What is the extreme sqrt(x^2+2x)-x ?

The extreme sqrt(x^2+2x)-x is Minimum(-2,2),Minimum(0,0)

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extreme sqrt(x^2+2x)-x

extreme points

x^{2} + 2 x - x

Minimum (- 2, 2), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{x + 1}{x ^{2} + 2 x} - 1

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

x^{2} + 2 x - x : 2

Minimum (- 2, 2)

Plug

x = 0

into

x^{2} + 2 x - x : 0

Minimum (0, 0)

Minimum (- 2, 2), Minimum (0, 0)

y = 4 (x - 2) (x + 1)

12 + 6 x^{2} - x^{3}

f (x, y) = x^{4} + x^{2} - 6 x y + 3 y^{2}

f (x, y) = 3 x + 3 y

f (x, y) = x y e^{- x - y}

What is the extreme sqrt(x^2+2x)-x ?
The extreme sqrt(x^2+2x)-x is Minimum(-2,2),Minimum(0,0)