What is the extreme y=sqrt(x^2+4x)-x ?

The extreme y=sqrt(x^2+4x)-x is Minimum(-4,4),Minimum(0,0)

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extreme y=sqrt(x^2+4x)-x

extreme points

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

Minimum (- 4, 4), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{x + 2}{x ^{2} + 4 x} - 1

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: 0 < x < \infty

Plug

x = - 4

into

x^{2} + 4 x - x : 4

Minimum (- 4, 4)

Plug

x = 0

into

x^{2} + 4 x - x : 0

Minimum (0, 0)

Minimum (- 4, 4), Minimum (0, 0)

f (x) = 4 cos (2 x) + 1

0.5 x^{3} - 3 x^{2} + 4.5 x

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

f (x) = 7 x - 7 sin (x), 0 \leq x \leq 2 π

f (x) = x^{4} + 12 x^{3} + 16 x^{2} + 4

What is the extreme y=sqrt(x^2+4x)-x ?
The extreme y=sqrt(x^2+4x)-x is Minimum(-4,4),Minimum(0,0)