We are given the function: \(y = x^3 - x\) To find the critical points, we set the derivative equal to zero: \(3x^2 - 1 = 0\) Solving this gives: \(x = \pm \frac{1}{\sqrt{3}}\) Using the second derivative test, we find that: \(\frac{d^2y}{dx^2} = 6x\) - For \( x = \frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} > 0 \) (minimum). - For \( x = -\frac{1}{\sqrt{3}} \), \( \frac{d^2y}{dx^2} < 0 \) (maximum). Thus, the function \( y = x^3 - x \) has a minimum value at: \(x = \frac{1}{\sqrt{3}} \quad \text{or approximately} \quad x \approx 0.577\)