What is the extreme x^2+12+18x^{-1} ?

The extreme x^2+12+18x^{-1} is Minimum(\sqrt[3]{9},9^{2/3}+12+(18)/(\sqrt[3]{9)})

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extreme x^2+12+18x^{-1}

extreme

Minimum (39, 9^{\frac{2}{3}} + 12 + \frac{18}{3 9})

Solution steps

f^{'} (x) = 2 x - \frac{18}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 39,

Increasing

: 39 < x < \infty

Plug

x = 39

into

x^{2} + 12 + 18 x^{- 1} : 9^{\frac{2}{3}} + 12 + \frac{18}{3 9}

Minimum (39, 9^{\frac{2}{3}} + 12 + \frac{18}{3 9})

What is the extreme x^2+12+18x^{-1} ?
The extreme x^2+12+18x^{-1} is Minimum(\sqrt[3]{9},9^{2/3}+12+(18)/(\sqrt[3]{9)})