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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 3 \frac{1}{4},

Increasing

: 3 \frac{1}{4} < x < \infty

Plug

x = \frac{1}{3 4}

into

x^{4} - x : - \frac{3}{4 3 4}

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

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)})