Question

the domain of f(x)= 1/ sqrt 1-x^2

Answer 1

I understand that a sqrt needs to be > or = to 0 and that the denominator needs to be 0, but I'm lost on how to combine the two concepts

Answer 2

Do I just have to make sqrt 1-x^2 equal to zero?

Answer 3

since the denominator must not be a zero, you can rule out the equality, thus, leaving you with solving \[1-x^2>0\]

Answer 4

ok, so is it \[\sqrt{1}>x?\]

Answer 5

nope. since you have not considered other values.

try doing \[1-x^2>0 \rightarrow x^2-1<0 \rightarrow (x-1)(x+1)<0\]

Answer 6

how do you get to the domain from that?

Answer 7

[-1<x>1]

Answer 8

I meant (-1<x>1)

Answer 9

From there, we have -1 < x < 1, and this is the range of values permissible for x in the function. So, the domain is -1 < x < 1.

Why -1 < x < 1? here's how...

\[(x-1)(x+1)<0\] implies that one of the factors x - 1 or x + 1 is negative, i.e. less than 0, and the other factor is positive, i.e. greater than 0.

So we would have 2 cases:

Case 1:
\[x-1>0 \rightarrow x>1\] and\[x+1 <0 \rightarrow x <-1\]however, this cannot be simultaneously true, so no solution for this case.

Case 2:
\[x-1<0 \rightarrow x<1\]and\[x+1>0 \rightarrow x>-1\]the two inequalities are true for -1 < x < 1, thus the answer.

Answer 10

I give myself an F for not practicing my algebra skills for a year. (-1<x<1) Thanks for all your help and patience

Answer 11

no problem... :D