What is the extreme f(x)=x^4-6x^{3/2},x>0 ?

The extreme f(x)=x^4-6x^{3/2},x>0 is Minimum((81/16)^{1/5},(81/16)^{4/5}-6(81/16)^{3/10})

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extreme f(x)=x^4-6x^{3/2},x>0

extreme points

f (x) = x^{4} - 6 x^{\frac{3}{2}}, x > 0

Minimum ((\frac{81}{16})^{\frac{1}{5}}, (\frac{81}{16})^{\frac{4}{5}} - 6 (\frac{81}{16})^{\frac{3}{10}})

Solution steps

Find the critical points:

x = (\frac{81}{16})^{\frac{1}{5}}

Domain of

x^{4} - 6 x^{\frac{3}{2}}, x > 0 : x > 0

Find intervals:

Decreasing

: 0 < x < (\frac{81}{16})^{\frac{1}{5}},

Increasing

: (\frac{81}{16})^{\frac{1}{5}} < x < \infty

Plug the extreme point

x = (\frac{81}{16})^{\frac{1}{5}}

into

x^{4} - 6 x^{\frac{3}{2}} \Rightarrow y = (\frac{81}{16})^{\frac{4}{5}} - 6 (\frac{81}{16})^{\frac{3}{10}}

Minimum ((\frac{81}{16})^{\frac{1}{5}}, (\frac{81}{16})^{\frac{4}{5}} - 6 (\frac{81}{16})^{\frac{3}{10}})

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f (x) = 2 x^{4} - x^{2} + 1

3 x^{3} + 2 y^{3} - 18 y^{2} - 9 x + 9

What is the extreme f(x)=x^4-6x^{3/2},x>0 ?
The extreme f(x)=x^4-6x^{3/2},x>0 is Minimum((81/16)^{1/5},(81/16)^{4/5}-6(81/16)^{3/10})