What is the extreme f(x)=2x-5/(x^2) ?

The extreme f(x)=2x-5/(x^2) is Maximum(-\sqrt[3]{5},-3\sqrt[3]{5})

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extreme f(x)=2x-5/(x^2)

extreme points

f (x) = 2 x - \frac{5}{x ^{2}}

Solution steps

f^{'} (x) = 2 + \frac{10}{x ^{3}}

Find intervals:

Increasing

Decreasing

Increasing

: 0 < x < \infty

Plug

into

(- 6, - 6), (3, 5)

lo g_{2} (2 x - 2) - 3

f (x) = 310 x^{3} - x^{2} - 8 x + 48

f (x) = x^{2} - 10 x

f (x) = 2 x + 5 cos (x)

What is the extreme f(x)=2x-5/(x^2) ?
The extreme f(x)=2x-5/(x^2) is Maximum(-\sqrt[3]{5},-3\sqrt[3]{5})