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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = \frac{10 x - 6}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{3}{5},

Increasing

: \frac{3}{5} < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = \frac{3}{5}

into

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

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

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

y = sin (x) cos (x), 0 \leq x \leq π

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

f (x, y) = x^{3} - y^{3} - 6 x y - 4

f (x) = x^{3} - 3 x^{2} - 9 x + 15

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

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