What is the extreme 2x^2+(36)/x ?

The extreme 2x^2+(36)/x is Minimum(\sqrt[3]{9},2*9^{2/3}+(36)/(\sqrt[3]{9)})

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extreme 2x^2+(36)/x

extreme

Minimum (39, 2 \cdot 9^{\frac{2}{3}} + \frac{36}{3 9})

Solution steps

f^{'} (x) = 4 x - \frac{36}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 39,

Increasing

: 39 < x < \infty

Plug

x = 39

into

2 x^{2} + \frac{36}{x} : 2 \cdot 9^{\frac{2}{3}} + \frac{36}{3 9}

Minimum (39, 2 \cdot 9^{\frac{2}{3}} + \frac{36}{3 9})

What is the extreme 2x^2+(36)/x ?
The extreme 2x^2+(36)/x is Minimum(\sqrt[3]{9},2*9^{2/3}+(36)/(\sqrt[3]{9)})