What is the extreme f(x)=x^{2/3}*(4-x^2) ?

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

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extreme f(x)=x^{2/3}*(4-x^2)

extreme points

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

Maximum (- 1, 3), Minimum (0, 0), Maximum (1, 3)

Solution steps

f^{'} (x) = - \frac{8 ( x + 1 ) ( x - 1 )}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, 3)

Plug

x = 0

into

x^{\frac{2}{3}} \cdot (4 - x^{2}) : 0

Minimum (0, 0)

Plug

x = 1

into

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

Maximum (1, 3)

Maximum (- 1, 3), Minimum (0, 0), Maximum (1, 3)

3 x^{2} + 2 y^{2} - 4 y

f (x) = 102 t + 0.5 t^{2} - t^{3}

f (x) = x^{3} - 8 x^{2} - 12 x + 5

f (x) = x^{3} - 8 x^{2} - 12 x + 9

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

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