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

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

Solution: Finding the derivative f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. Evaluating f(0) = 16, f(2) = 0, and f(-2) = 0, the minimum value is 0.

Steps: 8

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 tells us the slope of the function. The derivative is f'(x) = 4x^3 - 16x.
Step 3: Set the derivative equal to zero to find the 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.
Step 6: Now we have three critical points: x = 0, x = 2, and x = -2.
Step 7: Evaluate the original function f(x) at each critical point: f(0) = 16, f(2) = 0, and f(-2) = 0.
Step 8: Compare the values obtained: 16, 0, and 0. The minimum value is the smallest of these, which is 0.

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

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