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

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

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

extreme points

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

Maximum (0, 0), Minimum (\frac{4}{5}, (\frac{4}{5})^{\frac{5}{3}} - 2 (\frac{4}{5})^{\frac{2}{3}})

Solution steps

f^{'} (x) = \frac{5 x - 4}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{4}{5},

Increasing

: \frac{4}{5} < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = \frac{4}{5}

into

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

Minimum (\frac{4}{5}, (\frac{4}{5})^{\frac{5}{3}} - 2 (\frac{4}{5})^{\frac{2}{3}})

Maximum (0, 0), Minimum (\frac{4}{5}, (\frac{4}{5})^{\frac{5}{3}} - 2 (\frac{4}{5})^{\frac{2}{3}})

f (x) = x^{\frac{2}{3}} (x^{2} - 9)

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

f (x) = ∣ x ∣ + \frac{x}{7}

f (x) = \frac{1}{x ( 12 - x ^{2} )}

f (x) = 2 x^{3} + 12 x^{2} - 192 x - 45

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