How do you solve #(x-1)(x-2)(x-3)>=0#?

1 Answer
Aug 22, 2016

Answer:

#x in [1, 2] uu [3, oo)#

Explanation:

#f(x) = (x-1)(x-2)(x-3) = x^3-6x^2+11x-6#

is a cubic with positive leading coefficient and zeros of multiplicity #1# at #x=1#, #x=2# and #x=3#.

As a result:

  • #f(x)# is continuous.

  • #f(x)# is positive for large positive values of #x#.

  • #f(x)# is negative for large negative values of #x#.

  • #f(x)# changes sign at each of its zeros.

Hence #f(x)# is non-negative in the intervals #[1, 2]# and #[3, oo)#

graph{x^3-6x^2+11x-6 [-3.397, 6.603, -2.24, 2.76]}