Question

What the domain of? `(x+3/4)/sqrt(x^2-9)`

can you help to solve this question?

Original wording:

What the domain of? x+3/4-/`sqrtx^2-9`

Answer 1

The domain is `x in (-oo,-3)uu(3,+oo)`

Explanation:

The denominator must be `!=0` and for the square root sign, `>0`

Therefore,

`x^2-9>0`

`(x+3)(x-3)>0`

Let `g(x)=(x+3)(x-3)`

Solve this inequality with a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-3` `color(white)(aaaa)` `+3` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+3` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)`

`color(white)(aaaa)` `x-3` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)`

`color(white)(aaaa)` `g(x)` `color(white)(aaaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)`

Therefore,

`g(x)>0` when `x in (-oo,-3)uu(3,+oo)`

The domain is `x in (-oo,-3)uu(3,+oo)`

graph{(x+0.75)/(sqrt(x^2-9)) [-36.53, 36.57, -18.27, 18.27]}