Question

Question #21e52

Answer 1

The domain is `x in[-1,1]uu(3,+oo)`

Explanation:

Let's factorise the denominator of the function

`x^2-4x+3=(x-1)(x-3)`

Therefore,

`sqrt(((x-1)^2(x+1))/((x^2-4x+3)))=sqrt(((x-1)^2(x+1))/((x-1)(x-3)))`

`=sqrt(((x-1)(x+1))/((x-3)))`

Therefore,

`((x-1)(x+1))/((x-3))>=0`

Let `f(x)=((x-1)(x+1))/((x-3))`

Let's build a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-1` `color(white)(aaaaaaa)` `1` `color(white)(aaaaaa)` `3` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x+1` `color(white)(aaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaa)` `+` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-1` `color(white)(aaaa)` `-` `color(white)(aaaa)` ###color(white)(aaaa)#`-` `color(white)(aa)` `0` `color(white)(aa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-3` `color(white)(aaaa)` `-` `color(white)(aaaa)` ###color(white)(aaaa)#`-` `color(white)(aa)` ###color(white)(aaa)#`-` `color(white)(aa)` `||` `color(white)(a)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aaa)` `+` `color(white)(aa)` `0` `color(white)(aa)` `-` `color(white)(aa)` `||` `color(white)(a)` `+`

Therefore,

`f(x)>=0` when `x in[-1,1]uu(3,+oo)`

graph{sqrt(((x-1)(x+1))/(x-3)) [-12.76, 19.27, -3.04, 12.98]}