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

The extreme f(x)=x^{(6)}-x^{(3)}+3 is Saddle(0,3),Minimum(1/(\sqrt[3]{2)}, 11/4)

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extreme f(x)=x^{(6)}-x^{(3)}+3

extreme points

f (x) = x^{(6)} - x^{(3)} + 3

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

Increasing

Plug

x = 0

into

x^{6} - x^{3} + 3 : 3

Saddle (0, 3)

Plug

into

x^{6} - x^{3} + 3 : \frac{11}{4}

f (x) = 5 - 6 x - 4 x^{2}

f (x) = 0.002 x^{3} + 9 x + 7626

f (x, y) = (1 - x y)

f (x) = x^{3} - 15 x^{2} + 27 x - 8

f (x) = 2 x + 37 x^{\frac{2}{37}}

What is the extreme f(x)=x^{(6)}-x^{(3)}+3 ?
The extreme f(x)=x^{(6)}-x^{(3)}+3 is Saddle(0,3),Minimum(1/(\sqrt[3]{2)}, 11/4)