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}

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

into

p (x) = 12 x^{4} + 15 x^{3} + a x^{2} - 69 x - 30

f (x, y) = x + 2 \frac{y}{2} x + y

f (x, y) = 16 - (x - 2)^{2} - (y - 2)^{2}

f (x) = x^{4} - 18 x^{2} - 3

f (x) = 13 + 2 x - x^{2}, 0 \leq x \leq 5

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})