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Module 4 · L14 of 15 ~40 min ⚡ +95 XP available

Optimisation with Exponentials

Exponential functions appear in optimisation when we want to maximise profit, minimise cost, or find the peak of a growth curve. The key insight: $e^x$ is never zero, so it always factors out cleanly when you differentiate.

Today's hook — The function $y = xe^{-x}$ starts at zero, climbs to a peak, then falls back toward zero. Where is the maximum? Without calculus you can only guess. With it, the answer is exact — and elegantly simple.

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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question

Recall — your gut answer first

+5 XP warm-up

A function multiplies $e^{-x}$ (which shrinks) by $x$ (which grows). Sketch a guess of where its maximum lies — early in $x$, late, or somewhere in between? Write your guess before reading on.

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The two moves

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Optimisation with exponentials uses the same process as any calculus optimisation — the key trick is factoring out the exponential to simplify your equation.

Move 1 — Differentiate carefully: Apply the product/quotient rule as needed. Then factor out $e^x$ or $e^{-x}$ since it is never zero.
Move 2 — Solve the remaining factor: After factoring, you only need to solve the polynomial factor for zero. Check endpoints too.

$$\frac{d}{dx}\!\left[xe^{-x}\right] = e^{-x}(1 - x)$$

$e^x \neq 0$ ever

When solving $e^x f(x) = 0$, only solve $f(x) = 0$.

Check domain

If the problem restricts $x \ge 0$, a stationary point at $x = -2$ is irrelevant.

Verify nature

Use the second derivative test or sign analysis to confirm max or min.

What you'll master

Know

Key facts

$e^x > 0$ for all real $x$
Factoring out $e^x$ simplifies equations
Standard optimisation process: differentiate, solve, verify

Understand

Concepts

Why exponentials factor out cleanly at stationary points
How product/quotient rules apply to exponential functions
The interplay between algebraic growth and exponential decay

Can do

Skills

Set up and solve exponential optimisation problems
Classify stationary points using sign or second derivative
Interpret optimal solutions in real-world contexts

Key terms

Stationary pointA point where $f'(x) = 0$; may be a local maximum, minimum, or point of inflection.

Second derivative testIf $f'(c)=0$ and $f''(c) < 0$, the point is a local maximum; if $f''(c) > 0$, a local minimum.

Product rule$(uv)' = u'v + uv'$. Essential for functions like $f(x) = xe^{-x}$.

Domain restrictionA constraint on $x$ (e.g. $x \ge 0$) that limits which stationary points are valid.

Optimisation strategy for exponentials

core concept

Optimisation with exponentials follows the same process as general calculus optimisation: define the objective function, differentiate, find stationary points, and verify their nature. The key advantage is that exponential derivatives remain exponential — so expressions like $e^x f(x) = 0$ are solved by setting $f(x) = 0$ only, since $e^x$ can never equal zero.

The factoring trick. When you differentiate $f(x) = xe^{-x}$ using the product rule, you get $f'(x) = e^{-x} - xe^{-x} = e^{-x}(1 - x)$. Since $e^{-x} > 0$ always, you only need $1 - x = 0$, giving $x = 1$. This pattern — factor out the exponential, solve the polynomial — appears in virtually every HSC exponential optimisation problem.

Strategy: differentiate → factor out $e^x$ (always positive) → solve remaining factor; $e^x \neq 0$ ever — this is the key that unlocks the solution

Pause — copy the optimisation strategy — differentiate, factor out $e^x$ (always positive, never zero), then set the remaining factor to zero — into your book.

Did you get this? True or false: when solving $e^{-x}(3 - x) = 0$, you can set $e^{-x} = 0$ to find one solution.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · MAXIMUM OF xe⁻ˣ

Find the maximum value of $f(x) = xe^{-x}$ for $x \ge 0$.

Product rule: $f'(x) = e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$

Differentiate $u = x$, $v = e^{-x}$ using $u'v + uv'$.

PROBLEM 2 · PROFIT MAXIMISATION

The profit function is $P(x) = 100xe^{-0.1x}$ where $x$ is the price. Find the price that maximises profit.

$P'(x) = 100e^{-0.1x} + 100x(-0.1e^{-0.1x}) = 100e^{-0.1x}(1 - 0.1x)$

Product rule, then factor out $100e^{-0.1x}$.

PROBLEM 3 · MINIMUM OF eˣ + e⁻ˣ

Find the minimum value of $f(x) = e^x + e^{-x}$.

$f'(x) = e^x - e^{-x}$

Differentiate each term using standard rules.

Quick check: For $f(x) = x^2 e^{-x}$, using the product rule $f'(x) =$

Common errors · the 3 traps that cost marks

Trap 01

Forgetting that $e^x$ is never zero

When solving $e^x f(x) = 0$, the factor $e^x$ can never be zero. Only solve $f(x) = 0$. Students who try to set $e^x = 0$ will find no solution from that branch — it's a dead end.

Trap 02

Confusing $e^{-x}$ with $-e^x$

$e^{-x} = \frac{1}{e^x}$ which is always positive. $-e^x$ is always negative. These are very different. When differentiating $e^{-x}$, the chain rule gives $-e^{-x}$ — but the original function $e^{-x}$ remains positive.

Trap 03

Not checking the stationary point is in the domain

If the problem restricts $x$ to $x \ge 0$, a stationary point at $x = -2$ is outside the domain and irrelevant. For restricted domains, also check the value at the boundary.

Think through the logic: To find the maximum of $f(x) = 3xe^{-2x}$ for $x \ge 0$, list the three steps you would follow.

Work mode · how are you completing this lesson?

Quick-fire practice · 5 problems

Find the maximum of $f(x) = x^2 e^{-x}$ for $x \ge 0$.

Find the minimum of $f(x) = e^{2x} + e^{-2x}$.

The cost function is $C(x) = 50 + 10e^{0.1x}$. Find the average cost $\frac{C(x)}{x}$ and its minimum for $x > 0$.

Find the maximum of $f(x) = \dfrac{x}{e^x}$.

Find the minimum of $f(x) = x + \dfrac{1}{x}$ for $x > 0$.

Match up: Drag (or mentally match) each function to its stationary point $x$-value.

$f(x) = xe^{-x}$
$f(x) = x^2e^{-x}$
$f(x) = xe^{-2x}$

$x = \tfrac{1}{2}$
$x = 2$
$x = 1$

Revisit your thinking

Earlier you guessed where $y = xe^{-x}$ peaks. Setting $\frac{dy}{dx} = 0$ gives $x = 1$ — exactly where the linear growth of $x$ and the exponential decay of $e^{-x}$ balance. Many optimisation problems with exponentials share this structure: $e^x$ never vanishes, so it factors out cleanly, leaving a polynomial equation to solve. The peak is always earlier than you might intuit, because the exponential decay dominates for large $x$.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 43 marks

Q1. Find the maximum value of $f(x) = 2xe^{-x}$ for $x \ge 0$. (3 marks)

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ApplyBand 43 marks

Q2. A rectangle has area 100 cm$^2$. One side is $x$ and the other is $\frac{100}{x}$. The perimeter is $P = 2x + \frac{200}{x}$. Find the value of $x$ that minimises the perimeter. (3 marks)

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AnalyseBand 54 marks

Q3. Find the coordinates of the stationary points of $y = x^2 e^{-x}$ and determine their nature. (4 marks)

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Comprehensive answers (click to reveal)

Drill 1: $f'(x) = xe^{-x}(2-x)$; stationary at $x=0$ (min) and $x=2$ (max); $f(2)=4e^{-2}$

Drill 2: $f'(x) = 2e^{2x} - 2e^{-2x} = 0 \Rightarrow x=0$; $f(0)=2$ (minimum)

Drill 4: $f(x) = xe^{-x}$; max at $x=1$, value $1/e$

Drill 5: $f'(x) = 1 - 1/x^2 = 0 \Rightarrow x=1$; min value $= 2$

Q1 (3 marks): $f'(x) = 2e^{-x}(1 - x)$ [1]; $e^{-x}>0$, so $x=1$ [0.5]; $f(1) = 2/e$ [1]; maximum value is $\frac{2}{e}$ [0.5]

Q2 (3 marks): $\frac{dP}{dx} = 2 - \frac{200}{x^2}$ [0.5]; set to zero: $x^2=100$, $x=10$ [1.5]; $P''= \frac{400}{x^3}>0$, confirming minimum [1]

Q3 (4 marks): $\frac{dy}{dx} = xe^{-x}(2-x)$ [1]; $x=0$ and $x=2$ [0.5]; $y(0)=0$, $y(2)=4e^{-2}$ [0.5]; $f''$: at $x=0$, $f''>0$ (min); at $x=2$, $f''<0$ (max) [2]

Boss battle · The Optimiser

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Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

Enter the arena

Science Jump · platform challenge

Climb platforms by answering optimisation questions.

Mark lesson as complete

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Module overview · Maths Advanced