Find the values of x for which y=[x(x-2)]^{2} | vedclass

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(A) We have,$y = [x(x-2)]^2 = (x^2 - 2x)^2$. To find the intervals where the function is increasing,we calculate the derivative: $\frac{dy}{dx} = 2(x^

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$x \in (0, 1) \cup (2, \infty)$

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(A) We have,$y = [x(x-2)]^2 = (x^2 - 2x)^2$. To find the intervals where the function is increasing,we calculate the derivative: $\frac{dy}{dx} = 2(x^2 - 2x)(2x - 2) = 4x(x-2)(x-1)$. Setting $\frac{dy}{dx} = 0$,we get the critical points: $x = 0, 1, 2$. These points divide the real line into intervals $(-\infty, 0), (0, 1), (1, 2), (2, \infty)$. Testing the sign of $\frac{dy}{dx}$ in each interval: $1$. For $x \in (-\infty, 0)$,$\frac{dy}{dx} $2$. For $x \in (0, 1)$,$\frac{dy}{dx} > 0$ (increasing). $3$. For $x \in (1, 2)$,$\frac{dy}{dx} $4$. For $x \in (2, \infty)$,$\frac{dy}{dx} > 0$ (increasing). Thus,the function is increasing for $x \in (0, 1) \cup (2, \infty)$.

Find the values of $x$ for which $y=[x(x-2)]^{2}$ is an increasing function.

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