Answer
The inequality is valid for values less than 1 and values between 2 and 3 inclusive i.e. $(-\infty,1]\cap [2,3]$
Work Step by Step
First, we are going to find the x-intercepts by equating to zero:
$(x-1)(x-2)(x-3)=0$
$x_1=1$
$x_2=2$
$x_3=3$
These are the critical points. We are going to take four values: one less than 1; one between 1 and 2; one between 2 and 3; and one more than 3 to test in the original equation and check if the inequality is true or not:
First test with a value less than 1:
$(0-1)(0-2)(0-3)\leq0$
$(-1)(-2)(-3)\leq0$
$-6\leq0 \rightarrow \text{ TRUE}$
Second test with a value between 1 and 2:
$(1.5-1)(1.5-2)(1.5-3)\leq0$
$(0.5)(-0.5)(-1.5)\leq0$
$0.375\leq0 \rightarrow \text{ FALSE}$
Third test with a value between 2 and 3:
$(2.5-1)(2.5-2)(2.5-3)\leq0$
$(1.5)(0.5)(-0.5)\leq0$
$-0.375\leq0 \rightarrow \text{ TRUE}$
Fourth test with a value greater than 3:
$(4-1)(4-2)(4-3)\leq0$
$(3)(2)(1)\leq0$
$6\leq0 \rightarrow \text{ FALSE}$
These tests show that the inequality $(x-1)(x-2)(x-3)\leq0$ is valid for values less than 1 and values between 2 and 3 (including them) i.e. $(-\infty,1]\cap [2,3]$
