What is the extreme f(x)=(x^2-36)^{1/3} ?

The extreme f(x)=(x^2-36)^{1/3} is Minimum(0,-36^{1/3})

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extreme f(x)=(x^2-36)^{1/3}

extreme points

f (x) = (x^{2} - 36)^{\frac{1}{3}}

Minimum (0, - 3 6^{\frac{1}{3}})

Solution steps

f^{'} (x) = \frac{2 x}{3 ( x ^{2} - 36 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < 0,

Increasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

(x^{2} - 36)^{\frac{1}{3}} : - 3 6^{\frac{1}{3}}

Minimum (0, - 3 6^{\frac{1}{3}})

f (x) = 460 x - 11040

f (x) = 4 x^{3} - 3 x^{4}

3 - x^{2}

y = 2 x - 6

What is the extreme f(x)=(x^2-36)^{1/3} ?
The extreme f(x)=(x^2-36)^{1/3} is Minimum(0,-36^{1/3})