What is the extreme f(x)=x^2-1/x ?

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

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extreme f(x)=x^2-1/x

extreme points

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

Minimum (- \frac{1}{3 2}, \frac{1}{2 ^{\frac{2}{3}}} + 32)

Solution steps

f^{'} (x) = 2 x + \frac{1}{x ^{2}}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{3 2},

Increasing

: - \frac{1}{3 2} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{1}{3 2}

into

x^{2} - \frac{1}{x} : \frac{1}{2 ^{\frac{2}{3}}} + 32

Minimum (- \frac{1}{3 2}, \frac{1}{2 ^{\frac{2}{3}}} + 32)

f (x) = 2 x^{2} - 2 x^{4}

f (x) = 2 e^{x^{2} - 2 x + 1}

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

f (x) = \frac{2 x - 5}{x}

\frac{5}{x + 6}

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