Question

How do you solve and write the following in interval notation: `1+ 20/(x + 1) ≤ 20/x`?

Answer 1

The solution is `x in [-5,1) uu (0,4]`

Explanation:

We cannot do crossing over.

So, we rewrite, simplify and factorise the inequality

`1+(20)/(x+1)<=20/x`

`(x+1+20)/(x+1)<=20/x`

`(x+1+20)/(x+1)-20/x<=0`

`(x(x+21)-20(x+1))/(x(x+1))<=0`

`(x^2+21x-20x-20)/(x(x+1))<=0`

`(x^2+x-20)/(x(x+1))<=0`

`((x-4)(x+5))/(x(x+1))<=0`

Let `f(x)=((x-4)(x+5))/(x(x+1))`

We can construct a sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-5` `color(white)(aaaaa)` `-1` `color(white)(aaaaaa)` `0` `color(white)(aaaaaa)` `4` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+5` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `||` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x+1` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `||` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `-` `color(white)(aa)` `||` `color(white)(aaa)` `+` `color(white)(aaa)` `+`

`color(white)(aaaa)` `x-4` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `-` `color(white)(aa)` `||` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aa)` `||` `color(white)(aa)` `+` `color(white)(aa)` `||` `color(white)(aaa)` `-` `color(white)(aaa)` `+`

Therefore,

`f(x)<=0` when `x in [-5,1) uu (0,4]`