1. **Problem Statement:** Given the function $$y = x^4 - x^3 + 4x - 1,$$ find the second derivative $$\frac{d^2y}{dx^2}$$ and determine the values of $$x$$ for which $$\frac{d^2y}{dx^2} = 0.$$\n\n2. **Step 1: Find the first derivative $$\frac{dy}{dx}$$.**\nUse the power rule: $$\frac{d}{dx} x^n = n x^{n-1}$$.\n$$\frac{dy}{dx} = \frac{d}{dx}(x^4) - \frac{d}{dx}(x^3) + \frac{d}{dx}(4x) - \frac{d}{dx}(1)$$\n$$= 4x^3 - 3x^2 + 4 - 0 = 4x^3 - 3x^2 + 4$$\n\n3. **Step 2: Find the second derivative $$\frac{d^2y}{dx^2}$$.**\nDifferentiate $$\frac{dy}{dx}$$ again using the power rule:\n$$\frac{d^2y}{dx^2} = \frac{d}{dx}(4x^3) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(4)$$\n$$= 12x^2 - 6x + 0 = 12x^2 - 6x$$\n\n4. **Step 3: Find values of $$x$$ such that $$\frac{d^2y}{dx^2} = 0$$.**\nSet the second derivative equal to zero:\n$$12x^2 - 6x = 0$$\nFactor out common terms:\n$$6x \left(\cancel{2x} - \cancel{1}\right) = 0$$\nIntermediate step with cancellation:\n$$\cancel{6}x \left(\cancel{2}x - 1\right) = 0$$\nThis gives two factors:\n$$6x = 0 \quad \Rightarrow \quad x = 0$$\n$$2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2}$$\n\n**Final answer:**\n$$\frac{d^2y}{dx^2} = 12x^2 - 6x$$\nThe values of $$x$$ for which $$\frac{d^2y}{dx^2} = 0$$ are $$x = 0$$ and $$x = \frac{1}{2}$$.