1. The problem states that the second derivative of a function $f$ is given by $f''(x) = 6x$. 2. We want to find the original function $f(x)$ by integrating twice. 3. First, integrate $f''(x) = 6x$ to find the first derivative $f'(x)$: $$f'(x) = \int 6x \, dx = 3x^2 + C_1$$ where $C_1$ is the constant of integration. 4. Next, integrate $f'(x) = 3x^2 + C_1$ to find $f(x)$: $$f(x) = \int (3x^2 + C_1) \, dx = x^3 + C_1 x + C_2$$ where $C_2$ is another constant of integration. 5. Therefore, the general form of the function is: $$f(x) = x^3 + C_1 x + C_2$$ 6. Without initial conditions, $C_1$ and $C_2$ remain arbitrary constants. This completes the solution.