At which values of x does the function f(x) = x^4 – 8x^2 have a relative minimum?

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To determine the values of x at which the function f(x) = x^4 – 8x^2 has a relative minimum, we begin by finding the derivative of the function.

The first derivative, f'(x), gives us the critical points where the function can potentially have relative minima or maxima. Let’s calculate the derivative:

f'(x) = 4x^3 - 16x

Next, we’ll set the derivative equal to zero to find the critical points:

4x^3 - 16x = 0

Factoring out the common term:

4x(x^2 - 4) = 0

This gives us:

  • 4x = 0 ⟹ x = 0
  • x^2 – 4 = 0 ⟹ x = ±2

So, the critical points are x = -2, x = 0, and x = 2.

To determine which of these points is a relative minimum, we need to perform the second derivative test:

First, we calculate the second derivative:

f''(x) = 12x^2 - 16

Now evaluating the second derivative at each critical point:

  • For x = -2:
    • f''(-2) = 12(-2)^2 - 16 = 48 - 16 = 32 (positive)
  • This indicates a relative minimum at x = -2.
  • For x = 0:
    • f''(0) = 12(0)^2 - 16 = -16 (negative)
  • This indicates a relative maximum at x = 0.
  • For x = 2:
    • f''(2) = 12(2)^2 - 16 = 48 - 16 = 32 (positive)
  • This indicates a relative minimum at x = 2.

In conclusion, the function f(x) = x^4 – 8x^2 has relative minimum values at x = -2 and x = 2.

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