What is the extreme f(x)=x^2+(160)/x ?

The extreme f(x)=x^2+(160)/x is Minimum(2\sqrt[3]{10},12*10^{2/3})

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extreme f(x)=x^2+(160)/x

extreme points

f (x) = x^{2} + \frac{160}{x}

Solution steps

f^{'} (x) = 2 x - \frac{160}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

x^{2} + \frac{160}{x} : 12 \cdot 1 0^{\frac{2}{3}}

f (x) = sin^{2} (x)

f (x) = 4 x

(7, 2), (7, 7)

f (x) = \frac{2}{3} x + 1

- \frac{1}{x - 1}

What is the extreme f(x)=x^2+(160)/x ?
The extreme f(x)=x^2+(160)/x is Minimum(2\sqrt[3]{10},12*10^{2/3})