Math Help - find values such that f(x) is a real number

1. find values such that f(x) is a real number

ok so here goes...

let f(x) = arcsin(arcsin x). find the values of x in the interval -1< x< 1 such that f(x) is a real number. please note the less than greater than signs should be equal to as well, i don't know how to make that symbol on the computer.

can anyone suggest some software or program that is good for making or creating math symbols. thanks in advance.

real quick let me get back to the problem, what i basically did for this problem, was graph the sin function and the inverse of sin, and after looking at the graphs came to the conclusion that the x is defined in between the -1 and 1, or where sin (-1) and sin (1), after that the arcsin function doesn't exist. for some reason im not sure i could use that as a viable answer. is there a way i could prove this? thx

2. Originally Posted by slapmaxwell1
ok so here goes...

let f(x) = arcsin(arcsin x). find the values of x in the interval -1< x< 1 such that f(x) is a real number. please note the less than greater than signs should be equal to as well, i don't know how to make that symbol on the computer.

can anyone suggest some software or program that is good for making or creating math symbols. thanks in advance.

real quick let me get back to the problem, what i basically did for this problem, was graph the sin function and the inverse of sin, and after looking at the graphs came to the conclusion that the x is defined in between the -1 and 1, or where sin (-1) and sin (1), after that the arcsin function doesn't exist. for some reason im not sure i could use that as a viable answer. is there a way i could prove this? thx
1. The arcsin-function is defined for $-1\le x \le 1$.

2. Then the function

$f(x)=\arcsin(\arcsin(x))$ is defined for $-1 \le \arcsin(x) \le 1$

3. Use the inverse function of arcsin to solve the inequality for x:

$-1 \le \arcsin(x) \le 1~\implies~ \sin(-1) \le x \le \sin(1)$