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

1 Answer
Aug 25, 2017

Answer:

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]}