What is the extreme f(x)=\sqrt[3]{3x^3+24} ?

The extreme f(x)=\sqrt[3]{3x^3+24} is Saddle(0,2\sqrt[3]{3})

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extreme f(x)=\sqrt[3]{3x^3+24}

extreme points

Solution steps

f^{'} (x) = \frac{3 x ^{2}}{( 3 x ^{3} + 24 ) ^{\frac{2}{3}}}

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

f (x) = x^{3} + 15 x - 9

f (x) = x^{3} - 2 x^{2} - 15 x + 52.2

f (x) = 3 x^{2} - 5 y^{2} - 255 x + 70 y + 23

f (x) = 5 x e^{- x} + 9

What is the extreme f(x)=\sqrt[3]{3x^3+24} ?
The extreme f(x)=\sqrt[3]{3x^3+24} is Saddle(0,2\sqrt[3]{3})