How to find the domain in $f(x)=\frac{\arcsin x}{| \arcsin |x||}+\frac{|\arccos|x||}{\arccos|x|}+1$?

Question

The problem is as follows:

Find the range in the following function:

$f(x)=\frac{\arcsin x}{| \arcsin |x||}+\frac{|\arccos|x||}{\arccos|x|}+1$

$\begin{array}{ll} 1.&\{3,1\}\\ 2.&\left\{\frac{1}{3},1\right\}\\ 3.&\{1,2\}\\ 4.&\{3,2\}\\ \end{array}$

Gee. I'm confused with so many absolute values exactly on what should be done here to solve this problem. Can someone help me here?. I think the most misleading part here is the brackets used. I have no idea on its meaning.

Can someone help me on how to solve this thing without much fuss?.

Answer 1

First note that $\arccos |x|\in[0,\pi]$, and so is always non-negative. This means the second term is $1$. Secondly, $\arcsin |x| \ge 0$ as the $\arcsin$ of a positive number is positive, so $|\arcsin |x| | =\arcsin|x|$. $$f(x) =2+\frac{\arcsin x}{\arcsin|x|} $$ Now, when $x\le 0$, $\arcsin |x| =-\arcsin x$ and when $x\ge 0$, $\arcsin|x| =\arcsin x$. Therefore $f(x)$ can take either of the values $1$ and $3$.

Answer 2

ok so range: $\dfrac{a}{|a|} = \dfrac{|a|}{a} = 1 \text{ if } a>0, -1 \text{ if } a<0$

since $\arccos|x|> 0$ for all $x$ in its domain, $\dfrac{|\arccos|x||}{\arccos|x|} = 1$

as for $\dfrac{\arcsin(x)}{|\arcsin|x||}$, this can be either 1 or -1 depending of the value of x.

So your function has two cases:

$\{-1+1+1, 1+1+1\} = \{1,3\}$