What is the extreme f(x)=4.63x^2+(16000)/x ?

The extreme f(x)=4.63x^2+(16000)/x is Minimum((20\sqrt[3]{100})/(\sqrt[3]{463)},2000.05333…)

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extreme f(x)=4.63x^2+(16000)/x

extreme points

f (x) = 4.63 x^{2} + \frac{16000}{x}

Solution steps

f^{'} (x) = 9.26 x - \frac{16000}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

4.63 x^{2} + \frac{16000}{x} : 2000.05333 \dots

f (x) = (x + 4) (x - 2)^{2}

(- 2, - 5), (- 9, 4)

x^{5}

f (x) = 13 x - 13

2 x + 2 y = 11

What is the extreme f(x)=4.63x^2+(16000)/x ?
The extreme f(x)=4.63x^2+(16000)/x is Minimum((20\sqrt[3]{100})/(\sqrt[3]{463)},2000.05333…)