What is the extreme f(x)=x^2+(480)/x ?

The extreme f(x)=x^2+(480)/x is Minimum(2\sqrt[3]{30},12*30^{2/3})

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extreme f(x)=x^2+(480)/x

extreme points

f (x) = x^{2} + \frac{480}{x}

Solution steps

f^{'} (x) = 2 x - \frac{480}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

x^{2} + \frac{480}{x} : 12 \cdot 3 0^{\frac{2}{3}}

f (x) = 2 - 8 x

f (x) = 3

f (x) = \frac{2 x + 2}{x - 1}

7

f (x) = \frac{x ( x - 2 ) ^{2}}{( x + 3 ) ^{3}}

What is the extreme f(x)=x^2+(480)/x ?
The extreme f(x)=x^2+(480)/x is Minimum(2\sqrt[3]{30},12*30^{2/3})