18) when you factor the denominator, you get (x+3)((x-2). Since it doesn’t cancel out with any product in the numerator, the discontinuity would be non removable.
19) In this function, x can equal any value, since the denominator will never equal 0. So it’s continuous.
20) then for this one when you factor, you get 2(x-1)(x+1)/ (x+3)((x-1), so it has a removable at x=1 and a non removable one at x=-3.