What is the extreme f(x)=x^2+(360)/x ?

The extreme f(x)=x^2+(360)/x is Minimum(\sqrt[3]{180},180^{2/3}+(72*5^{2/3})/(\sqrt[3]{36)})

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extreme f(x)=x^2+(360)/x

extreme

Minimum (3180, 18 0^{\frac{2}{3}} + \frac{72 \cdot 5 ^{\frac{2}{3}}}{3 36})

Solution steps

f^{'} (x) = 2 x - \frac{360}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3180,

Increasing

: 3180 < x < \infty

Plug

x = 3180

into

x^{2} + \frac{360}{x} : 18 0^{\frac{2}{3}} + \frac{72 \cdot 5 ^{\frac{2}{3}}}{3 36}

Minimum (3180, 18 0^{\frac{2}{3}} + \frac{72 \cdot 5 ^{\frac{2}{3}}}{3 36})

What is the extreme f(x)=x^2+(360)/x ?
The extreme f(x)=x^2+(360)/x is Minimum(\sqrt[3]{180},180^{2/3}+(72*5^{2/3})/(\sqrt[3]{36)})