What is the extreme f(x)=x^{2/3}(x-5) ?

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

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extreme f(x)=x^{2/3}(x-5)

extreme points

f (x) = x^{\frac{2}{3}} (x - 5)

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

Solution steps

f^{'} (x) = \frac{5 ( x - 2 )}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x^{\frac{2}{3}} (x - 5) : 0

Maximum (0, 0)

Plug

x = 2

into

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

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

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

f (x) = \frac{7 x}{x ^{2} + 5}

(8, 10), (6, 4)

- 5 x + 5 y = 10

f (x) = \frac{x + 1}{x - 1}

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