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))

Popular >

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

extreme points

f (x) = x^{3} - 2 x

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})

f (x) = 90 x^{2} - 0.2 x^{4}

f (x \cdot y) = 3 x^{2} - 4 x y + 3 y^{2} + 8 x - 17 y + 5

x^{3} - 16 x y + 4 y^{2} + 16 x^{2} - 12 x + 5

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

y = x^{3} - 3 x^{2} + 4 x - 4

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))