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

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

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extreme x-8/(x^2)

extreme points

x - \frac{8}{x ^{2}}

Maximum (- 232, - 232 - \frac{2}{2 ^{\frac{2}{3}}})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 232,

Decreasing

: - 232 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 232

into

x - \frac{8}{x ^{2}} : - 232 - \frac{2}{2 ^{\frac{2}{3}}}

Maximum (- 232, - 232 - \frac{2}{2 ^{\frac{2}{3}}})

f (x) = - e^{- x}

f (x) = a^{x}

f (x) = x + \frac{16}{x}

y = ln (3 x)

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

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