QPaperGen

We start with:
\[
y = [x(x - 2)]^2 = \left(x^2 - 2x\right)^2
\]

Differentiate using chain rule:
\[
\frac{dy}{dx} = 2(x^2 - 2x) \cdot \frac{d}{dx}(x^2 - 2x) = 2(x^2 - 2x)(2x - 2)
\]
\[
\Rightarrow \frac{dy}{dx} = 4x(x - 2)(x - 1)
\]

Now set the derivative equal to zero:
\[
\frac{dy}{dx} = 0 \Rightarrow x = 0,\ 1,\ 2
\]

These critical points divide the real number line into the intervals:
\[
(-\infty, 0),\ (0, 1),\ (1, 2),\ (2, \infty)
\]

Now test the sign of \(\frac{dy}{dx}\) in each interval:
\[\]
In \((- \infty, 0)\), choose \(x = -1\):
\[
\frac{dy}{dx} = 4(-1)(-3)(-2) = -24 < 0
\]
So, \(y\) is decreasing.
\[\]

In \((0, 1)\), choose \(x = 0.5\):
\[
\frac{dy}{dx} = 4(0.5)(-1.5)(-0.5) = 1.5 > 0
\]
So, \(y\) is increasing.
\[\]

In \((1, 2)\), choose \(x = 1.5\):
\[
\frac{dy}{dx} = 4(1.5)(-0.5)(0.5) = -1.5 < 0
\]
So, \(y\) is decreasing.
\[\]

In \((2, \infty)\), choose \(x = 3\):
\[
\frac{dy}{dx} = 4(3)(1)(2) = 24 > 0
\]
So, \(y\) is increasing.
\[\]

\(\textbf{Conclusion:}\)
The function \( y = [x(x - 2)]^2 \) is increasing in the intervals:
\[
(0, 1) \quad \text{and} \quad (2, \infty)
\]