1. **State the problem:** We are given the limit $$\lim_{x \to 1} \frac{f^3(x) - 8}{2x^3 - x - 1} = 16$$ and the function $$y = f(x)$$. We want to find information about the function at $$x=1$$, such as the value of $$f(1)$$ and possibly the derivative $$f'(1)$$. 2. **Recall the limit definition and factorization:** The numerator is $$f^3(x) - 8$$ which is a difference of cubes. Recall the formula: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ Here, $$a = f(x)$$ and $$b = 2$$ because $$8 = 2^3$$. 3. **Apply the difference of cubes factorization:** $$f^3(x) - 8 = (f(x) - 2)(f^2(x) + 2f(x) + 4)$$ 4. **Analyze the denominator at $$x=1$$:** Calculate $$2(1)^3 - 1 - 1 = 2 - 1 - 1 = 0$$. 5. **Since both numerator and denominator approach 0 at $$x=1$$, this is an indeterminate form $$\frac{0}{0}$$, so the limit can be evaluated by factoring or using derivatives (L'Hôpital's Rule). But first, note that for the numerator to be zero at $$x=1$$, we must have: $$f^3(1) - 8 = 0 \implies f^3(1) = 8 \implies f(1) = 2$$. 6. **Rewrite the limit using the factorization:** $$\lim_{x \to 1} \frac{(f(x) - 2)(f^2(x) + 2f(x) + 4)}{2x^3 - x - 1} = 16$$ 7. **Since $$f^2(x) + 2f(x) + 4$$ is continuous at $$x=1$$, substitute $$f(1) = 2$$: $$f^2(1) + 2f(1) + 4 = 2^2 + 2 \times 2 + 4 = 4 + 4 + 4 = 12$$ 8. **Define $$g(x) = 2x^3 - x - 1$$ and note $$g(1) = 0$$. The limit becomes: $$\lim_{x \to 1} \frac{(f(x) - 2) \times 12}{g(x)} = 16$$ 9. **Rearranged:** $$\lim_{x \to 1} \frac{f(x) - 2}{g(x)} = \frac{16}{12} = \frac{4}{3}$$ 10. **Since $$f(1) = 2$$ and $$g(1) = 0$$, the limit of $$\frac{f(x) - f(1)}{g(x) - g(1)}$$ as $$x \to 1$$ is $$\frac{4}{3}$$. This is the definition of the derivative of $$f$$ with respect to $$g$$ at $$x=1$$: $$\frac{df}{dg}\bigg|_{x=1} = \frac{4}{3}$$ 11. **Find $$g'(x)$$:** $$g'(x) = 6x^2 - 1$$ At $$x=1$$: $$g'(1) = 6(1)^2 - 1 = 6 - 1 = 5$$ 12. **Use the chain rule:** $$\frac{df}{dx} = \frac{df}{dg} \times \frac{dg}{dx}$$ At $$x=1$$: $$f'(1) = \frac{4}{3} \times 5 = \frac{20}{3}$$ **Final answers:** - $$f(1) = 2$$ - $$f'(1) = \frac{20}{3}$$