What is the extreme x^2+(405000)/x ?

The extreme x^2+(405000)/x is Minimum(15\sqrt[3]{60},225*60^{2/3}+(1800*15^{2/3})/(\sqrt[3]{4)})

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extreme x^2+(405000)/x

extreme points

x^{2} + \frac{405000}{x}

Solution steps

f^{'} (x) = 2 x - \frac{405000}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

\frac{2 x ^{2} - 14 x + 24}{x ^{2} + 6 x + 40}

f (x) = x (5 - x) (8 - x), 0 < x < 5

f (x) = \frac{( x ^{2} + 121 )}{2 x}

f (x) = - 7 + 12 x - 3 x^{2}

2 x^{3} - 6

What is the extreme x^2+(405000)/x ?
The extreme x^2+(405000)/x is Minimum(15\sqrt[3]{60},225*60^{2/3}+(1800*15^{2/3})/(\sqrt[3]{4)})