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

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

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

extreme points

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

Minimum (0, 0)

Plug

x = 4

into

x^{\frac{2}{3}} (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} - 4)^{7}

y = x^{3} - 27 x + 3, - 2 \leq x \leq 4

f (x) = 2 x^{4} - 8 x + 3

2 x^{3} + 3 x^{2} - 12 x + 5, 0 \leq x \leq 2

f (x, y) = x^{2} + x y + y^{2} - 28 y + 261

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