3]{2}+1/(2^{2/3)})