Question

Question #d7eb4

Answer 1

The solution is `x in (-oo,-2/3] uu (2,+oo)`

Explanation:

Let `f(x)=(3x+2)/(2x-4)`

Build a sign chart

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

`color(white)(aaaa)` `3x+2` `color(white)(aaaa)` `-` `color(white)(aaaa)` `0` `color(white)(aa)` ###color(white)(aa)#`+` `color(white)(aaaa)` ###color(white)(aaaaaaa)#`+`

`color(white)(aaaa)` `2x-4` `color(white)(aaaa)` `-` `color(white)(aaaa)` ###color(white)(aaaaa)#`-` `color(white)(aaaa)` `||` `color(white)(aaaa)` ###color(white)(aa)#`+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaa)` `+` `color(white)(aaaa)` `0` `color(white)(aaaa)` `-` `color(white)(aaaa)` `||` `color(white)(aaaa)` ###color(white)(aa)#`+`

Therefore,

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

graph{(3x+2)/(2x-4) [-18.02, 18.03, -9.01, 9.01]}