What is the extreme f(x)=x^2+(400)/x ?

The extreme f(x)=x^2+(400)/x is Minimum(2\sqrt[3]{25},4*25^{2/3}+(200)/(\sqrt[3]{25)})

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extreme f(x)=x^2+(400)/x

extreme points

f (x) = x^{2} + \frac{400}{x}

Solution steps

f^{'} (x) = 2 x - \frac{400}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

f (x) = x + 3 - 4

f (x) = - 8 x + 7

y = 3 lo g_{2} (x - 4)

f (x) = ln (2 \cdot x + 1)

f (x) = \frac{x - 4}{x + 6}

What is the extreme f(x)=x^2+(400)/x ?
The extreme f(x)=x^2+(400)/x is Minimum(2\sqrt[3]{25},4*25^{2/3}+(200)/(\sqrt[3]{25)})