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

A function increases where its derivative is positive.

Step 1 — Differentiate. With $y=[x(x-2)]^2=(x^2-2x)^2$,

Step 2 — Critical points: $x=0,\,1,\,2$. Test the sign of $y'=4x(x-1)(x-2)$ on each interval:

• $x<0:\ (-)(-)(-)=-$ → decreasing …

The idea

A differentiable function is increasing on any interval where its derivative is positive. So we differentiate, find where $y'=0$ (the critical points), and read off the sign of $y'$ between them.

Set up

Work the steps

1. Differentiate (chain rule with $u=x^2-2x$, $y=u^2$):

Factor: $x^2-2x=x(x-2)$ and $2x-2=2(x-1)$, so

2. Critical points where $y'=0$: $x=0,\ 1,\ 2$. These split the number line into four intervals.
3. Sign chart of $y'=4x(x-1)(x-2)$:

Method: Finding Increasing Intervals for a Composite (Chain-Rule) Polynomial

When the function is written as something raised to a power, like $[g(x)]^n$, differentiate with the chain rule first — this typically produces an extra factor from $g(x)$ itself, giving you more critical points than you'd expect from just looking at the outer power.

Steps

Step 1: Differentiate using the chain rule

For $y = [g(x)]^n$, $y' = n[g(x)]^{n-1} \cdot g'(x)$. Expand $g(x)$ first if it will make the subsequent differentiation of $g'(x)$ easier.

Step 2: Factor $y'$ completely into its simplest pieces

Combine the chain-rule factor with the derivative of the inner function, and factor everything down to individual linear terms — don't leave any part as an unfactored quadratic or higher expression.

Step 3: List every critical point from every factor

Set each linear factor to zero; every one of these values is a genuine critical point of $y$ (even the ones that came from the inner function $g(x)$, not just the ones visible before differentiating), and together they split the domain into several sub-intervals.

Step 4: Build a sign table across all the critical points …

Common Mistakes

Mistake 1: Missing the factor of $x$ as a critical point

Why it's wrong: Since $y'=4x(x-1)(x-2)$, a student who only sets the original function's visible roots ($x=0$ and $x=2$, from $[x(x-2)]^2$) to zero, without actually factoring the derivative, can miss that $x=1$ is also a critical point contributed by the chain rule — leading to only two intervals tested instead of four. Correct approach: always factor the actual derivative $y'$ completely, not the original function, to find every critical point.

Mistake 2: Sign error in the four-interval sign chart …