What is the extreme f(x)=x^2+(520)/x ?

The extreme f(x)=x^2+(520)/x is Minimum(\sqrt[3]{260},260^{2/3}+(8*65^{2/3})/(\sqrt[3]{4)})

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extreme f(x)=x^2+(520)/x

extreme points

f (x) = x^{2} + \frac{520}{x}

Minimum (3260, 26 0^{\frac{2}{3}} + \frac{8 \cdot 6 5 ^{\frac{2}{3}}}{3 4})

Solution steps

f^{'} (x) = 2 x - \frac{520}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3260,

Increasing

: 3260 < x < \infty

Plug

x = 3260

into

x^{2} + \frac{520}{x} : 26 0^{\frac{2}{3}} + \frac{8 \cdot 6 5 ^{\frac{2}{3}}}{3 4}

Minimum (3260, 26 0^{\frac{2}{3}} + \frac{8 \cdot 6 5 ^{\frac{2}{3}}}{3 4})

f (x, y) = x y + 2 x - ln (x^{2} y)

2 x^{3} - 2 x^{2} - 2 x + 3

f (x) = 6 x^{2} - 12, - 4 \leq x \leq 1

f (y, x) = 0.6 x + 0.1 y - 0.01 x y

f (x) = 3 x^{4} - 8 x^{3} + 10

What is the extreme f(x)=x^2+(520)/x ?
The extreme f(x)=x^2+(520)/x is Minimum(\sqrt[3]{260},260^{2/3}+(8*65^{2/3})/(\sqrt[3]{4)})