What is the extreme f(x)=2x^2+(91.2)/x ?

The extreme f(x)=2x^2+(91.2)/x is Minimum(\sqrt[3]{114/5},48.24376…)

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extreme f(x)=2x^2+(91.2)/x

extreme points

f (x) = 2 x^{2} + \frac{91.2}{x}

Solution steps

f^{'} (x) = 4 x - \frac{91.2}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2.83559 \dots,

Increasing

: 2.83559 \dots < x < \infty

Plug

into

2 x^{2} + \frac{91.2}{x} : 48.24376 \dots

P (a) = y^{2} - 4 r^{2} + r + e^{2}

f (t) = 25 cos (2 t)

f (x) = - 2 x^{3} + 24 x^{2} - 42 x + 7

f (x) = - 0.01 x^{2} + 120 x + 200000

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

What is the extreme f(x)=2x^2+(91.2)/x ?
The extreme f(x)=2x^2+(91.2)/x is Minimum(\sqrt[3]{114/5},48.24376…)