For the function f(x) = x^3 - 6x^2 + 9x, find the local minima. (2022)

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Q: For the function f(x) = x^3 - 6x^2 + 9x, find the local minima. (2022)

Solution: f'(x) = 3x^2 - 12x + 9. Setting f'(x) = 0 gives x = 1, 3. f(2) = 1 is a local minimum.

Steps: 11

Step 1: Start with the function f(x) = x^3 - 6x^2 + 9x.
Step 2: Find the derivative of the function, which is f'(x) = 3x^2 - 12x + 9.
Step 3: Set the 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: This gives us the critical points x = 1 and x = 3.
Step 7: To find if these points are local minima or maxima, we can use the second derivative test.
Step 8: Find the second derivative: f''(x) = 6x - 12.
Step 9: Evaluate the second derivative at the critical points: f''(1) = 6(1) - 12 = -6 (which is less than 0, so x = 1 is a local maximum) and f''(3) = 6(3) - 12 = 6 (which is greater than 0, so x = 3 is a local minimum).
Step 10: To find the value of the local minimum, calculate f(3): f(3) = 3^3 - 6(3^2) + 9(3) = 27 - 54 + 27 = 0.
Step 11: Therefore, the local minimum occurs at x = 3 with a value of f(3) = 0.