What is the extreme f(x)=x^{2/3}*(6-x)^{1/3} ?

The extreme f(x)=x^{2/3}*(6-x)^{1/3} is Minimum(0,0),Maximum(4,2*2^{2/3})

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extreme f(x)=x^{2/3}*(6-x)^{1/3}

extreme points

f (x) = x^{\frac{2}{3}} \cdot (6 - x)^{\frac{1}{3}}

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

Solution steps

f^{'} (x) = \frac{4 - x}{x ^{\frac{1}{3}} ( - x + 6 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = 0

into

x^{\frac{2}{3}} \cdot (6 - x)^{\frac{1}{3}} : 0

Minimum (0, 0)

Plug

x = 4

into

x^{\frac{2}{3}} \cdot (6 - x)^{\frac{1}{3}} : 2 \cdot 2^{\frac{2}{3}}

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

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

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

f (x) = sin (\frac{x}{2})

f (x, y) = x^{3} + y^{3} - 3 x - 3 y + 20

f (x, y) = x^{3} + y^{2} - 4 x y + 17 x - 10 y + 8

f (x) = 2 x^{3} + 45 x^{2} + 84 x + 29

What is the extreme f(x)=x^{2/3}*(6-x)^{1/3} ?
The extreme f(x)=x^{2/3}*(6-x)^{1/3} is Minimum(0,0),Maximum(4,2*2^{2/3})