What is the extreme f(x)=x^4-x ?

The extreme f(x)=x^4-x is Minimum(1/(\sqrt[3]{4)},-3/(4\sqrt[3]{4)})

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extreme f(x)=x^4-x

extreme points

f (x) = x^{4} - x

Solution steps

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

Find intervals:

Decreasing

Increasing

Plug

into

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

f (t) = 10 e^{a t}

f (x) = 2 x^{3} - 6 x^{2} - 90 x + 1

f (x) = (11 - x) (x + 1)^{2}

f (x, y) = 400 - 16 x^{2} - 49 y^{2}

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