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

The extreme f(x)=sqrt(x^2+x)-x is Minimum(-1,1),Minimum(0,0)

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

extreme points

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

Minimum (- 1, 1), Minimum (0, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: 0 < x < \infty

Plug

x = - 1

into

x^{2} + x - x : 1

Minimum (- 1, 1)

Plug

x = 0

into

x^{2} + x - x : 0

Minimum (0, 0)

Minimum (- 1, 1), Minimum (0, 0)

f (x) = 3 x + 3 sin (x)

\frac{6}{x ^{2} + 3}

f (x) = 0.25 x^{2} + x + 2, (x \geq 0)

f (x) = \frac{x ^{3}}{3} - x

f (x) = x^{4} - 50 x^{2} + 1, - 4 \leq x \leq 11

What is the extreme f(x)=sqrt(x^2+x)-x ?
The extreme f(x)=sqrt(x^2+x)-x is Minimum(-1,1),Minimum(0,0)