Answer
$$(-∞, -1] ∪ [3,+∞)$$
Work Step by Step
Let $x^2 - 2x -3 \geq 0$ be a function $f$.
Find the $x$-intercepts by solving $x^2 - 2x -3=0$
Factor: $x^2 - 2x -3$ = $(x-3)(x+1)$
$x-3=0$ or $x+1=0$
$x = 3$ or $x = -1$
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:
$(-∞, -1)(-1,3)(3, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -1)$
Test Value = $-2$
$f=(x-3)(x+1)$
$f=(-2-3)(-2+1)$
$f=5$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -1)$.
$(-1,3)$
Test Value = $0$
$f=(x-3)(x+1)$
$f=(0-3)(0+1)$
$f=-3$
Conclusion: $f (x) < 0$ for all $x$ in $(-1,3)$.
$(3,+∞)$
Test Value = $4$
$f=(x-3)(x+1)$
$f=(4-3)(4+1)$
$f=5$
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 $f (x) = x^2 - 2x -3$. Based on the solution above, $f(x)\gt 0$ for all $x$ in $(-∞, -1)$ or $(3, +∞)$. However, because the inequality involves $\geq$ (greater than or equal to), we must also include the solutions of $x^2 - 2x -3 = 0$, namely $3$ and $-1$, in the solution set.
Thus, the solution set of the given inequality is:
$$(-∞, -1] ∪ [3,+∞)$$
