Rational Inequality B7Bb44

1. **State the problem:** Solve the inequality $$\frac{x-2}{x-8} > 0$$ and express the solution set in interval notation. 2. **Rewrite the inequality:** Define $$f(x) = \frac{x-2}{x-8}$$ and analyze where $$f(x) > 0$$. 3. **Find critical points:** The numerator $$x-2=0$$ at $$x=2$$ and the denominator $$x-8=0$$ at $$x=8$$ (where $$f(x)$$ is undefined). 4. **Determine intervals:** The critical points divide the real line into three intervals: - $$(-\infty, 2)$$ - $$(2, 8)$$ - $$(8, \infty)$$ 5. **Test signs on each interval:** - For $$x < 2$$, pick $$x=0$$: $$f(0) = \frac{0-2}{0-8} = \frac{-2}{-8} = \frac{1}{4} > 0$$ (positive) - For $$2 < x < 8$$, pick $$x=5$$: $$f(5) = \frac{5-2}{5-8} = \frac{3}{-3} = -1 < 0$$ (negative) - For $$x > 8$$, pick $$x=9$$: $$f(9) = \frac{9-2}{9-8} = \frac{7}{1} = 7 > 0$$ (positive) 6. **Sign chart summary:** | Interval | Sign of $$f(x)$$ | |----------------|-----------------| | $$(-\infty, 2)$$ | Positive | | $$(2, 8)$$ | Negative | | $$(8, \infty)$$ | Positive | 7. **Include critical points:** - At $$x=2$$, numerator zero, $$f(2)=0$$, inequality is strict $$>0$$, so exclude $$2$$. - At $$x=8$$, denominator zero, $$f(x)$$ undefined, exclude $$8$$. 8. **Solution set:** $$(-\infty, 2) \cup (8, \infty)$$ --- **Final answer:** The solution to $$\frac{x-2}{x-8} > 0$$ is $$(-\infty, 2) \cup (8, \infty)$$.