Set the monotone interval of function $f(x)=x+\sqrt{x^2+2x}$

Question

$$f\colon D\to \mathbb R, \qquad f(x)=x+\sqrt{x^2+2x}$$ 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

Answer 1

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 $[0,\infty)$.

Answer 2

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

$$\sqrt{x^2+2x}\ge -1-x\iff x^2+2x\ge1+x^2+2x\implies0\ge 1$$