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

The extreme f(x)=(x+2)^{2/3}(x-2)^{1/3} is Maximum(-2,0),Minimum(2/3 ,-(4*4^{1/3})/3)

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

extreme points

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

Maximum (- 2, 0), Minimum (\frac{2}{3}, - \frac{4 \cdot 4 ^{\frac{1}{3}}}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < \frac{2}{3},

Increasing

: \frac{2}{3} < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

(x + 2)^{\frac{2}{3}} (x - 2)^{\frac{1}{3}} : 0

Maximum (- 2, 0)

Plug

x = \frac{2}{3}

into

(x + 2)^{\frac{2}{3}} (x - 2)^{\frac{1}{3}} : - \frac{4 \cdot 4 ^{\frac{1}{3}}}{3}

Minimum (\frac{2}{3}, - \frac{4 \cdot 4 ^{\frac{1}{3}}}{3})

Maximum (- 2, 0), Minimum (\frac{2}{3}, - \frac{4 \cdot 4 ^{\frac{1}{3}}}{3})

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

\frac{x ^{2} - 1}{x ^{2} + 4}

f (x, y) = y - x

11 x^{2} + 2 \cdot (- 25) x (1 - x) + 24 (1 - x)^{2}

f (x, y) = x^{2} + y^{2} + \frac{16}{x} + \frac{2}{y}

What is the extreme f(x)=(x+2)^{2/3}(x-2)^{1/3} ?
The extreme f(x)=(x+2)^{2/3}(x-2)^{1/3} is Maximum(-2,0),Minimum(2/3 ,-(4*4^{1/3})/3)