1. **State the problem:** Find the points of discontinuity of the function $$f(x) = \frac{5x - 5}{x^4 - 12x^3 + 36x^2}$$ and evaluate the one-sided limits at each discontinuity. 2. **Find the points of discontinuity:** Points of discontinuity occur where the denominator is zero. Solve: $$x^4 - 12x^3 + 36x^2 = 0$$ 3. **Factor the denominator:** $$x^2(x^2 - 12x + 36) = 0$$ 4. **Factor the quadratic:** $$x^2(x - 6)^2 = 0$$ 5. **Set each factor equal to zero:** - $$x^2 = 0 \implies x = 0$$ - $$(x - 6)^2 = 0 \implies x = 6$$ So, the points of discontinuity are at $$x = 0$$ and $$x = 6$$. 6. **Evaluate one-sided limits at $$x=0$$:** $$f(x) = \frac{5x - 5}{x^2 (x - 6)^2}$$ As $$x \to 0^-$$: - Numerator: $$5(0^-) - 5 = -5$$ (negative) - Denominator: $$x^2 (x - 6)^2$$ is positive since squares are positive Therefore, $$\lim_{x \to 0^-} f(x) = \frac{-5}{\text{positive}} = -\infty$$ As $$x \to 0^+$$: - Numerator: $$5(0^+) - 5 = -5$$ (negative) - Denominator: positive Therefore, $$\lim_{x \to 0^+} f(x) = -\infty$$ 7. **Evaluate one-sided limits at $$x=6$$:** As $$x \to 6^-$$: - Numerator: $$5(6^-) - 5 = 30 - 5 = 25$$ (positive) - Denominator: $$x^2 (x - 6)^2$$, near 6, $$x - 6$$ is small and negative, but squared, so positive Therefore, $$\lim_{x \to 6^-} f(x) = \frac{25}{\text{positive}} = +\infty$$ As $$x \to 6^+$$: - Numerator: $$5(6^+) - 5 = 25$$ (positive) - Denominator: positive Therefore, $$\lim_{x \to 6^+} f(x) = +\infty$$ **Final answers:** **a.** Value: $$x = a = 0$$ - $$\lim_{x \to 0^-} f(x) = -\infty$$ - $$\lim_{x \to 0^+} f(x) = -\infty$$ **b.** Value: $$x = b = 6$$ - $$\lim_{x \to 6^-} f(x) = +\infty$$ - $$\lim_{x \to 6^+} f(x) = +\infty$$ **c.** No other discontinuities, so enter DNE.