Rational Inequality Aeb540

1. **State the problem:** Solve the inequality $$\frac{x}{x - 6} > 0$$ and express the solution set in interval notation. 2. **Rewrite the inequality:** Move all terms to the left side and define $$f(x) = \frac{x}{x - 6}$$. 3. **Find critical points:** These are values where the numerator or denominator is zero. - Numerator zero: $$x = 0$$ - Denominator zero (undefined): $$x - 6 = 0 \Rightarrow x = 6$$ 4. **Determine intervals based on critical points:** The real line is divided into three intervals: - $$(-\infty, 0)$$ - $$(0, 6)$$ - $$(6, \infty)$$ 5. **Test the sign of $$f(x)$$ in each interval:** - For $$x < 0$$, pick $$x = -1$$: $$f(-1) = \frac{-1}{-1 - 6} = \frac{-1}{-7} = \frac{1}{7} > 0$$ (positive) - For $$0 < x < 6$$, pick $$x = 1$$: $$f(1) = \frac{1}{1 - 6} = \frac{1}{-5} = -\frac{1}{5} < 0$$ (negative) - For $$x > 6$$, pick $$x = 7$$: $$f(7) = \frac{7}{7 - 6} = \frac{7}{1} = 7 > 0$$ (positive) 6. **Check critical points:** - At $$x=0$$, numerator zero, $$f(0) = 0$$, but inequality is strict ($>0$), so exclude 0. - At $$x=6$$, denominator zero, undefined, exclude 6. 7. **Solution set:** Intervals where $$f(x) > 0$$ are $$(-\infty, 0)$$ and $$(6, \infty)$$. 8. **Express in interval notation:** $$(-\infty, 0) \cup (6, \infty)$$ 9. **Summary table:** | Interval | Sign of $$f(x)$$ | |----------------|-----------------| | $$(-\infty, 0)$$ | Positive | | $$(0, 6)$$ | Negative | | $$(6, \infty)$$ | Positive |