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Find the domain of $y=\cos^{-1}\left(\dfrac{1-2|x|}{3}\right)+\log_{|x-1|}x$.

Since domain of $\cos^{-1}x$ is $[-1,1]$, therefore $-1 \leq (\frac{1-2|x|}{3}) \leq 1$

$$-3 \leq (1-2|x|) \leq 3 \Rightarrow -4 \leq -2|x| \leq 2 \Rightarrow -2 \leq -|x| \leq 1 $$ Please guide further on this..

Also $\log_{|x-1|}x$: $x >0$ and $|x-1| > 1$ .

StubbornAtom
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Sachin
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2 Answers2

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$$-2\le -|x|\le 1\iff 2\ge |x|\ge -1\implies |x|\le 2\implies -2\le x\le 2$$

For $\log_{|x-1|}x, |x-1|>0$ and $|x-1|\ne1$ and $x>0$

The first condition is always true, so we need $|x-1|\ne1$

As $x$ is real, $x-1\ne \pm1\implies x\ne 0,2$

So, the domain is $0<x<2$

lab bhattacharjee
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Use: $$\cos\in[-1,1]\ \text{So, }\cos^{-1}(A) \text{ exists }\forall A\in[-1,1]$$

$$\log_a b \ \text{ exists} \ \ \forall a\neq 1\ , a>0 \ \text{ and} \ \ b>0 $$

$$|x|\ge0$$

See for $cos^{-1}$ and similarly you can calculate for log function. Answer is the intersection of both.

ABC
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  • If you are typing text inside the math environment, instead of messing with extra backslashes all over, just wrap it in \text{...}. That way it is upright, not italicized. – Zev Chonoles Apr 09 '13 at 03:28
  • I think you mean $\forall a\ne1,a>0$, yes? – Cameron Buie Apr 09 '13 at 03:35
  • @CameronBuie Yes.Thanx. – ABC Apr 09 '13 at 03:38
  • I don't think that $\cos \in [-1, 1]$ is correct, because $\cos$ is a function. It should be $\operatorname{Im}\cos = [-1, 1]$, where $\operatorname{Im}$ denotes the image of a function. – rubik Dec 19 '14 at 11:25