What is the extreme f(x)=x^2+(120)/x ?

The extreme f(x)=x^2+(120)/x is Minimum(\sqrt[3]{60},60^{2/3}+(8*15^{2/3})/(\sqrt[3]{4)})

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extreme f(x)=x^2+(120)/x

extreme

Minimum (360, 6 0^{\frac{2}{3}} + \frac{8 \cdot 1 5 ^{\frac{2}{3}}}{3 4})

Solution steps

f^{'} (x) = 2 x - \frac{120}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 360,

Increasing

: 360 < x < \infty

Plug

x = 360

into

x^{2} + \frac{120}{x} : 6 0^{\frac{2}{3}} + \frac{8 \cdot 1 5 ^{\frac{2}{3}}}{3 4}

Minimum (360, 6 0^{\frac{2}{3}} + \frac{8 \cdot 1 5 ^{\frac{2}{3}}}{3 4})

What is the extreme f(x)=x^2+(120)/x ?
The extreme f(x)=x^2+(120)/x is Minimum(\sqrt[3]{60},60^{2/3}+(8*15^{2/3})/(\sqrt[3]{4)})