<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] $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0\iff 2x+1-2\sqrt{x^2+x}>0\iff 2x+1>2\sqrt{x^2+x}\iff \begin{cases}2x+1>0\\  x^2+x> 0\\ (2x+1)^2>4(x^2+x)\end{cases}\iff \begin{cases}x>\frac{-1}{2}\\x> 0 \vee x< -1\\ \forall x \end{cases}\iff x> 0 $$ Be careful to exclude $0$ in order to have the denominator not equal to $0$
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