# Chapter 8 - Section 8.5 - Polynomial and Rational Inequalities - Exercise Set - Page 649: 10

$$(-∞, -1] ∪ [3,+∞)$$ 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: $$(-∞, -1] ∪ [3,+∞)$$ 