For the function f(x) = 2x^3 - 9x^2 + 12x, find the intervals where the function is increasing.

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Q: For the function f(x) = 2x^3 - 9x^2 + 12x, find the intervals where the function is increasing.

Solution: f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 3. Testing intervals shows f is increasing on (1, 3).

Steps: 12

Step 1: Start with the function f(x) = 2x^3 - 9x^2 + 12x.
Step 2: Find the derivative of the function, which tells us the rate of change. The derivative is f'(x) = 6x^2 - 18x + 12.
Step 3: Set the derivative equal to zero to find critical points: 6x^2 - 18x + 12 = 0.
Step 4: Simplify the equation by dividing everything by 6: x^2 - 3x + 2 = 0.
Step 5: Factor the quadratic equation: (x - 1)(x - 2) = 0.
Step 6: Solve for x to find the critical points: x = 1 and x = 2.
Step 7: Determine the intervals to test: (-∞, 1), (1, 2), and (2, ∞).
Step 8: Choose a test point from each interval and plug it into the derivative f'(x) to see if it's positive (increasing) or negative (decreasing).
Step 9: For the interval (-∞, 1), test x = 0: f'(0) = 12 (positive, so increasing).
Step 10: For the interval (1, 2), test x = 1.5: f'(1.5) = -1.5 (negative, so decreasing).
Step 11: For the interval (2, ∞), test x = 3: f'(3) = 12 (positive, so increasing).
Step 12: Combine the results: The function is increasing on the intervals (-∞, 1) and (2, ∞).