1. **Problem 8: Find the removable point of discontinuity (RPOD) for** $f(x) = \frac{x+1}{x^2 - 1}$. 2. **Recall the formula and rules:** - The function is undefined where the denominator is zero. - Factor the denominator: $x^2 - 1 = (x-1)(x+1)$. - If a factor cancels with the numerator, the corresponding $x$-value is a removable discontinuity (RPOD). 3. **Factor numerator and denominator:** - Numerator: $x+1$ - Denominator: $(x-1)(x+1)$ 4. **Simplify the function:** $$f(x) = \frac{\cancel{x+1}}{(x-1)\cancel{(x+1)}} = \frac{1}{x-1}, \quad x \neq -1$$ 5. **Identify RPOD:** - The factor $x+1$ cancels, so $x = -1$ is a removable discontinuity. - To find the RPOD point, substitute $x = -1$ into the simplified function: $$f(-1) = \frac{1}{-1 - 1} = \frac{1}{-2} = -\frac{1}{2}$$ 6. **Answer for problem 8:** RPOD is at $(-1, -\frac{1}{2})$, which corresponds to option b). --- 7. **Problem 9: Find the x-intercept for** $f(x) = \frac{x+2}{x^2 - 4}$. 8. **Recall the formula and rules:** - The x-intercept occurs where $f(x) = 0$, i.e., numerator equals zero and denominator is not zero. - Factor denominator: $x^2 - 4 = (x-2)(x+2)$. 9. **Set numerator equal to zero:** $$x + 2 = 0 \implies x = -2$$ 10. **Check denominator at $x = -2$:** $$(-2)^2 - 4 = 4 - 4 = 0$$ - Denominator is zero, so $x = -2$ is not in the domain, no x-intercept here. 11. **Check other possible zeros:** - Numerator zero only at $x = -2$. - No other zeros, so no x-intercept. 12. **Answer for problem 9:** No x-intercept, option d).