What is the extreme f(x)=x^{3/4}-2x^{1/4} ?

The extreme f(x)=x^{3/4}-2x^{1/4} is Maximum(0,0),Minimum(4/9 ,(4/9)^{3/4}-2(4/9)^{1/4})

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extreme f(x)=x^{3/4}-2x^{1/4}

extreme points

f (x) = x^{\frac{3}{4}} - 2 x^{\frac{1}{4}}

Maximum (0, 0), Minimum (\frac{4}{9}, (\frac{4}{9})^{\frac{3}{4}} - 2 (\frac{4}{9})^{\frac{1}{4}})

Solution steps

f^{'} (x) = \frac{3}{4 x ^{\frac{1}{4}}} - \frac{1}{2 x ^{\frac{3}{4}}}

Find intervals:

Decreasing

: 0 < x < \frac{4}{9},

Increasing

: \frac{4}{9} < x < \infty

Plug

x = 0

into

x^{\frac{3}{4}} - 2 x^{\frac{1}{4}} : 0

Maximum (0, 0)

Plug

x = \frac{4}{9}

into

x^{\frac{3}{4}} - 2 x^{\frac{1}{4}} : (\frac{4}{9})^{\frac{3}{4}} - 2 (\frac{4}{9})^{\frac{1}{4}}

Minimum (\frac{4}{9}, (\frac{4}{9})^{\frac{3}{4}} - 2 (\frac{4}{9})^{\frac{1}{4}})

Maximum (0, 0), Minimum (\frac{4}{9}, (\frac{4}{9})^{\frac{3}{4}} - 2 (\frac{4}{9})^{\frac{1}{4}})

f (t) = t^{2} - t^{2} h (t - 2) + 3 h (t - 2)

f (x) = 3 (x^{2} + 4) (x^{2} + 8)^{2}

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

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

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

What is the extreme f(x)=x^{3/4}-2x^{1/4} ?
The extreme f(x)=x^{3/4}-2x^{1/4} is Maximum(0,0),Minimum(4/9 ,(4/9)^{3/4}-2(4/9)^{1/4})