What is the extreme f(x)=x^2+(14)/x ?

The extreme f(x)=x^2+(14)/x is Minimum(\sqrt[3]{7},3*7^{2/3})

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extreme f(x)=x^2+(14)/x

extreme points

f (x) = x^{2} + \frac{14}{x}

Solution steps

f^{'} (x) = 2 x - \frac{14}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

x^{2} + \frac{14}{x} : 3 \cdot 7^{\frac{2}{3}}

f (x) = e^{x} (2 x^{2} + x - 8)

f (x) = x^{\frac{1}{3}} (x - 7)^{2}

f (x) = 4 - 3 x - x^{3}

f (x) = 11 - 5 x + x^{2}

f (x) = \frac{x ^{3}}{3} - 2 x^{2} + 4

What is the extreme f(x)=x^2+(14)/x ?
The extreme f(x)=x^2+(14)/x is Minimum(\sqrt[3]{7},3*7^{2/3})