Finding natural domain of real functions involving square root and absolute value I am trying to find the natural domain of the following 2 functions:


*

*$ \sqrt{\frac{x+1}{x-2}+3} $

*$ \sqrt{|x-5| - |x+1|} $ 
For question 1 I got $[-4,\infty) \cap (-\infty,1]$, but I'm not sure that it is correct. However, I cannot get a solution for the second question. 
Steps for question 1:
$\sqrt{\frac{x+1}{x-2}+3} $. 
The formula will be well defined if:
$\frac{x+1}{x-2}+3 \ge 0 $
$\frac{x+1}{x-2} \ge -3 $
Hence,
$(x+1 \ge -3)\land(x-2>-3)$ or $(x+1\le-3)\land(x-2<-3)$
Their intersection means the domain is $[-4,\infty) \cap (-\infty,1]$ but I assume that that is wrong.
 A: For the first question, note that $x \ne 2$,
$$\frac{x+1}{x-2} \ge -3 $$
multiply both sides by $(x-2)^2$,
$$(x+1)(x-2) \ge -3 (x-2)^2$$
$$(x-2)((x+1)+3(x-2)) \ge 0$$
$$(x-2)(4x-5) \ge 0$$
Try to complete it?
For the second question,
We need $$|x-5|-|x+1| \ge 0$$
$$|x-5| \ge |x+1|$$
squaring booth sides
$$x^2-10x+25 \ge x^2+2x +1$$
Can you continue from here?
A: For the first part, you need $\frac{x+1}{x-2} \ge 0$ and $x \neq 2$. You need same sign on numerator and denominator, this is possible only in the interval $x \in (-\infty, -1] \cup (2, \infty)$
For the second part, you can say that distance of $x$ from $5$ must be greater or equal to distance from $-1$, or that $x \in (-\infty, 2]$ because $x=2$ is the midpoint of $-1$ and $5$. Other way is that you simplify the expression in different intervals.
A: For your first question: we get $$\frac{x+1}{x-2}+3=\frac{4x-5}{x-2}$$
This gives: $$x\geq \frac{5}{4}$$ or $$-\infty<x<2$$
and let $$y=\sqrt{\frac{x+1}{x-2}+3}$$ then the domain is given by $$0<y<2$$ or $$y>2$$
For your second question:
We have $$|x-5|\geq |x+1|$$ after squaring we get $$-\infty<x\le 2$$
and the range is given by $$0\le y\le \sqrt{6}$$
