Q.The interval in which $y = x^2 e^{-x}$ is increasing is (A) $(-\infty, \infty)$ (B) $(-2, 0)$ (C) $(2, \infty)$ (D) $(0, 2)$

Concept: Derivative Sign Analysis — a function increases where its derivative is positive.

Step 1: Differentiate

$y = x^2 e^{-x}$

Using product rule:

$y' = 2x e^{-x} + x^2 (-e^{-x}) = e^{-x}(2x - x^2) = e^{-x} x (2 - x)$

Step 2: Sign of $y'$

Since $e^{-x} > 0$ for all $x$, the sign of $y'$ depends only on $x(2 - x)$. …

Why derivative sign analysis works

A function is increasing on an interval if its slope — the derivative — is positive there. For $y = x^2 e^{-x}$, we need to find where $y' > 0$. The derivative is a product of factors; its sign depends on the sign of each factor. By factoring $y'$ completely, we can test intervals between its zeros and determine where it is positive.

Step-by-step solution

1. Compute the derivative using the product rule.

Let $u = x^2$ and $v = e^{-x}$. Then $u' = 2x$ and $v' = -e^{-x}$.

Factor out $x$:

2. Identify where the derivative is zero or undefined.

$e^{-x}$ is always positive (never zero). So $y' = 0$ when $x = 0$ or $x = 2$. The derivative is defined for all real $x$.

These two points split the real line into three intervals: $(-\infty, 0)$, $(0, 2)$, and $(2, \infty)$.

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

Pick a test point in each interval and evaluate the sign of each factor.

Method: Increasing/Decreasing Intervals for a Product of a Polynomial and an Exponential

Use this method whenever the function is a product like $x^n e^{-x}$ (or $x^n e^{x}$) — the product rule generates a derivative that factors into a polynomial part times the same exponential, and since the exponential is always positive, the polynomial factor alone controls the sign.

Steps

Step 1: Differentiate using the product rule.

For $y=x^2 e^{-x}$, with $u=x^2$ and $v=e^{-x}$: $y' = u'v+uv' = 2xe^{-x} - x^2e^{-x}$.

Step 2: Factor out the exponential and simplify the remaining polynomial factor.

The exponential factor $e^{-x}$ is positive for every real $x$, so it never affects the sign — it only scales the magnitude.

Step 3: Find the zeros of the remaining polynomial factor; these are the critical points.

Here $x(2-x)=0$ gives $x=0$ and $x=2$, splitting the real line into three intervals. …

Common Mistakes

Mistake 1: Dropping the negative sign when differentiating $e^{-x}$.

Why it's wrong: $\dfrac{d}{dx}e^{-x} = -e^{-x}$, not $e^{-x}$ — missing this sign flips the entire derivative and leads to the wrong increasing interval. Correct approach: always apply the chain rule carefully to $e^{-x}$, remembering the extra factor of $-1$ from the inner derivative of $-x$.

Mistake 2: Treating $e^{-x}$ as a factor that could be zero or negative and needs to be included in the sign analysis.

Why it's wrong: $e^{-x}>0$ for every real $x$, no exceptions — treating it as a factor that could change sign leads to an unnecessarily complicated (and potentially wrong) sign chart. Correct approach: recognize $e^{-x}$ as an always-positive scaling factor and focus the sign analysis entirely on the remaining polynomial part. …