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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

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

Increasing

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

Decreasing

: 0 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, 0)

Plug

x = - \frac{1}{2}

into

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

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

Plug

x = 0

into

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

Maximum (0, 0)

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

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

f (x) = 98 y^{2} + x^{2} - x^{2} y

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

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

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

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