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

The extreme f(x)=x^{2/3}(x+2) is Maximum(-4/5 ,(6(-4/5)^{2/3})/5),Minimum(0,0)

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

extreme points

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

Maximum - \frac{4}{5}, \frac{6 ( - \frac{4}{5} ) ^{\frac{2}{3}}}{5}, Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{5 x + 4}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < - \frac{4}{5},

Decreasing

: - \frac{4}{5} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{4}{5}

into

x^{\frac{2}{3}} (x + 2) : \frac{6 ( - \frac{4}{5} ) ^{\frac{2}{3}}}{5}

Maximum - \frac{4}{5}, \frac{6 ( - \frac{4}{5} ) ^{\frac{2}{3}}}{5}

Plug

x = 0

into

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

Minimum (0, 0)

Maximum - \frac{4}{5}, \frac{6 ( - \frac{4}{5} ) ^{\frac{2}{3}}}{5}, Minimum (0, 0)

y

f (x) = 2 x y - 5 x^{2} - 2 y^{2} + 4 x - 4 y + 4

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

f (x, y) = x^{3} - 3 x y + y^{2} + y - 5

f (x) = \frac{( 8 x ^{3} - x ^{2} - 108 )}{4 x ^{2}}

What is the extreme f(x)=x^{2/3}(x+2) ?
The extreme f(x)=x^{2/3}(x+2) is Maximum(-4/5 ,(6(-4/5)^{2/3})/5),Minimum(0,0)