What is the extreme f(x)=(16x^2-16)^{1/3} ?

The extreme f(x)=(16x^2-16)^{1/3} is Minimum(0,-2*2^{1/3})

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extreme f(x)=(16x^2-16)^{1/3}

extreme points

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

Minimum (0, - 2 \cdot 2^{\frac{1}{3}})

Solution steps

f^{'} (x) = \frac{32 x}{3 ( 16 x ^{2} - 16 ) ^{\frac{2}{3}}}

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}{3}} : - 2 \cdot 2^{\frac{1}{3}}

Minimum (0, - 2 \cdot 2^{\frac{1}{3}})

f (x) = \frac{x + 2}{24 - x ^{2} - 49}

x + \frac{12}{x}

y = x^{2} + 1

\frac{1}{x}

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

What is the extreme f(x)=(16x^2-16)^{1/3} ?
The extreme f(x)=(16x^2-16)^{1/3} is Minimum(0,-2*2^{1/3})