Answer
$$[1,2]∪[3+∞)$$
Work Step by Step
Let $(x - 1)(x - 2)(x - 3) \geq 0$ be a function $f$.
Find the $x$-intercepts by solving $(x - 1)(x - 2)(x - 3)=0$
$x-1=0$ or $x-2=0$ or $(x-3)$
$x = 1$ or $x =2$ or $x=3$
These $x$-intercepts serve as boundary points that separate the number line into intervals.
The boundary points of this equation therefore divide the number line into three intervals:
$(-∞, 1)(1,2),(2,3),(3, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, 1)$
Test Value = $0$
$f=(x - 1)(x - 2)(x - 3)$
$f=(0 - 1)(0 - 2)(0- 3)$
$f=-6$
Conclusion: $f (x) < 0$ for all $x$ in $(-∞, 1)$.
$(1, 2)$
Test Value = $\frac{3}{2}$
$f=(x - 1)(x - 2)(x - 3)$
$f=(\frac{3}{2} - 1)(\frac{3}{2} - 2)(\frac{3}{2} - 3)$
$f=\frac{3}{8}$
Conclusion: $f (x) > 0$ for all $x$ in $(1, 2)$.
$(2, 3)$
Test Value = $\frac{5}{2}$
$f=(x - 1)(x - 2)(x - 3)$
$f=(\frac{5}{2} - 1)(\frac{5}{2} - 2)(\frac{5}{2} - 3)$
$f=-\frac{3}{8}$
Conclusion: $f (x) < 0$ for all $x$ in $(2,3)$.
$(3,+∞)$
Test Value = $4$
$f=(x - 1)(x - 2)(x - 3)$
$f=(4 - 1)(4 - 2)(4 - 3)$
$f=6$
Conclusion: $f (x) > 0$ for all $x$ in $(3,+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \geq 0$, where $f (x) = (x - 1)(x - 2)(x - 3)$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(1,2)$ or $(3+∞)$. However, because the inequality involves $\geq$ (greater than or equal to), we must also include the solutions of $(x - 1)(x - 2)(x - 3) = 0$, namely $1$, $2$ and $3$, in the solution set.
Thus, the solution set of the given inequality is:
$$[1,2]∪[3+∞)$$