What is the extreme f(x)=\sqrt[3]{x^2+27} ?

The extreme f(x)=\sqrt[3]{x^2+27} is Minimum(0,3)

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extreme f(x)=\sqrt[3]{x^2+27}

extreme

Minimum (0, 3)

Solution steps

f^{'} (x) = \frac{2 x}{3 ( x ^{2} + 27 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

3 x^{2} + 27 : 3

Minimum (0, 3)

What is the extreme f(x)=\sqrt[3]{x^2+27} ?
The extreme f(x)=\sqrt[3]{x^2+27} is Minimum(0,3)