To find the relative minimum of the function f(x) = x^4 – 8x^2, we first need to compute its derivative and find the critical points.
The first derivative of f(x) is:
f'(x) = 4x^3 – 16x
We can factor this as follows:
f'(x) = 4x(x^2 – 4) = 4x(x – 2)(x + 2)
Setting the derivative equal to zero gives us the critical points:
4x(x – 2)(x + 2) = 0
This results in:
- x = 0
- x = 2
- x = -2
Next, we need to determine which of these points correspond to a relative minimum. We do this by using the second derivative test. The second derivative of f(x) is:
f”(x) = 12x^2 – 16
Now, we evaluate the second derivative at each critical point:
- f”(0) = 12(0)^2 – 16 = -16 (negative, so x = 0 is a local maximum)
- f”(2) = 12(2)^2 – 16 = 32 (positive, so x = 2 is a local minimum)
- f”(-2) = 12(-2)^2 – 16 = 32 (positive, so x = -2 is also a local minimum)
Thus, the function f(x) = x^4 – 8x^2 has relative minima at x = 2 and x = -2.