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

f (x, y) = x^{3} - 12 x + y^{3} + 3 y^{2} - 9 y

f (x, y) = 4 x^{2} - 2 y^{2} + 2 x y

f (x) = a e^{2 x} - e^{3 x}

f (x) = x^{4} + 8 x^{3} + 18 x^{2} + 4

f (x) = x^{4} + 8 x^{3} + 18 x^{2} - 8

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})