What is the extreme f(x)=x^2+(12)/x ?

The extreme f(x)=x^2+(12)/x is Minimum(\sqrt[3]{6},3*6^{2/3})

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extreme f(x)=x^2+(12)/x

extreme

Minimum (36, 3 \cdot 6^{\frac{2}{3}})

Solution steps

f^{'} (x) = 2 x - \frac{12}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 36,

Increasing

: 36 < x < \infty

Plug

x = 36

into

x^{2} + \frac{12}{x} : 3 \cdot 6^{\frac{2}{3}}

Minimum (36, 3 \cdot 6^{\frac{2}{3}})

What is the extreme f(x)=x^2+(12)/x ?
The extreme f(x)=x^2+(12)/x is Minimum(\sqrt[3]{6},3*6^{2/3})