Find the values of x for which y = [x(x-2)]² is an increasing function.
The function y = [x(x-2)]² is an increasing function for x in (0, 1) ∪ (2, ∞).
Explanation
To find where the function is increasing, we first differentiate y with respect to x. Using the chain rule, the derivative is found to be 4x(x-1)(x-2). A function is strictly increasing when its derivative is positive. We solve the inequality 4x(x-1)(x-2) > 0 by identifying critical points at x=0, x=1, and x=2. By testing the sign of the derivative in the intervals formed by these points, we determine that the derivative is positive in the intervals (0, 1) and (2, ∞).
Solution Steps
Step 1: Let y = [x(x-2)]² = (x² - 2x)².
Step 2: Differentiate with respect to x: dy/dx = 2(x² - 2x) * d/dx(x² - 2x) = 2(x² - 2x)(2x - 2).
Step 3: Simplify the derivative: dy/dx = 4(x² - 2x)(x - 1) = 4x(x - 2)(x - 1).
Step 4: For an increasing function, dy/dx > 0. So, 4x(x - 1)(x - 2) > 0, which implies x(x - 1)(x - 2) > 0.
Step 5: The critical points are x = 0, 1, 2. Testing intervals: the product is positive when x ∈ (0, 1) and when x ∈ (2, ∞).