Determine the local maxima of f(x) = x^4 - 8x^2 + 16. (2021)

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Q: Determine the local maxima of f(x) = x^4 - 8x^2 + 16. (2021)

Solution: Find f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 12 is a local maximum.

Steps: 10

Step 1: Write down the function f(x) = x^4 - 8x^2 + 16.
Step 2: Find the derivative of the function, f'(x). 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: This gives us x = 0, x = 2, and x = -2.
Step 6: To determine if these points are local maxima or minima, we need to evaluate the second derivative, f''(x).
Step 7: Find the second derivative: f''(x) = 12x^2 - 16.
Step 8: Evaluate the second derivative at each critical point: f''(2) = 12(2^2) - 16 = 32 (positive, so x=2 is a local minimum), f''(0) = 12(0^2) - 16 = -16 (negative, so x=0 is a local maximum), f''(-2) = 12((-2)^2) - 16 = 32 (positive, so x=-2 is a local minimum).
Step 9: Calculate the function value at the local maximum: f(0) = 0^4 - 8(0^2) + 16 = 16.
Step 10: The local maximum is at x = 0 with a value of f(0) = 16.