Set the monotone interval of function $f(x)=x+\sqrt{x^2+2x}$ $$f\colon D\to \mathbb R, \qquad f(x)=x+\sqrt{x^2+2x}$$
I have determinated the derivative:
$$f'(x)=\frac{x+1+\sqrt{x^2+2x}}{\sqrt{x^2+2x}}$$
I'm trying to use the known theorem that says : if $f'(x)\geq0 $ then $f$ is  increasing, the opposite for $f'(x)\leq 0$
The thing is I'm supposed to find the critical points by getting the solution of $f'(x)=0$ And I don't know how to do that.Please help, explain like I'm five
 A: HINT
We need to consider the following systems


*

*$x^2+2x\ge 0 \implies x(x+2)\ge 0 \implies x\le -2 \land x\ge 0 $

*$f'(x)\ge 0 \implies x+1+\sqrt{x^2+2x}\ge 0$


and


*

*$x^2+2x\ge 0\implies x\le -2 \land x\ge 0 $

*$f'(x)\le 0 \implies x+1+\sqrt{x^2+2x}\le 0$
In each case we need to consider two different cases for example for


*

*$f'(x)\ge 0 \implies x+1+\sqrt{x^2+2x}\ge 0\implies \sqrt{x^2+2x}\ge -1-x$


we have


*

*for $-1-x<0$ the inequality is always true for $x\le -2 \land x\ge 0$

*for $-1-x\ge 0$ we can square both side and obtain
$$\sqrt{x^2+2x}\ge -1-x\iff x^2+2x\ge1+x^2+2x\implies0\ge 1$$
A: You can write the function as
$$
f(x)=(x+1)+\sqrt{(x+1)^2-1}
$$
so you can study the easier function
$$
g(x)=x+\sqrt{x^2-1}
$$
which is defined for $x\le-1$ or $x\ge1$ (so $f$ is defined for $x\le-2$ or $x\ge0$).
You have
$$
g'(x)=1+\frac{x}{\sqrt{x^2-1}}=\frac{x+\sqrt{x^2-1}}{\sqrt{x^2-1}}
$$
(note that $g$ is not differentiable at $-1$ and $1$). This is clearly positive for $x>1$. For $x<-1$, note that
$$
x+\sqrt{x^2-1}=\frac{1}{x-\sqrt{x^2-1}}<0
$$
Thus $g$ is increasing over $[1,\infty)$ and decreasing over $(-\infty,-1]$. Therefore $f$ is decreasing over $(-\infty,-2]$ and increasing over $[0,\infty)$.

You can of course work directly with $f'(x)$. Note that the denominator is positive in the domain for $f'$, so you can study $x+1+\sqrt{x^2+2x}$. This is clearly positive for $x>0$; for $x<-2$ we have
$$
x+1+\sqrt{x^2+2x}=\frac{1}{x+1-\sqrt{x^2+2x}}<0
$$
confirming that $f$ is decreasing over $(-\infty,-2]$ and increasing over $[0,\infty)$.
