Mathematics Advanced • Year 11 • Module 4 • Lesson 14
Optimisation with Exponentials
Build procedural fluency in differentiating exponential functions, factoring out the positive exponential, and finding and classifying stationary points.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete each derivative:
d/dx ( e^x ) = ____________
d/dx ( e^(−x) ) = ____________ (chain rule)
Q1.2 Why can we safely "factor out e^(−x) " when solving e^(−x) · g(x) = 0? (One sentence.)
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Q1.3 For a function f with f'(x₀) = 0, state the second-derivative test for maximum and for minimum:
If f''(x₀) ____________, then x₀ is a maximum.
If f''(x₀) ____________, then x₀ is a minimum.
2. Worked example — maximum of f(x) = x e^(−x) on x ≥ 0
Follow each line. Every step has a reason on the right.
Problem. Find the maximum value of f(x) = x e^(−x) for x ≥ 0.
Step 1 — Differentiate using the product rule.
f'(x) = 1 · e^(−x) + x · (−e^(−x)) = e^(−x) · (1 − x)
Reason: u = x, v = e^(−x); v' = −e^(−x) (chain rule).
Step 2 — Solve f'(x) = 0 by factoring out e^(−x).
e^(−x) > 0 for all x, so f'(x) = 0 ⇔ 1 − x = 0 ⇔ x = 1
Reason: the exponential factor is never zero.
Step 3 — Classify using sign of f'.
For 0 ≤ x < 1: 1 − x > 0, so f'(x) > 0 (increasing)
For x > 1: 1 − x < 0, so f'(x) < 0 (decreasing)
Reason: sign goes + then −, so x = 1 is a maximum.
Step 4 — Compute the maximum value.
f(1) = 1 · e^(−1) = 1/e ≈ 0.368
Conclusion. Maximum at x = 1 with value 1/e.
3. Faded example — minimum of f(x) = e^x + e^(−x)
Fill in the missing steps. 4 marks
Step 1 — Differentiate each term:
f'(x) = ____________ − ____________
Step 2 — Set f'(x) = 0 and combine into one exponential equation:
e^x − e^(−x) = 0 ⇒ e^x = ____________ ⇒ e^(2x) = ____________
⇒ 2x = ____________ ⇒ x = ____________
Step 3 — Compute the value at the stationary point:
f(0) = ________ + ________ = ____________
Step 4 — Classify using f''(x):
f''(x) = ____________ + ____________ which is always ____________ (sign?), so x = 0 is a ____________.
Conclusion. Minimum value of f is ____________ at x = 0.
4. Graduated practice — find and classify stationary points
For each function, find dy/dx, locate stationary points, classify them, and (where asked) state the optimum value. Show all working.
Foundation — direct rule (4 questions)
| Q | Function | dy/dx |
|---|---|---|
| 4.1 1 | y = e^(2x) | |
| 4.2 1 | y = e^(−3x) | |
| 4.3 1 | y = x e^x | |
| 4.4 1 | y = e^x − x |
Standard — typical HSC difficulty (6 questions)
For each, find the stationary x, evaluate f, and state if it is max or min.
4.5 Find the maximum of f(x) = 2x e^(−x) for x ≥ 0. 3 marks
4.6 Find the maximum of f(x) = x² e^(−x) for x ≥ 0. 3 marks
4.7 Find the minimum of f(x) = e^(2x) + e^(−2x). 3 marks
4.8 Find the maximum of f(x) = x / e^x. 3 marks
4.9 The profit function is P(x) = 100x · e^(−0.1 x). Find the value of x that maximises P. 3 marks
4.10 Find the minimum of f(x) = x + 1/x for x > 0 (no exponentials here — included as a contrast). 3 marks
Extension — combine ideas (2 questions)
4.11 Find the stationary point of y = x² e^(−x) and classify it. (You should find two stationary x-values; one is at the endpoint of the natural domain.) 4 marks
4.12 Show that for any a > 0, the function f(x) = x e^(−ax) on x ≥ 0 has its maximum at x = 1/a, with maximum value 1/(a e). 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Derivatives
d/dx ( e^x ) = e^x. d/dx ( e^(−x) ) = −e^(−x).
Q1.2 — Why factor e^(−x)
e^(−x) is positive for every real x, so it can never be zero; therefore e^(−x) · g(x) = 0 iff g(x) = 0.
Q1.3 — Second-derivative test
Maximum: f''(x₀) < 0. Minimum: f''(x₀) > 0.
Q3 — Faded example f(x) = e^x + e^(−x)
Step 1: f'(x) = e^x − e^(−x).
Step 2: e^x = e^(−x) ⇒ e^(2x) = 1 ⇒ 2x = 0 ⇒ x = 0.
Step 3: f(0) = 1 + 1 = 2.
Step 4: f''(x) = e^x + e^(−x), always positive, so x = 0 is a minimum.
Conclusion: minimum value = 2.
Q4.1 — y = e^(2x)
dy/dx = 2 e^(2x).
Q4.2 — y = e^(−3x)
dy/dx = −3 e^(−3x).
Q4.3 — y = x e^x
Product rule: dy/dx = 1 · e^x + x · e^x = e^x (1 + x).
Q4.4 — y = e^x − x
dy/dx = e^x − 1. (Stationary at x = 0; minimum value 1 since e^0 − 0 = 1.)
Q4.5 — Max of f(x) = 2x e^(−x), x ≥ 0
f'(x) = 2 e^(−x)(1 − x). e^(−x) > 0 ⇒ stationary at x = 1. Sign goes + then −, so maximum. f(1) = 2/e. Max value = 2/e ≈ 0.736.
Q4.6 — Max of f(x) = x² e^(−x), x ≥ 0
f'(x) = 2x e^(−x) + x²(−e^(−x)) = e^(−x) · x (2 − x). Stationary at x = 0 (endpoint) and x = 2. At x = 0: f = 0 (minimum). At x = 2: f = 4 e^(−2) ≈ 0.541. Sign of f' goes + then − around x = 2, so max at x = 2, value 4/e².
Q4.7 — Min of f(x) = e^(2x) + e^(−2x)
f'(x) = 2 e^(2x) − 2 e^(−2x) = 2 (e^(2x) − e^(−2x)). Stationary when e^(2x) = e^(−2x) ⇒ e^(4x) = 1 ⇒ x = 0. f(0) = 1 + 1 = 2. f''(x) = 4 e^(2x) + 4 e^(−2x) > 0, so minimum. Min value = 2 at x = 0.
Q4.8 — Max of f(x) = x/e^x = x e^(−x)
Same as Worked Example: max at x = 1, value 1/e ≈ 0.368.
Q4.9 — Max of P(x) = 100 x e^(−0.1 x)
P'(x) = 100 e^(−0.1 x)(1 − 0.1 x). e^(−0.1x) > 0 ⇒ stationary when 1 = 0.1 x ⇒ x = 10. Maximum value P(10) = 1000 e^(−1) = 1000/e ≈ 368.
Q4.10 — Min of f(x) = x + 1/x for x > 0
f'(x) = 1 − 1/x². Stationary when x² = 1, so x = 1 (positive only). f(1) = 2. f''(x) = 2/x³ > 0 for x > 0, so minimum. Min value = 2 at x = 1.
Q4.11 — Stationary points of y = x² e^(−x)
dy/dx = e^(−x) · x · (2 − x). Two stationary x-values: x = 0 and x = 2. At x = 0, f = 0 (minimum, sign goes − to +). At x = 2, f = 4/e² (maximum, sign goes + to −).
Q4.12 — Max of x e^(−ax), a > 0, x ≥ 0
f'(x) = e^(−ax) (1 − a x). e^(−ax) > 0 ⇒ stationary at x = 1/a. f(1/a) = (1/a) e^(−1) = 1/(a e). Sign goes + then − (since 1 − a x changes sign), so this is a maximum.