Find the x-coordinate of the point where the function f(x) = 2x^3 - 9x^2 + 12x has a local maximum.

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Q: Find the x-coordinate of the point where the function f(x) = 2x^3 - 9x^2 + 12x has a local maximum.

Solution: f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f''(1) < 0 indicates a local maximum at x = 1.

Steps: 12

Step 1: Write down the function f(x) = 2x^3 - 9x^2 + 12x.
Step 2: Find the derivative of the function, which is f'(x). This tells us the slope of the function.
Step 3: Calculate the derivative: f'(x) = 6x^2 - 18x + 12.
Step 4: Set the derivative equal to zero to find critical points: 6x^2 - 18x + 12 = 0.
Step 5: Simplify the equation by dividing everything by 6: x^2 - 3x + 2 = 0.
Step 6: Factor the quadratic equation: (x - 1)(x - 2) = 0.
Step 7: Solve for x: This gives us x = 1 and x = 2 as critical points.
Step 8: To determine if these points are local maxima or minima, find the second derivative f''(x).
Step 9: Calculate the second derivative: f''(x) = 12x - 18.
Step 10: Evaluate the second derivative at x = 1: f''(1) = 12(1) - 18 = -6.
Step 11: Since f''(1) < 0, this indicates a local maximum at x = 1.
Step 12: Therefore, the x-coordinate of the point where the function has a local maximum is x = 1.