Question

How do you solve `(x(4-x))/(x+2)>=0` using a sign chart?

Answer 1

The solution is `x in (-oo, -2) uu [0, 4]`

Explanation:

Let

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

Let's build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaaa)` `-2` `color(white)(aaaaaa)` `0` `color(white)(aaaaaa)` `4` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `x+2` `color(white)(aaaaaa)` `-` `color(white)(aa)` `||` `color(white)(aaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `-` `color(white)(aa)` `0` `color(white)(aa)` `+` `color(white)(aaaaa)` `+`

`color(white)(aaaa)` `4-x` `color(white)(aaaaaaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aa)` ###color(white)(aaa)#`+` `color(white)(aa)` `0` `color(white)(aa)` `-`

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

Therefore,

`f(x)>=0` when `x in (-oo, -2) uu [0, 4]`

graph{(x(4-x))/(x+2) [-38.63, 43.6, -7.9, 33.2]}