Final answer:
To find the relative extrema of the function f(x) = x^4 - 8x^2 + 16, we set the first derivative equal to zero and solve for x. We find critical points at x = 0 and x = ±2. Using the second derivative test, we determine that there is a relative maximum at x = 0 (value 16) and relative minima at x = ±2 (value 0).
Step-by-step explanation:
To find the x-values of all points where the function has any relative extrema, we look at the function f(x) = x⁴ - 8x² + 16. To find the extrema, we need to find the first derivative of the function and set it equal to zero. The first derivative of f(x) is f'(x) = 4x³ - 16x. Setting this equal to zero gives us the equation 4x³ - 16x = 0. Factoring out 4x, we get 4x(x² - 4) = 0, which gives us the solutions x = 0 and x = ±2. These are the x-values where the function could potentially have extrema. Next, we use the second derivative test to determine the nature of each critical point. The second derivative of f(x) is f"(x) = 12x² - 16. Plugging the critical points into the second derivative, we find that f"(0) is negative, indicating a relative maximum at x = 0, and f"(±2) is positive, indicating relative minima at x = ±2. Finally, we calculate the value of each extremum by plugging the x-values back into the original function: f(0) = 16 (maximum value), f(2) = 0 (minimum value), and f(-2) = 0 (minimum value).