To find the value of x for which the function y = \(x^3 - x\) has a minimum value, we can take the derivative of the function and set it equal to zero to find the critical points:

y' = 3x\(^2\) - 1

Setting y' equal to zero, we get:

3x\(^2\) - 1 = 0

Solving for x, we get:

3x\(^2\) = 1

x\(^2\) = 1/3

x = ± \(\sqrt{1/3}\)

So, the critical points are x = \(\sqrt{1/3}\) and x = -\(\sqrt{1/3}\).

To determine whether these critical points are minima, maxima, or neither, we can use the second derivative test. The second derivative of the function is:

y'' = 6x

At x = \(\sqrt{1/3}\), y'' is positive, which means that this critical point is a local minimum. At x = -\(\sqrt{1/3}\), y'' is negative, which means that this critical point is a local maximum.

So, the value of x for which the function y = \(x^3 - x\) has a minimum value is x = \(\sqrt{1/3}\).