[tex]f(x)=\frac{(x-1)(x+2)(x+4)}{(x+1)(x-2)(x-4)}[/tex]
The denominator of a fraction can't be equal to 0.
[tex](x+1)(x-2)(x-4) \not= 0 \\
x+1 \not=0 \ \land \ x-2 \not= 0 \ \land \ x-4 \not= 0 \\
x \not= -1 \ \land \ x \not = 2 \ \land \ x \not= 4[/tex]
The function is undefined at x=-1, x=2, x=4, because for these values the denominator of the function would equal 0, and it's impossible to divide by 0.
Edward is correct.