Chapter 6 · Application of Derivatives

Exercise 6.2 · Q19 (MCQ): Where is $x^2e^{-x}$ increasing?

Exercise 6.2Increasing and decreasing functions · Section 6.3

The question

19. The interval in which $y = x^2e^{-x}$ is increasing is

(A) $(-\infty, \infty)$
(B) $(-2, 0)$
(C) $(2, \infty)$
(D) $(0, 2)$

Exam answer

By the product rule,

\begin{aligned} y' &= 2xe^{-x} - x^2e^{-x} \\ &= x(2 - x)e^{-x} \end{aligned}

As $e^{-x} > 0$ ⁠, $y' > 0$ on $(0, 2)$ ⁠. And $y' < 0$ on $(-\infty, 0)$ and $(2, \infty)$ ⁠.

Answer: (D) $(0, 2)$

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The product rule gives $y' = 2xe^{-x} - x^2e^{-x} = x(2 - x)e^{-x}$ ⁠. The exponential factor is always positive, so the sign of $y'$ is the sign of $x(2 - x)$ ⁠: negative before 0, positive between 0 and 2, negative after 2. So $y$ rises only on $(0, 2)$ ⁠.

Given, to find, and the plan

Given: $y = x^2e^{-x}$ ⁠, defined for every real $x$ ⁠.
To find: Which of the four intervals $y$ is increasing on.
Plan: (1) Differentiate with the product and chain rules. (2) Factorise and find the zeros of $y'$ ⁠. (3) Sign chart. (4) Read off where $y$ increases and decreases (Theorem 1). (5) Examine every option, with a counterexample for each wrong one.

Basics you need

1Increasing (Definition 1, Section 6.3)

$y$ is increasing on I if $x_1 \lt x_2$ in I gives $y(x_1) \le y(x_2)$ ⁠. One pair in I with $y(x_1) \gt y(x_2)$ shows it is not. Lesson: increasing and decreasing functions

2Theorem 1 on an open interval

For any $x_1 \lt x_2$ in the interval, apply Theorem 1 on $[x_1, x_2]$ ⁠; $y' \gt 0$ throughout gives $y(x_1) \lt y(x_2)$ (Mean Value Theorem step of the proof). Lesson: increasing and decreasing functions

3Product rule and the derivative of $e^{-x}$

$(uv)' = u'v + uv'$ ⁠. By the chain rule (inner $-x$ ⁠, outer $e^u$ ⁠), $\frac{d}{dx}e^{-x} = e^{-x}\cdot(-1) = -e^{-x}$ ⁠. Basics: the chain rule

4 $e^{-x} \gt 0$ always

Every power of $e$ is positive, so $e^{-x}$ never changes the sign of a product. Basics: sign of a product

Solution

Differentiate with the product rule

Take $u = x^2$ and $v = e^{-x}$ ⁠. Then $u' = 2x$ and, by the chain rule, $v' = -e^{-x}$ ⁠. So
$\begin{aligned} y' &= u'v + uv'\\ &= 2x\,e^{-x} + x^2\left(-e^{-x}\right)\\ &= 2xe^{-x} - x^2e^{-x}. \end{aligned}$
$y$ is a product of two functions of $x$ ⁠, so the product rule is the tool. The minus sign in $v'$ comes from differentiating the inner function $-x$ ⁠.

Product rule (Class XI, Limits and Derivatives); $\frac{d}{dx}e^x = e^x$ (Theorem 5 (1), Section 5.4); chain rule (Theorem 4, Section 5.3.1). Basics: the chain rule
Factorise and find the zeros
$y' = xe^{-x}(2 - x) = x(2 - x)e^{-x}.$
Since $e^{-x} \gt 0$ for every $x$ ⁠, $y' = 0$ exactly when $x = 0$ or $2 - x = 0$ ⁠, that is, at $x = 0$ and $x = 2$ ⁠. These two points split $\R$ into $(-\infty, 0)$ ⁠, $(0, 2)$ and $(2, \infty)$ ⁠.

Factorising turns the sign question into the signs of three simple factors. The exponential factor never vanishes, so it contributes no zeros.

A product is zero exactly when one factor is zero. Basics: factorising and the discriminant
Sign chart of $y'$
- $x \lt 0$ ⁠: $x \lt 0$ ⁠, $2 - x \gt 0$ ⁠, $e^{-x} \gt 0$ ⁠, so $y' \lt 0$ ⁠.
- $0 \lt x \lt 2$ ⁠: $x \gt 0$ ⁠, $2 - x \gt 0$ ⁠, $e^{-x} \gt 0$ ⁠, so $y' \gt 0$ ⁠.
- $x \gt 2$ ⁠: $x \gt 0$ ⁠, $2 - x \lt 0$ ⁠, $e^{-x} \gt 0$ ⁠, so $y' \lt 0$ ⁠.
Sign chart of $y' = x(2 - x)e^{-x}$ (not to scale). Only on $(0, 2)$ are all three factors positive.

On each piece every factor keeps one sign, because each factor can only change sign where it is zero, and those points are the dividers $0$ and $2$ ⁠.

The sign of a product is decided by the number of negative factors. Basics: sign of a product
Where $y$ increases and decreases

Applying Theorem 1 on $[x_1, x_2]$ for any $x_1 \lt x_2$ inside one piece: $y$ is (strictly) increasing on $(0, 2)$ , and strictly decreasing on $(-\infty, 0)$ and on $(2, \infty)$ ⁠. Since $y$ is continuous at $0$ and $2$ ⁠, these extend to $[0, 2]$ ⁠, $(-\infty, 0]$ and $[2, \infty)$ ⁠.

$y = x^2e^{-x}$ for $-0.8 \le x \le 7$ (the axes use different scales). It falls to $(0, 0)$ ⁠, rises on the shaded band $0 \le x \le 2$ to $\left(2, \frac{4}{e^2}\right) \approx (2, 0.5413)$ ⁠, and then falls again towards $0$ ⁠.

The increasing piece is bounded by the two zeros of $y'$ ⁠: the lowest point $(0, 0)$ and the hump at $x = 2$ ⁠.

Theorem 1, Section 6.3. Lesson: increasing and decreasing functions

Examine every option

Option	Sign of $y'$	Verdict
(A)	changes sign at $0$ and $2$	Wrong. $-1 \lt 0$ but $y(-1) = e \approx 2.71828 \gt 0 = y(0)$ ⁠.
(B)	$y' \lt 0$ on $(-2, 0)$	Wrong: $y$ is decreasing there. $-1 \lt -\frac12$ but $y(-1) \approx 2.71828$ is greater than $y\left(-\frac12\right) = \frac{\sqrt e}{4} \approx 0.41218$ ⁠.
(C)	$y' \lt 0$ on $(2, \infty)$	Wrong: $y$ is decreasing there. $3 \lt 4$ but $y(3) = 9e^{-3} \approx 0.44808$ is greater than $y(4) = 16e^{-4} \approx 0.29305$ ⁠.
(D)	$y' \gt 0$ on $(0, 2)$	Correct. $y$ is increasing on $(0, 2)$ (Step 4).

(B) and (C) are the two intervals where

y

decreases; they are what you get by misreading the sign chart. (A) fails because

y

decreases on both outer pieces, so a pair taken from either of them breaks "increasing".

Result

The correct answer is (D) $(0, 2)$ ⁠. $y' = x(2 - x)e^{-x}$ is positive exactly on $(0, 2)$ ⁠, so $y = x^2e^{-x}$ is increasing there (and on $[0, 2]$ ⁠), and decreasing on $(-\infty, 0]$ and $[2, \infty)$ ⁠. (Textbook answer: D.)

Check it with numbers

Inside $(0, 2)$ ⁠: $y(0.5) \approx 0.151633$ ⁠, $y(1) = e^{-1} \approx 0.367879$ ⁠, $y(1.5) \approx 0.502043$ ⁠, $y(2) = 4e^{-2} \approx 0.541341$ ⁠: rising. ✓
Derivative: $y'(1) = 1\cdot1\cdot e^{-1} \approx 0.367879 \gt 0$ ⁠, $y'(-1) = (-1)(3)e \approx -8.154845 \lt 0$ ⁠, $y'(3) = 3(-1)e^{-3} \approx -0.149361 \lt 0$ ⁠. ✓

Common mistakes

$\frac{d}{dx}e^{-x} = e^{-x}$ ⁠. The chain rule gives $-e^{-x}$ ⁠; without the minus sign you get $y' = (2x + x^2)e^{-x}$ and the wrong intervals.
Forgetting the factor $x$ when factorising, which loses the zero at $x = 0$ ⁠.
Worrying about the sign of $e^{-x}$ ⁠. It is positive for every $x$ ⁠, negative $x$ included.

What to take away

For $y = (\text{polynomial})\times e^{kx}$ ⁠, factor out the exponential: it is positive, so only the polynomial part decides the sign of $y'$ ⁠.
Draw the sign chart factor by factor; each factor changes sign only at its own zero.
In an MCQ about intervals, test each wrong option with an explicit pair of points.

Every fact used, and where it comes from

Fact	Used in	Source
Increasing and decreasing on an interval	Steps 4, 5	Definition 1, Section 6.3 (Lesson: increasing and decreasing functions)
Sign of $y'$ decides increasing or decreasing	Step 4	Theorem 1, Section 6.3, and the Mean Value Theorem step of its proof
Product rule	Step 1	Class XI, Limits and Derivatives
$\frac{d}{dx}e^x = e^x$	Step 1	Theorem 5 (1), Section 5.4
$e^{-x} \gt 0$ for every $x$ (the range of an exponential function is the positive reals)	Steps 2, 3	Feature (2) after Definition 3, Section 5.4
Chain rule	Step 1	Theorem 4, Section 5.3.1
Interval method with a sign table	Step 3	Example 11, Section 6.3

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 6.2 · Q19 (MCQ): Where is x2e−xx^2e^{-x}x2e−x increasing?

Basics you need

Solution

Differentiate with the product rule

Factorise and find the zeros

Sign chart of y′y'y′

Where yyy increases and decreases

Examine every option

Every fact used, and where it comes from

Exercise 6.2 · Q19 (MCQ): Where is $x^2e^{-x}$ increasing?

Sign chart of $y'$

Where $y$ increases and decreases