What is the extreme f(t)=t^2+1/t ?

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

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extreme f(t)=t^2+1/t

extreme points

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

Solution steps

f^{'} (t) = 2 t - \frac{1}{t ^{2}}

Find intervals:

Decreasing

: - \infty < t < 0,

Decreasing

Increasing

Plug

into

y = \frac{1}{3} x^{3} + 2 x^{2} + 3 x

f (x) = y \cdot (e^{x} - 1)

f (x, y) = 2 x^{2} + 8 x - 3 y^{2} + 24 y + 7

f (x) = 2 x^{3} - 6 x^{2} - 48 x

f (x) = x^{2} - 6 x y + y^{2} + 16 y + 9

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