What is the extreme f(x)=(4x^2-64)^{1/5} ?

The extreme f(x)=(4x^2-64)^{1/5} is Minimum(0,-2*2^{1/5})

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extreme f(x)=(4x^2-64)^{1/5}

extreme points

f (x) = (4 x^{2} - 64)^{\frac{1}{5}}

Minimum (0, - 2 \cdot 2^{\frac{1}{5}})

Solution steps

f^{'} (x) = \frac{8 x}{5 ( 4 x ^{2} - 64 ) ^{\frac{4}{5}}}

Find intervals:

Decreasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

(4 x^{2} - 64)^{\frac{1}{5}} : - 2 \cdot 2^{\frac{1}{5}}

Minimum (0, - 2 \cdot 2^{\frac{1}{5}})

f (x, y) = e^{y^{2} + x^{3} - 6 x y - 6 x + 2 y}

\frac{2 x + 4}{x ^{2} - 16}

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

f (x) = x [18 - 2 (x)]^{2}

f (x, y) = x^{2} - 3 y^{2} - 8 x + 9 y + 3 x y

What is the extreme f(x)=(4x^2-64)^{1/5} ?
The extreme f(x)=(4x^2-64)^{1/5} is Minimum(0,-2*2^{1/5})