What is the extreme f(x)=x-8/(x^2) ?

The extreme f(x)=x-8/(x^2) is Maximum(-2\sqrt[3]{2},-2\sqrt[3]{2}-2/(2^{2/3)})

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extreme f(x)=x-8/(x^2)

extreme points

f (x) = x - \frac{8}{x ^{2}}

Solution steps

f^{'} (x) = 1 + \frac{16}{x ^{3}}

Find intervals:

Increasing

Decreasing

Increasing

: 0 < x < \infty

Plug

into

3 x^{5} - 20 x^{3}

f (s) = \frac{4 - s ^{2}}{2 s x ^{2} - 7 s - 4}

L (x, y) = 5 x + 4 y - (8 + 2 x^{2} + y^{2} - x y)

f (x) = e^{2 x} (x + y^{2} + 2 y)

f (x) = - 2 x - 5

What is the extreme f(x)=x-8/(x^2) ?
The extreme f(x)=x-8/(x^2) is Maximum(-2\sqrt[3]{2},-2\sqrt[3]{2}-2/(2^{2/3)})