What is the extreme f(x)=(16x^2-16)^{1/5} ?

The extreme f(x)=(16x^2-16)^{1/5} is Minimum(0,-16^{1/5})

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extreme f(x)=(16x^2-16)^{1/5}

extreme points

f (x) = (16 x^{2} - 16)^{\frac{1}{5}}

Minimum (0, - 1 6^{\frac{1}{5}})

Solution steps

f^{'} (x) = \frac{32 x}{5 ( 16 x ^{2} - 16 ) ^{\frac{4}{5}}}

Find intervals:

Decreasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

(16 x^{2} - 16)^{\frac{1}{5}} : - 1 6^{\frac{1}{5}}

Minimum (0, - 1 6^{\frac{1}{5}})

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

f (x) = \frac{1}{\frac{2}{x + 5} - 2}

\frac{1}{x - 4}

x^{2} - y^{2} = 4

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

What is the extreme f(x)=(16x^2-16)^{1/5} ?
The extreme f(x)=(16x^2-16)^{1/5} is Minimum(0,-16^{1/5})