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

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

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extreme x^2+(400)/x

extreme points

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

Minimum (2325, 4 \cdot 2 5^{\frac{2}{3}} + \frac{200}{3 25})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2325,

Increasing

: 2325 < x < \infty

Plug

x = 2325

into

x^{2} + \frac{400}{x} : 4 \cdot 2 5^{\frac{2}{3}} + \frac{200}{3 25}

Minimum (2325, 4 \cdot 2 5^{\frac{2}{3}} + \frac{200}{3 25})

0.05 x + 20 + \frac{125}{x}

f (x) = \frac{9 ( t - 2 ) ^{2}}{t ^{2}}

x y - 5 x - 5 y - x^{2} - y^{2}

e^{2 x}

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 12 y + 13

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