Question

Find the domain of
f(x) =
$\sqrt{ |2x - 1 \div {x}^{2} - 1| } - 1$
​

Answer 1

Given:

$f(x) = \sqrt{ | \frac{2x - 1}{ {x}^{2} - 1} | } - 1$

Everything in the under root should be positive

So,

f(x) ≥ p

| 2x - 1 / x² -1 | - 1 ≥ 0

| 2x - 1 / (x + 1)(x - 1) | - 1 ≥ 0

| 2x - 1 / (x + 1)(x - 1) | ≥ 1

Case - 1

==> 2x - 1 / (x + 1)(x - 1) ≥ 1

==> 2x - 1 / (x + 1)(x - 1) - 1 ≥ 0

==> 2x - 1 - x² + 1 / (x + 1)(x - 1) ≥ 0

==> 2x - x² / (x + 1)(x - 1) ≥ 0

==> x ( 2 - x ) / (x + 1)(x - 1) ≥ 0

==> x ( x - 2) / (x + 1)(x - 1) ≤ 0

Critical Points : 0, 2, -1, 1

Here,

x € ( -1, 0 ] U ( 1, 2 ]

Case - 2

==> 2x - 1 / (x + 1)(x - 1) ≤ - 1

==> 2x - 1 + x² - 1 / (x + 1)(x - 1) ≤ 0

==> x² + 2x - 2 / (x + 1)(x - 1) ≤ 0

==> (x - √3 + 1)(x + 1 + √3) / (x + 1)(x - 1) ≤ 0

Critical Points : √3 - 1 , -√3 - 1 , -1, 1

Here,

x € [ -√3 - 1, -1 ) U [ √3 - 1, 1 )

Total:

x € [ - √3 - 1 , - 1 ) U ( 1 , 2 ]