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

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

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

extreme points

f (x) = 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) = - x^{3} + 9 x^{2} - 53

G (t) = - \frac{13}{( 4 + t ) ^{2}}

(- 5, - 1), (4, 2)

4 x - 5 y = - 10

f (x) = 4^{x}

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