1. **Problem Statement:** Show that the equation $$x^3 + 3x + 1 = 0$$ has exactly one real solution. 2. **Define the function:** Let $$f(x) = x^3 + 3x + 1$$. 3. **Find the derivative:** The derivative is $$f'(x) = 3x^2 + 3$$. 4. **Analyze the derivative:** Since $$3x^2 \geq 0$$ for all real $$x$$, we have $$f'(x) = 3x^2 + 3 \geq 3 > 0$$. This means $$f'(x)$$ is always positive and never zero. 5. **Implication of derivative sign:** Because $$f'(x) > 0$$ everywhere, $$f(x)$$ is strictly increasing on $$\mathbb{R}$$. 6. **Use Rolle's Theorem:** If there were two distinct real roots $$a$$ and $$b$$ with $$a < b$$ such that $$f(a) = f(b) = 0$$, then by Rolle's Theorem there must be some $$c \in (a,b)$$ where $$f'(c) = 0$$. But since $$f'(x) > 0$$ for all $$x$$, this is impossible. 7. **Conclusion on number of roots:** Therefore, $$f(x)$$ can have at most one real root. 8. **Existence of a root:** Evaluate $$f(-1) = (-1)^3 + 3(-1) + 1 = -1 - 3 + 1 = -3 < 0$$ and $$f(0) = 0 + 0 + 1 = 1 > 0$$. 9. **Intermediate Value Theorem:** Since $$f(x)$$ is continuous and changes sign between $$x = -1$$ and $$x = 0$$, there exists at least one root in $$(-1,0)$$. 10. **Final conclusion:** Combining the above, $$f(x) = 0$$ has exactly one real solution.