1. **Problem 8: Find the R P O D (Removable Point of Discontinuity) for $f(x) = \frac{x+1}{x^2 - 1}$.** 2. The function is a rational function where the denominator is $x^2 - 1 = (x-1)(x+1)$. Points where the denominator is zero are potential discontinuities. 3. Set denominator equal to zero: $$x^2 - 1 = 0 \implies (x-1)(x+1) = 0 \implies x = 1 \text{ or } x = -1.$$ These are points where the function is undefined. 4. Check if numerator also zero at these points to identify removable discontinuities: - At $x = -1$: numerator $= (-1)+1 = 0$ - At $x = 1$: numerator $= 1+1 = 2 \neq 0$ 5. Since numerator and denominator both zero at $x = -1$, there is a removable discontinuity there. 6. To find the R P O D value, simplify the function by factoring and canceling common factors: $$f(x) = \frac{x+1}{(x-1)(x+1)} = \frac{\cancel{x+1}}{(x-1)\cancel{(x+1)}} = \frac{1}{x-1}, \quad x \neq -1.$$ 7. Evaluate the simplified function at $x = -1$ to find the R P O D value: $$f(-1) = \frac{1}{-1 - 1} = \frac{1}{-2} = -\frac{1}{2}.$$ 8. So the R P O D is at $(-1, -\frac{1}{2})$, which corresponds to option b). 9. **Problem 9: Find the x-intercept for $f(x) = \frac{x+2}{x^2 - 4}$.** 10. The x-intercept occurs where $f(x) = 0$, which means numerator equals zero (and denominator not zero). 11. Set numerator equal to zero: $$x + 2 = 0 \implies x = -2.$$ 12. Check denominator at $x = -2$: $$(-2)^2 - 4 = 4 - 4 = 0,$$ so denominator is zero, function undefined at $x = -2$. 13. Since denominator zero at $x = -2$, no x-intercept there. 14. Check if numerator zero anywhere else: no. 15. Therefore, there is no x-intercept, option d). **Final answers:** - Problem 8: b) $(-1, -\frac{1}{2})$ - Problem 9: d) none