1. **Problem statement:** Evaluate the integral $$\int_{-1}^1 f(x) \, dx$$ where $$f(x) = e^{\sqrt[3]{x}} - 4x$$, given that $$\int_0^1 f(x) \, dx = 3e - 8$$ and $$\int_0^{-1} f(x) \, dx = \frac{15}{e} - 8$$. 2. **Part (a): Evaluate $$\int_{-1}^1 f(x) \, dx$$.** Recall the property of definite integrals: $$\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx$$. We can write: $$\int_{-1}^1 f(x) \, dx = \int_{-1}^0 f(x) \, dx + \int_0^1 f(x) \, dx$$. Given: $$\int_0^1 f(x) \, dx = 3e - 8$$ and $$\int_0^{-1} f(x) \, dx = \frac{15}{e} - 8$$. Using the property: $$\int_{-1}^0 f(x) \, dx = -\int_0^{-1} f(x) \, dx = -\left( \frac{15}{e} - 8 \right) = 8 - \frac{15}{e}$$. Therefore: $$\int_{-1}^1 f(x) \, dx = \left(8 - \frac{15}{e}\right) + (3e - 8) = 3e - \frac{15}{e}$$. 3. **Part (b): Show there exists $$c \in (-1,1)$$ such that $$f(c) = \frac{3e}{2} - \frac{15}{2e}$$.** By the Mean Value Theorem for integrals, if $$f$$ is continuous on $$[-1,1]$$, then there exists $$c \in (-1,1)$$ such that: $$f(c) = \frac{1}{1 - (-1)} \int_{-1}^1 f(x) \, dx = \frac{1}{2} \int_{-1}^1 f(x) \, dx$$. From part (a), we have: $$\int_{-1}^1 f(x) \, dx = 3e - \frac{15}{e}$$. Thus: $$f(c) = \frac{1}{2} \left(3e - \frac{15}{e}\right) = \frac{3e}{2} - \frac{15}{2e}$$. Since $$f$$ is continuous (as it is composed of exponential and root functions which are continuous on $$[-1,1]$$), the Mean Value Theorem guarantees such a $$c$$ exists. **Final answers:** $$\int_{-1}^1 f(x) \, dx = 3e - \frac{15}{e}$$ and there exists $$c \in (-1,1)$$ such that $$f(c) = \frac{3e}{2} - \frac{15}{2e}$$.