What is the extreme 2x^2+(400)/x+10 ?

The extreme 2x^2+(400)/x+10 is Minimum(\sqrt[3]{100},2*100^{2/3}+(400)/(\sqrt[3]{100)}+10)

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extreme 2x^2+(400)/x+10

extreme points

2 x^{2} + \frac{400}{x} + 10

Solution steps

f^{'} (x) = 4 x - \frac{400}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

\frac{6 x}{5 x - 6}

f (x) = 4 - x

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

x^{2} + 8 x + 9

f (x) = 5 - 6 x

What is the extreme 2x^2+(400)/x+10 ?
The extreme 2x^2+(400)/x+10 is Minimum(\sqrt[3]{100},2*100^{2/3}+(400)/(\sqrt[3]{100)}+10)