What is the extreme f(x)=x^2+(240)/x ?

The extreme f(x)=x^2+(240)/x is Minimum(2\sqrt[3]{15},12*15^{2/3})

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extreme f(x)=x^2+(240)/x

extreme

Minimum (2315, 12 \cdot 1 5^{\frac{2}{3}})

Solution steps

f^{'} (x) = 2 x - \frac{240}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2315,

Increasing

: 2315 < x < \infty

Plug

x = 2315

into

x^{2} + \frac{240}{x} : 12 \cdot 1 5^{\frac{2}{3}}

Minimum (2315, 12 \cdot 1 5^{\frac{2}{3}})

What is the extreme f(x)=x^2+(240)/x ?
The extreme f(x)=x^2+(240)/x is Minimum(2\sqrt[3]{15},12*15^{2/3})