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

The extreme x^{2/3}(x-4) is Maximum(0,0),Minimum(8/5 ,-(48)/(5*5^{2/3)})

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

extreme points

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

Maximum (0, 0), Minimum (\frac{8}{5}, - \frac{48}{5 \cdot 5 ^{\frac{2}{3}}})

Solution steps

f^{'} (x) = \frac{5 x - 8}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{8}{5},

Increasing

: \frac{8}{5} < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = \frac{8}{5}

into

x^{\frac{2}{3}} (x - 4) : - \frac{48}{5 \cdot 5 ^{\frac{2}{3}}}

Minimum (\frac{8}{5}, - \frac{48}{5 \cdot 5 ^{\frac{2}{3}}})

Maximum (0, 0), Minimum (\frac{8}{5}, - \frac{48}{5 \cdot 5 ^{\frac{2}{3}}})

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

f (x) = - 0.04 x^{2} + 42 x - 500

f (x) = x^{5} - 10 x^{4} + 2 = x^{4} (x - 10) + 2

f (x) = x^{9} - x^{7}

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

What is the extreme x^{2/3}(x-4) ?
The extreme x^{2/3}(x-4) is Maximum(0,0),Minimum(8/5 ,-(48)/(5*5^{2/3)})