What is the extreme (x^2+x)/(|x|+1) ?

The extreme (x^2+x)/(|x|+1) is Minimum(1-sqrt(2),(4-3sqrt(2))/(sqrt(2)))

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extreme (x^2+x)/(|x|+1)

extreme points

\frac{x ^{2} + x}{∣ x ∣ + 1}

Minimum (1 - 2, \frac{4 - 3 2}{2})

Solution steps

Find the critical points:

x = 1 - 2

Find intervals:

Decreasing

: - \infty < x < 1 - 2,

Increasing

: 1 - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug the extreme point

x = 1 - 2

into

\frac{x ^{2} + x}{∣ x ∣ + 1} \Rightarrow y = \frac{4 - 3 2}{2}

Minimum (1 - 2, \frac{4 - 3 2}{2})

18 x y - 3 x^{2} y - 2 x y^{2}

f (x) = x^{2} - 8 x - 9, 1 \leq x \leq 6

f (x, y) = (y^{2} - 4) (x - x^{2})

f (x) = x^{2} - 8 x + 19

f (x) = - 0.02 x^{2} + 4 x - 40

What is the extreme (x^2+x)/(|x|+1) ?
The extreme (x^2+x)/(|x|+1) is Minimum(1-sqrt(2),(4-3sqrt(2))/(sqrt(2)))