Rational Inequality 301Daf

1. **State the problem:** Solve the inequality $$\frac{x^2 - x - 6}{x + 1} \leq 0$$ and represent the solution on a number line and in union notation. 2. **Factor the numerator:** The quadratic $x^2 - x - 6$ factors as $$(x - 3)(x + 2)$$ 3. **Rewrite the inequality:** $$\frac{(x - 3)(x + 2)}{x + 1} \leq 0$$ 4. **Identify critical points:** These are values where numerator or denominator is zero: - $x = 3$ (numerator zero) - $x = -2$ (numerator zero) - $x = -1$ (denominator zero, undefined) 5. **Determine intervals:** The critical points divide the number line into four intervals: - $(-\infty, -2)$ - $(-2, -1)$ - $(-1, 3)$ - $(3, \infty)$ 6. **Test each interval:** Choose test points in each interval to check the sign of the expression. - For $x = -3$ in $(-\infty, -2)$: $$\frac{(-3 - 3)(-3 + 2)}{-3 + 1} = \frac{(-6)(-1)}{-2} = \frac{6}{-2} = -3 < 0$$ (True) - For $x = -1.5$ in $(-2, -1)$: $$\frac{(-1.5 - 3)(-1.5 + 2)}{-1.5 + 1} = \frac{(-4.5)(0.5)}{-0.5} = \frac{-2.25}{-0.5} = 4.5 > 0$$ (False) - For $x = 0$ in $(-1, 3)$: $$\frac{(0 - 3)(0 + 2)}{0 + 1} = \frac{(-3)(2)}{1} = -6 < 0$$ (True) - For $x = 4$ in $(3, \infty)$: $$\frac{(4 - 3)(4 + 2)}{4 + 1} = \frac{(1)(6)}{5} = \frac{6}{5} > 0$$ (False) 7. **Check equality points:** - At $x = 3$, numerator zero, denominator $3 + 1 = 4 \neq 0$, so expression equals zero (included). - At $x = -2$, numerator zero, denominator $-2 + 1 = -1 \neq 0$, expression equals zero (included). - At $x = -1$, denominator zero, expression undefined (excluded). 8. **Write solution set:** $$(-\infty, -2] \cup (-1, 3]$$ 9. **Number line representation:** - Shade from $-\infty$ to $-2$ including $-2$. - Open circle at $-1$ (excluded). - Shade from $-1$ to $3$ including $3$. This represents all $x$ values where the expression is less than or equal to zero.