1. **State the problem:** We are given the second derivative of a function $f''(x) = 3x^2 - 2$ and the condition $f'(-1) = 0$. We need to find the original function $f(x)$. 2. **Recall the formulas:** The second derivative $f''(x)$ is the derivative of the first derivative $f'(x)$, and $f'(x)$ is the derivative of $f(x)$. So we integrate $f''(x)$ to find $f'(x)$, then integrate $f'(x)$ to find $f(x)$. 3. **Integrate $f''(x)$ to find $f'(x)$:** $$f'(x) = \int (3x^2 - 2) \, dx = \int 3x^2 \, dx - \int 2 \, dx = x^3 - 2x + C_1$$ where $C_1$ is a constant of integration. 4. **Use the condition $f'(-1) = 0$ to find $C_1$:** $$f'(-1) = (-1)^3 - 2(-1) + C_1 = -1 + 2 + C_1 = 1 + C_1 = 0$$ So, $$C_1 = -1$$ Thus, $$f'(x) = x^3 - 2x - 1$$ 5. **Integrate $f'(x)$ to find $f(x)$:** $$f(x) = \int (x^3 - 2x - 1) \, dx = \int x^3 \, dx - \int 2x \, dx - \int 1 \, dx = \frac{x^4}{4} - x^2 - x + C_2$$ where $C_2$ is another constant of integration. 6. **Final answer:** The function $f(x)$ is $$f(x) = \frac{x^4}{4} - x^2 - x + C_2$$ where $C_2$ can be any real number since no initial condition for $f(x)$ was given.