What is the extreme f(x)=x^{5/3}-6x^{2/3} ?

The extreme f(x)=x^{5/3}-6x^{2/3} is Maximum(0,0),Minimum(12/5 ,(12/5)^{5/3}-6(12/5)^{2/3})

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extreme f(x)=x^{5/3}-6x^{2/3}

extreme points

f (x) = x^{\frac{5}{3}} - 6 x^{\frac{2}{3}}

Maximum (0, 0), Minimum (\frac{12}{5}, (\frac{12}{5})^{\frac{5}{3}} - 6 (\frac{12}{5})^{\frac{2}{3}})

Solution steps

f^{'} (x) = \frac{5 x - 12}{3 x ^{\frac{1}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{12}{5},

Increasing

: \frac{12}{5} < x < \infty

Plug

x = 0

into

x^{\frac{5}{3}} - 6 x^{\frac{2}{3}} : 0

Maximum (0, 0)

Plug

x = \frac{12}{5}

into

x^{\frac{5}{3}} - 6 x^{\frac{2}{3}} : (\frac{12}{5})^{\frac{5}{3}} - 6 (\frac{12}{5})^{\frac{2}{3}}

Minimum (\frac{12}{5}, (\frac{12}{5})^{\frac{5}{3}} - 6 (\frac{12}{5})^{\frac{2}{3}})

Maximum (0, 0), Minimum (\frac{12}{5}, (\frac{12}{5})^{\frac{5}{3}} - 6 (\frac{12}{5})^{\frac{2}{3}})

f (r, s) = \frac{2 r}{r ^{4} + 7 s}

f (x) = 1 + \frac{7}{x} - \frac{2}{x ^{2}}

f (x) = x^{\frac{1}{3}} (x + 3)^{\frac{2}{3}}

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

f (x) = x^{4} - 8 x^{2} + 8

What is the extreme f(x)=x^{5/3}-6x^{2/3} ?
The extreme f(x)=x^{5/3}-6x^{2/3} is Maximum(0,0),Minimum(12/5 ,(12/5)^{5/3}-6(12/5)^{2/3})