# Find the domain of the following function

The given function is:

$$f(x)=\sqrt{\log_{|x|-1}(x^2 + 4x +4)}$$

My approach:

The argument $$x^2+4x+4>0$$ for all $$x\neq-2$$

Also, the base $$|x|-1$$ should be greater than 0 and not equal to 1.

$$\therefore |x|-1>0$$
$$\implies |x|>1$$
$$\implies x>1$$ or $$x<-1$$

And $$|x|-1\neq1$$
$$\implies x\neq2,-2$$
Taking all this in consideration, my answer is $$D_f= (-\infty,-2)\cup(-2,-1)\cup(1,\infty)$$

However, the given answer is $$(-\infty,-3]\cup(-2,-1)\cup(2,\infty)$$

Where did I go wrong? Thanks in advance.

EDIT:
My new approach:

Case I:
$$0<|x|-1<1\implies 1<|x|<2$$

Then, $$x^2 + 4x+4\leq1\implies x^2+4x+3\leq0$$
or $$-3\leq x\leq-1$$
$$\therefore x \in (-2,-1)$$---(1)

Case II:
$$|x|-1>1 \implies |x|>2$$
Then, $$x^2+4x+4\geq1 \implies x^2=4x+3\geq0$$
or $$x\geq-1$$ or $$x\leq-3$$
$$\therefore x \in (-\infty,-3]\cup(2,\infty)$$---(2)

From (1) and (2), $$x\in(-\infty,-3]\cup(-2,-1)\cup(2,\infty)$$

P.S.: Thanks @gimusi

• I am dumb. I just saw that there was a root over the function. So, the argument has to be greater than the base. Nonetheless, please answer my question as I want to know the different methods to solve this. – Vaibhav Oct 17 '18 at 13:11

HINT

Recall that

• for $$a>1 \quad \log_a x \ge0 \iff x\ge 1$$

• for $$0

therefore in order to have $$\log_{|x|-1}(x^2 + 4x +4) \ge 0$$ we need to consider two cases

• $$|x|-1>1 \implies x^2 + 4x +4\ge 1$$

• $$0<|x|-1<1 \implies 0

• Thanks. Now, I can proceed on my own – Vaibhav Oct 17 '18 at 14:59
• @Vaibhav That's nice! You are welcome. If you want show your work editing your own question I can take a look to it. Bye – user Oct 17 '18 at 15:00

You need $$\sqrt{\bullet}$$ to be fed a positive real number, which you did not consider.