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

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

Solution: f'(x) = 3x^2 - 3. Setting f'(x) = 0 gives x = ±1. f''(1) = 6 > 0 (min), f''(-1) = 6 > 0 (min). Local maxima at (0, 0).

Steps: 9

Step 1: Write down the function f(x) = x^3 - 3x.
Step 2: Find the first derivative f'(x) to determine where the function's slope is zero. The first derivative is f'(x) = 3x^2 - 3.
Step 3: Set the first derivative equal to zero to find critical points: 3x^2 - 3 = 0.
Step 4: Solve for x by factoring or using algebra: 3(x^2 - 1) = 0, which gives x^2 - 1 = 0, leading to x = ±1.
Step 5: Now, find the second derivative f''(x) to determine the concavity. The second derivative is f''(x) = 6x.
Step 6: Evaluate the second derivative at the critical points x = 1 and x = -1: f''(1) = 6(1) = 6 and f''(-1) = 6(-1) = -6.
Step 7: Determine the nature of the critical points: Since f''(1) > 0, x = 1 is a local minimum. Since f''(-1) < 0, x = -1 is a local maximum.
Step 8: Find the function values at the critical points: f(1) = 1^3 - 3(1) = -2 and f(-1) = (-1)^3 - 3(-1) = 2.
Step 9: Identify the local maxima and minima: Local maximum at (-1, 2) and local minimum at (1, -2).