What is the extreme f(x)=4x^2+(3000)/x ?

The extreme f(x)=4x^2+(3000)/x is Minimum(5\sqrt[3]{3},300*3^{2/3})

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extreme f(x)=4x^2+(3000)/x

extreme points

f (x) = 4 x^{2} + \frac{3000}{x}

Solution steps

f^{'} (x) = 8 x - \frac{3000}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

4 x^{2} + \frac{3000}{x} : 300 \cdot 3^{\frac{2}{3}}

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

f (x, y) = x^{3} + 8 y^{3} - 12 x y

f (x) = 4 x e^{- 4 x + 1}

f (x) = \frac{( x ^{2} - 9 x + 50 )}{x - 7}

f (x) = x^{4} - 32 x + 9

What is the extreme f(x)=4x^2+(3000)/x ?
The extreme f(x)=4x^2+(3000)/x is Minimum(5\sqrt[3]{3},300*3^{2/3})