What is the extreme f(x)=2x^2+(10800)/x ?

The extreme f(x)=2x^2+(10800)/x is Minimum(3\sqrt[3]{100},18*100^{2/3}+(3600)/(\sqrt[3]{100)})

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extreme f(x)=2x^2+(10800)/x

extreme points

f (x) = 2 x^{2} + \frac{10800}{x}

Solution steps

f^{'} (x) = 4 x - \frac{10800}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

f (x, y) = x^{2} + y^{2} + 6 x - 4 y + 13

f (x) = (3 - 5 x) e^{2 x}

f (x) = 3 - 5 x^{2}

f (x) = x^{2} - 10 x - 2, 2 \leq x \leq 7

f (x, y) = x^{2} + x y - 5 x - 2 y^{2} + 2 y

What is the extreme f(x)=2x^2+(10800)/x ?
The extreme f(x)=2x^2+(10800)/x is Minimum(3\sqrt[3]{100},18*100^{2/3}+(3600)/(\sqrt[3]{100)})