If f(x) = x^4 - 8x^2 + 16, what is the minimum value of f(x)? (2023)

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Question: If f(x) = x^4 - 8x^2 + 16, what is the minimum value of f(x)? (2023)

Options:

Correct Answer: 0

Exam Year: 2023

Solution:

Finding the derivative f\'(x) = 4x^3 - 16x. Setting it to zero gives x = 0, ±2. The minimum value occurs at x = 2, f(2) = 0.

If f(x) = x^4 - 8x^2 + 16, what is the minimum value of f(x)? (2023)

If f(x) = x^4 - 8x^2 + 16, what is the minimum value of f(x)? (2023)

Step 1: Write down the function f(x) = x^4 - 8x^2 + 16.
Step 2: Find the derivative of the function, which 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: This gives us x = 0, x = 2, and x = -2.
Step 6: To find the minimum value, evaluate f(x) at the critical points: f(0), f(2), and f(-2).
Step 7: Calculate f(0) = 16, f(2) = 0, and f(-2) = 0.
Step 8: Compare the values: f(0) = 16, f(2) = 0, f(-2) = 0.
Step 9: The minimum value of f(x) is 0, which occurs at x = 2 and x = -2.

Finding Minimum Values – The question tests the ability to find the minimum value of a polynomial function using calculus, specifically by finding critical points through derivatives.
Critical Points – Understanding how to set the derivative to zero to find critical points where the function may have minimum or maximum values.
Second Derivative Test – Although not explicitly required, knowing how to apply the second derivative test can confirm whether a critical point is a minimum or maximum.