What is the minimum value of the function f(x) = x^4 - 8x^2 + 16?

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Q: What is the minimum value of the function f(x) = x^4 - 8x^2 + 16?

Solution: f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x(x^2 - 4) = 0, so x = 0, 2, -2. f(0) = 16, f(2) = 0, f(-2) = 0. Minimum value is 0.

Steps: 9

Step 1: Write down the function we want to analyze: f(x) = x^4 - 8x^2 + 16.
Step 2: Find the derivative of the function, which helps us find the points where the function's slope is zero. The derivative is f'(x) = 4x^3 - 16x.
Step 3: Set the derivative equal to zero to find critical points: 4x^3 - 16x = 0.
Step 4: Factor the equation: 4x(x^2 - 4) = 0.
Step 5: Solve for x by setting each factor to zero: 4x = 0 gives x = 0, and x^2 - 4 = 0 gives x = 2 and x = -2.
Step 6: Now we have three critical points: x = 0, x = 2, and x = -2.
Step 7: Calculate the function value at each critical point: f(0) = 16, f(2) = 0, and f(-2) = 0.
Step 8: Compare the function values: 16 (at x = 0), 0 (at x = 2), and 0 (at x = -2).
Step 9: The minimum value of the function is the smallest of these values, which is 0.