Find the value of x for which the function y = \(x^3 - x\) has a minimum value.
The correct answer is C. \(\sqrt{\frac{1}{3}}\)
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 = \(\sqrt{1/3}\).
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