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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, 0)

Plug

x = - \frac{1}{2}

into

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

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

Plug

x = 0

into

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

Minimum (0, 0)

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

(x^{2} + 12) (4 - x^{2})

f (x y) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6

f (x) = 2 x - 4 x

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

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

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