Question

How do you solve and write the following in interval notation: `x^3-3x^2-4x<0`?

Answer 1

The solution is `x in (-oo,-1) uu (0,4)`

Explanation:

We start by factorising the inequality

`x^3-3x^2-4x<0`

`x(x^2-3x+4)<0`

`x(x+1)(x-4)<0`

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

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

`color(white)(aaaa)` `x+1` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x` `color(white)(aaaaaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-4` `color(white)(aaaaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `-` `color(white)(aaaa)` `+` `color(white)(aaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<0` when `x in (-oo,-1) uu (0,4)`

graph{x^3-3x^2-4x [-20, 20.55, -14.88, 5.4]}