Q.$y = x(x - 3)^2$ decreases for the values of $x$ given by:
(A) $1 < x < 3$
(B) $x < 0$
(C) $x > 0$
(D) $0 < x < \dfrac{3}{2}$

Concept: Derivative Sign Analysis — a function decreases where its derivative is negative.

Step 1: Differentiate

$y = x(x-3)^2$

Using the product rule:

$y' = (x-3)^2 + x \cdot 2(x-3) = (x-3)[(x-3) + 2x] = (x-3)(3x-3)$

So $y' = 3(x-3)(x-1)$.

Step 2: Find critical points

Set $y' = 0$: $x = 1$ and $x = 3$.

Step 3: Sign analysis

• For $x < 1$: $(x-3)$ negative, $(x-1)$ negative → product positive → $y' > 0$ (increasing). …

The key to solving this is understanding that a function's increase or decrease is governed by the sign of its first derivative. If $y' > 0$, the function is increasing; if $y' < 0$, it is decreasing. So we need to find where $y'$ is negative.

Let’s work through it step by step.

1. Find the derivative.

   $y = x(x-3)^2$ is a product. Use the product rule: $y' = (1)(x-3)^2 + x \cdot 2(x-3)(1)$.

   Simplify:

   $y' = (x-3)^2 + 2x(x-3) = (x-3)[(x-3) + 2x] = (x-3)(3x-3) = 3(x-3)(x-1)$.

   So $y' = 3(x-1)(x-3)$.

2. Find the critical points.

   Set $y' = 0$: $3(x-1)(x-3) = 0$ gives $x = 1$ and $x = 3$. These are the points where the derivative changes sign.

3. Analyze the sign of $y'$ in each interval.

   The critical points divide the real line into three intervals: $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$.

   Pick a test point in each:

   • For $x < 1$, say $x = 0$: $y' = 3(0-1)(0-3) = 3(-1)(-3) = 9 > 0$. So $y$ is increasing here.
   • For $1 < x < 3$, say $x = 2$: $y' = 3(2-1)(2-3) = 3(1)(-1) = -3 < 0$. So $y$ is decreasing here.
   • For $x > 3$, say $x = 4$: $y' = 3(4-1)(4-3) = 3(3)(1) = 9 > 0$. So $y$ is increasing here.

   Therefore, $y$ decreases only on the interval $(1, 3)$.

Method: Sign-Chart Method for Intervals of Increase or Decrease

Use this whenever you are asked to find the interval(s) on which a polynomial function is increasing or decreasing.

Steps

Step 1: Differentiate using the correct rule.

For a product of factors, apply the product rule: for $y=uv$, $y'=u'v+uv'$. Expand and simplify afterward.

Step 2: Factor the derivative completely and solve $y'=0$.

Write $y'$ as a product of simple linear (or otherwise sign-determinable) factors and find every value of $x$ that makes it zero — these are the critical points.

Step 3: Split the number line at the critical points and test one value in each interval. …

Common Mistakes

Mistake 1: Forgetting the product rule and differentiating only one factor.

Why it's wrong: treating $(x-3)^2$ as a constant or ignoring the leading $x$ gives a wrong derivative entirely. Correct approach: always identify both factors of the product ($x$ and $(x-3)^2$) and apply $y'=u'v+uv'$ carefully, using the chain rule on $(x-3)^2$.

Mistake 2: Testing the sign of $y'$ at the critical points themselves instead of between them.

Why it's wrong: $y'=0$ exactly at the critical points, which tells you nothing about the sign on either side. Correct approach: always pick a test value strictly inside each sub-interval created by the critical points, never the critical point itself. …