To find the value of \( x \) for which the function \( y = x^3 - x \) has a minimum value, we will follow these steps: ### Step 1: Find the derivative of the function To locate the minimum or maximum points of a function, we first need to find its derivative and set it to zero. The derivative of \( y \) with respect to \( x \) is: \[ \frac{dy}{dx} = \frac{d}{dx}(x^3 - x) = 3x^2 - 1 \] ### Step 2: Set the derivative to zero Next, we set the derivative equal to zero to find the critical points: \[ 3x^2 - 1 = 0 \] ### Step 3: Solve for \( x \) Now, we solve for \( x \): \[ 3x^2 = 1 \\ x^2 = \frac{1}{3} \\ x = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3} \] This gives us two critical points: \( x = \frac{\sqrt{3}}{3} \) and \( x = -\frac{\sqrt{3}}{3} \). ### Step 4: Determine whether these points are minima or maxima To determine whether these critical points are minima or maxima, we can use the second derivative test. We first find the second derivative of \( y \): \[ \frac{d^2y}{dx^2} = \frac{d}{dx}(3x^2 - 1) = 6x \] Now we evaluate the second derivative at our critical points: 1. For \( x = \frac{\sqrt{3}}{3} \): \[ \frac{d^2y}{dx^2} = 6 \left(\frac{\sqrt{3}}{3}\right) = 2\sqrt{3} > 0 \] This indicates that \( x = \frac{\sqrt{3}}{3} \) is a local minimum. 2. For \( x = -\frac{\sqrt{3}}{3} \): \[ \frac{d^2y}{dx^2} = 6 \left(-\frac{\sqrt{3}}{3}\right) = -2\sqrt{3} < 0 \] This indicates that \( x = -\frac{\sqrt{3}}{3} \) is a local maximum. ### Conclusion Since we are looking for the minimum value of the function, the correct value of \( x \) that gives the minimum is: \[ x = \frac{\sqrt{3}}{3} \] ### Option Analysis Now, let's analyze the options provided: - **Option A: \( -\sqrt{3} \)** - This is incorrect because it does not correspond to either critical point we found. - **Option B: \( -\frac{\sqrt{3}}{3} \)** - This is incorrect because it corresponds to a local maximum, not a minimum. - **Option C: \( \frac{\sqrt{3}}{3} \)** - This is correct as it corresponds to the local minimum we found. - **Option D: \( \sqrt{3} \)** - This is incorrect because it does not correspond to either critical point we found. ### Summary - The function \( y = x^3 - x \) has critical points at \( x = \pm \frac{\sqrt{3}}{3} \). - The second derivative test shows that \( x = \frac{\sqrt{3}}{3} \) is a local minimum. - The correct answer is \( x = \frac{\sqrt{3}}{3} \) (Option C). ### Revision Summary - Find the derivative to locate critical points. - Set the derivative to zero and solve for \( x \). - Use the second derivative test to determine if critical points are minima or maxima. - The minimum value occurs at \( x = \frac{\sqrt{3}}{3} \).