<s>[INST] studying the function $f(x)= \sqrt{x^2+x}-x$ I'm studying the function $$f(x)= \sqrt{x^2+x}-x$$
The domain is $(- \infty, -1] \cup [0, + \infty)$.
In particular I'm making some mistakes about the sign of the
first derivative $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}$$
The first derivative is never zero.
$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0 \Rightarrow   2x+1-2 \sqrt{x^2+x} >0 \Rightarrow   2x+1>2 \sqrt{x^2+x}  \Rightarrow   1>0 $ . So it seems that the derivative is always positive in the domain.
But comparing with the result in wolfram alpha the function is decreasing in $(-\infty, -1]$
 [/INST] Note that the domain of the derivative
$$f^\prime(x)\:=\: \frac{2x+1}{2 \sqrt{x^2+x}}-1$$
is not equal to the domain of $f$ since the denominator zeros $\{-1,0\}\,$ have to be excluded.
The sign analysis is quickly achieved using the AGM inequality:
If $0<x$ then
$$\sqrt{x(x+1)}\:<\:\frac{2x+1}{2}\quad\implies\; 0\;<\;\frac{2x+1}{2 \sqrt{x^2+x}}-1$$
For $x<-1$ one obtains
$$\sqrt{(-x)(-(x+1))}\:<\:\frac{-x-(x+1)}2 \;=\; -\frac{2x+1}{2}\quad\implies\;
f^\prime(x)<-2$$
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