1. The problem gives the function $$y = 2x^4 - x^2$$ and asks to find the value of $$x$$ for which the derivative $$y' = 0$$. 2. Find the derivative of $$y$$ with respect to $$x$$. Using the power rule, $$\frac{d}{dx} x^n = nx^{n-1}$$, we get: $$y' = \frac{d}{dx}(2x^4) - \frac{d}{dx}(x^2) = 2 \cdot 4x^{3} - 2x = 8x^{3} - 2x$$ 3. Set the derivative equal to zero to find critical points: $$8x^{3} - 2x = 0$$ 4. Factor out the common term $$2x$$: $$2x(4x^{2} - 1) = 0$$ 5. Set each factor equal to zero: $$2x = 0 \to x = 0$$ $$4x^{2} - 1 = 0 \to 4x^{2} = 1 \to x^{2} = \frac{1}{4} \to x = \pm \frac{1}{2}$$ 6. So the values of $$x$$ where $$y' = 0$$ are $$x = 0, \frac{1}{2}, -\frac{1}{2}$$. 7. Among the given options, the positive value $$\frac{1}{2}$$ corresponds to Option C. Final answer: $$x = \frac{1}{2}$$.