What is the extreme f(x,y)=(x^3+9)/(x^2) ?

The extreme f(x,y)=(x^3+9)/(x^2) is Minimum(\sqrt[3]{18},(27)/(18^{2/3)})

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extreme f(x,y)=(x^3+9)/(x^2)

extreme

Minimum (318, \frac{27}{1 8 ^{\frac{2}{3}}})

Solution steps

f^{'} (x) = \frac{x ^{3} - 18}{x ^{3}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 318,

Increasing

: 318 < x < \infty

Plug

x = 318

into

\frac{x ^{3} + 9}{x ^{2}} : \frac{27}{1 8 ^{\frac{2}{3}}}

Minimum (318, \frac{27}{1 8 ^{\frac{2}{3}}})

What is the extreme f(x,y)=(x^3+9)/(x^2) ?
The extreme f(x,y)=(x^3+9)/(x^2) is Minimum(\sqrt[3]{18},(27)/(18^{2/3)})