What is the extreme f(x)=|x^3|-2x ?

The extreme f(x)=|x^3|-2x is Minimum(sqrt(2/3),(2/3)^{3/2}-2sqrt(2/3))

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extreme f(x)=|x^3|-2x

extreme

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

Solution steps

Find the critical points:

x = \frac{2}{3}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{2}{3},

Increasing

: \frac{2}{3} < x < \infty

Plug the extreme point

x = \frac{2}{3}

into

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

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

What is the extreme f(x)=|x^3|-2x ?
The extreme f(x)=|x^3|-2x is Minimum(sqrt(2/3),(2/3)^{3/2}-2sqrt(2/3))