Question

What is the domain of `(g @f)(x)` where `f(x)=(x-1)/(2-x)` and `g(x) = sqrt(x+2)`?

Answer 1

The domain is `x in ] -oo,2 [ uu [ 3, +oo[`

Explanation:

`f(x)=(x-1)/(2-x)`

`g(x)=sqrt(x+2)`

`(gof)(x)=g(f(x))`

`=g((x-1)/(2-x))`

`=sqrt((x-1)/(2-x)+2)`

`=sqrt(((x-1)+2(2-x))/(2-x))`

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

`=sqrt((3-x)/(2-x))`

Therefore,

`(3-x)/(2-x)>=0` and `x!=0`

To solve this inequality, we do a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaaaa)` `2` `color(white)(aaaaaaa)` `3` `color(white)(aaaaaa)` `+oo`

`color(white)(aaaa)` `2-x` `color(white)(aaaaa)` `+` `color(white)(aaa)` `∥` `color(white)(aaa)` `-` `color(white)(aaaaa)` `-`

`color(white)(aaaa)` `3-x` `color(white)(aaaaa)` `+` `color(white)(aaa)` `∥` `color(white)(aaa)` `+` `color(white)(aaaaa)` `-`

`color(white)(aaaa)` `g(f(x))` `color(white)(aaaa)` `+` `color(white)(aaa)` `∥` `color(white)(aaa)` `O/` `color(white)(aaaaaa)` `+`

Therefore,

`g(f(x)>=0)`, when `x in ] -oo,2 [ uu [ 3, +oo[`

The domain is `D_g(f(x))` is `x in ] -oo,2 [ uu [ 3, +oo[`