Determine the local maxima and minima of the function f(x) = x^3 - 6x^2 + 9x.

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Q: Determine the local maxima and minima of the function f(x) = x^3 - 6x^2 + 9x.

Solution: f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f''(1) > 0 (min), f''(3) < 0 (max).

Steps: 9

Step 1: Write down the function f(x) = x^3 - 6x^2 + 9x.
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 - 12x + 9.
Step 3: Set the first derivative equal to zero to find critical points: 3x^2 - 12x + 9 = 0.
Step 4: Simplify the equation by dividing everything by 3: x^2 - 4x + 3 = 0.
Step 5: Factor the quadratic equation: (x - 1)(x - 3) = 0.
Step 6: Solve for x to find critical points: x = 1 and x = 3.
Step 7: Find the second derivative f''(x) to determine the concavity. The second derivative is f''(x) = 6x - 12.
Step 8: Evaluate the second derivative at the critical points: f''(1) = 6(1) - 12 = -6 and f''(3) = 6(3) - 12 = 6.
Step 9: Determine the nature of the critical points: Since f''(1) < 0, x = 1 is a local maximum. Since f''(3) > 0, x = 3 is a local minimum.