Domain of function $\frac{1}{x + \sqrt{x+1}}$ I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$
I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
$$x^2 \neq |x+1| \quad \wedge \quad x\geq -1$$
Since $x \geq -1$ the expression $x+1$ is never negative, therefore:
$$x^2 \neq x+1 \quad \wedge \quad x\geq -1$$
$$x^2 - x - 1 \neq 0 \quad \wedge \quad x\geq -1$$
$$x \neq \frac{1\pm \sqrt{5}}{2} \quad \wedge \quad x\geq -1$$
Now both $\frac{1 + \sqrt{5}}{2}$ and $\frac{1 - \sqrt{5}}{2}$ are greater than $-1$. So why isn't my domain $$\mathbb{R} \setminus \{\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\}$$
When I plot the graph I can see that $\frac{1 + \sqrt{5}}{2}$ is part of the domain, but analitically I can't see why or derive from these equations that conclusion.

A value of $x$ satisfying the equation
$$
x=-\sqrt{x+1}
$$
must be nonpositive since $\sqrt{x+1}$ is nonnegative.

Squaring both sides is legal but of the two solutions you found, one of them is positive so doesn't satisfy the given equation, hence should not excluded from the domain.

Thus the domain is $[-1,r)\cup(r,\infty)\;$where $r={\large{\frac{1 - \sqrt{5}}{2}}}$. 
