3
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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

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  • $\begingroup$ 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. $\endgroup$ – Vaibhav Oct 17 '18 at 13:11
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HINT

Recall that

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

  • for $0<a<1 \quad \log_a x \ge 0 \iff 0<x\le 1$

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

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

  • $0<|x|-1<1 \implies 0<x^2 + 4x +4\le 1$

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  • $\begingroup$ Thanks. Now, I can proceed on my own $\endgroup$ – Vaibhav Oct 17 '18 at 14:59
  • $\begingroup$ @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 $\endgroup$ – user Oct 17 '18 at 15:00
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You need $\sqrt{\bullet}$ to be fed a positive real number, which you did not consider.

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