Domain of $\sqrt{\frac{|x-2|-3|2x+2|}{|x-1|-2}}$

Question

I need to find the domain of

$$\sqrt{\frac{|x-2|-3|2x+2|}{|x-1|-2}}$$ But there are a ton of different cases. What is the smart way to do this? Or rather, what are the steps I need to do in order to tackle this?

Thank you very much.

Answer 1

To solve this, first notice that the denominator cannot be $0$.So:

$$|x-1|-2\neq0$$ $$\implies x\neq-1,3\implies x\in(-\infty,-1)\cup(-1,3)\cup(3,+\infty)$$ For the square root notation, it requires that $$\dfrac{|x-2|-3|2x+2|}{|x-1|-2}\ge0$$ It requires that the denominator and the nominator must both be positive or negative$$|x-2|-3|2x+2|\ge0\implies x\in[-\dfrac{8}{5},-\dfrac{4}{7}]$$and$$|x-1|-2>0\implies x\in(-\infty,-1)\cup(3,+\infty)$$ So the first one is $$x\in[-\dfrac{8}{5},-1)$$ Then, $$|x-2|-3|2x+2|\le0\implies x\in(-\infty,-\dfrac{8}{5}]\cup[-\dfrac{4}{7},+\infty)$$and$$|x-1|-2<0\implies x\in(-1,3)$$ So the second one is $$x\in[-\dfrac{4}{7},3)$$

So to find the domain, we need to get the union of these two sets, which is: $$D=[-\dfrac{8}{5},-1)\cup[-\dfrac{4}{7},3)$$ Hint: Don't rush, analyze step by step you'll finally get the correct answer.

Answer 2

No free lunch here we need to distinguish the cases and solve for the inequality

$$\frac{|x-2|-3|2x+2|}{|x-1|-2}\ge 0$$

with the condition $|x-1|-2\neq 0$ on the intervals $(-\infty,-1)$, $(-1,1]$, $(1,2]$ and$ (2,\infty)$.

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Edit

Notably we have

• for $x\in (-\infty,-1)$

$$\frac{|x-2|-3|2x+2|}{|x-1|-2}\ge 0 \iff \frac{5x+8}{-x-1}\ge 0$$

$$\iff x\in\left[-\frac85,-1\right)$$

• for $x\in (-1,1]$

$$\frac{|x-2|-3|2x+2|}{|x-1|-2}\ge 0 \iff \frac{-7x-4}{-x-1}\ge 0$$

$$\iff x\in\left[-\frac47,1\right]$$

• for $x\in (1,2]$

$$\frac{|x-2|-3|2x+2|}{|x-1|-2}\ge 0 \iff \frac{-7x-4}{x-3}\ge 0$$

$$\iff x\in\left(1,2\right]$$

• for $x\in (2,\infty)$

$$\frac{|x-2|-3|2x+2|}{|x-1|-2}\ge 0 \iff \frac{-5x-8}{x-3}\ge 0$$

$$\iff x\in\left(2,3\right)$$

therefore

$$x\in\left[-\frac85,-1\right)\cup \left[-\frac47,3\right)$$