What is the extreme f(x)=\sqrt[3]{x^2-4} ?

The extreme f(x)=\sqrt[3]{x^2-4} is Minimum(0,-\sqrt[3]{4})

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extreme f(x)=\sqrt[3]{x^2-4}

extreme points

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

What is the extreme f(x)=\sqrt[3]{x^2-4} ?
The extreme f(x)=\sqrt[3]{x^2-4} is Minimum(0,-\sqrt[3]{4})