Step 1: Understanding the Question:
A function is increasing where its derivative $f'(x) > 0$.
Step 3: Detailed Explanation:
$f(x) = (x^2 - 2x)^2$
$f'(x) = 2(x^2 - 2x) \cdot \frac{d}{dx}(x^2 - 2x)$
$f'(x) = 2(x(x-2))(2x-2) = 4x(x-2)(x-1)$
Find roots of $f'(x)$: $x = 0, 1, 2$.
Using sign scheme (Wavy curve method):
- For $x > 2$, $f'(x)$ is $(+)(+)(+) = +$. (Increasing)
- For $1 < x < 2$, $f'(x)$ is $(+)(-)(+) = -$. (Decreasing)
- For $0 < x < 1$, $f'(x)$ is $(+)(-)(-) = +$. (Increasing)
- For $x < 0$, $f'(x)$ is $(-)(-)(-) = -$. (Decreasing)
Increasing intervals: $(0, 1) \cup (2, \infty)$.
Step 4: Final Answer:
The function is increasing on $(0, 1) \cup (2, \infty)$.