What is the extreme f(x)=\sqrt[3]{x^2+125} ?

The extreme f(x)=\sqrt[3]{x^2+125} is Minimum(0,5)

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extreme f(x)=\sqrt[3]{x^2+125}

extreme points

Minimum (0, 5)

Solution steps

f^{'} (x) = \frac{2 x}{3 ( x ^{2} + 125 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

Minimum (0, 5)

f (x) = x e^{\frac{- x ^{2}}{162}}

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

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

f (x, y) = 3 x^{2} + 10 x y + 7 y^{2} + 3 x y^{2} + 8 x^{2} y

f (x, y) = y e^{- x}

What is the extreme f(x)=\sqrt[3]{x^2+125} ?
The extreme f(x)=\sqrt[3]{x^2+125} is Minimum(0,5)