1. **State the problem:** Determine if the function $$f(x) = \frac{x^3 - 2x^2 - 49x + 98}{x - 7}$$ is discontinuous at $$x=7$$. 2. **Recall the rule for continuity:** A function is continuous at $$x=a$$ if $$\lim_{x \to a} f(x) = f(a)$$ and $$f(a)$$ is defined. 3. **Check if $$f(7)$$ is defined:** The denominator $$x-7$$ is zero at $$x=7$$, so $$f(7)$$ is undefined. 4. **Simplify the numerator:** Factor the numerator to see if the factor $$x-7$$ cancels. The numerator is $$x^3 - 2x^2 - 49x + 98$$. Use polynomial division or factor by grouping: Group terms: $$(x^3 - 2x^2) + (-49x + 98)$$ Factor each group: $$x^2(x - 2) - 49(x - 2)$$ Factor out $$(x - 2)$$: $$(x - 2)(x^2 - 49)$$ Note $$x^2 - 49 = (x - 7)(x + 7)$$. So numerator factors as $$ (x - 2)(x - 7)(x + 7)$$. 5. **Rewrite the function:** $$f(x) = \frac{(x - 2)(x - 7)(x + 7)}{x - 7}$$ 6. **Cancel the common factor:** $$f(x) = \frac{(x - 2)\cancel{(x - 7)}(x + 7)}{\cancel{(x - 7)}} = (x - 2)(x + 7)$$ for $$x \neq 7$$. 7. **Evaluate the limit:** $$\lim_{x \to 7} f(x) = \lim_{x \to 7} (x - 2)(x + 7) = (7 - 2)(7 + 7) = 5 \times 14 = 70$$. 8. **Conclusion:** Since $$f(7)$$ is undefined but the limit exists and equals 70, the function has a removable discontinuity at $$x=7$$. If we define $$f(7) = 70$$, the function becomes continuous at $$x=7$$. **Final answer:** The function is discontinuous at $$x=7$$ due to a removable discontinuity.