What is the extreme f(x)=\sqrt[5]{x^2}-5 ?

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

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extreme f(x)=\sqrt[5]{x^2}-5

extreme points

Minimum (0, - 5)

Solution steps

f^{'} (x) = \frac{2 x}{5 ( x ^{2} ) ^{\frac{4}{5}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

Minimum (0, - 5)

\frac{3}{1 - 2 ( \frac{1}{x} )}

f (x) = 6 + 2 x^{2} - x^{4}

f (x) = \frac{( 16 x ^{2} - 8 x + 1 )}{4 x - 3}

f (x) = 16 - t

f (x) = \frac{e ^{x^{8}}}{2}

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