Determine the local maxima and minima of the function f(x) = x^4 - 4x^3 + 4x.

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Q: Determine the local maxima and minima of the function f(x) = x^4 - 4x^3 + 4x.

Solution: f'(x) = 4x^3 - 12x^2 + 4. Setting f'(x) = 0 gives x = 0 and x = 2. f''(0) = 4 > 0 (min), f''(2) = -8 < 0 (max).

Steps: 7

Step 1: Write down the function f(x) = x^4 - 4x^3 + 4x.
Step 2: Find the first derivative f'(x) to determine where the function's slope is zero. The first derivative is f'(x) = 4x^3 - 12x^2 + 4.
Step 3: Set the first derivative equal to zero to find critical points: 4x^3 - 12x^2 + 4 = 0.
Step 4: Factor the equation or use the quadratic formula to solve for x. This gives us the critical points x = 0 and x = 2.
Step 5: Find the second derivative f''(x) to determine the concavity of the function. The second derivative is f''(x) = 12x^2 - 24x.
Step 6: Evaluate the second derivative at the critical points. First, calculate f''(0) = 12(0)^2 - 24(0) = 0. Then calculate f''(2) = 12(2)^2 - 24(2) = -8.
Step 7: Determine the nature of the critical points. Since f''(0) > 0, x = 0 is a local minimum. Since f''(2) < 0, x = 2 is a local maximum.