What is the extreme f(x)=(8-x^3)/(x^2) ?

The extreme f(x)=(8-x^3)/(x^2) is Minimum(-2\sqrt[3]{2},3*2^{1/3})

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extreme f(x)=(8-x^3)/(x^2)

extreme

Minimum (- 232, 3 \cdot 2^{\frac{1}{3}})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 232,

Increasing

: - 232 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 232

into

\frac{8 - x ^{3}}{x ^{2}} : 3 \cdot 2^{\frac{1}{3}}

Minimum (- 232, 3 \cdot 2^{\frac{1}{3}})

What is the extreme f(x)=(8-x^3)/(x^2) ?
The extreme f(x)=(8-x^3)/(x^2) is Minimum(-2\sqrt[3]{2},3*2^{1/3})