Difficulty: Intermediate | Prerequisites: Critical points and extrema (Doc 1 of this series), basic differentiation rules.
The first derivative tells you whether a function is increasing or decreasing. The second derivative tells you how the rate of change itself is changing, which is the idea behind concavity. This section also covers the second derivative test (a shortcut for classifying critical points) and optimisation, where you build an objective function from a word problem and find its maximum or minimum. Optimisation is one of the most heavily tested applied topics in brief calculus courses.
The second derivative measures concavity: f''(x) > 0 means concave up (bowl shape), f''(x) < 0 means concave down (arch shape). You can use it to classify critical points without a sign chart. Optimisation problems ask you to maximise or minimise a function subject to a constraint, and the procedure is: set up the equation, reduce to one variable, differentiate, solve.
Second derivative, f''(x)
The derivative of the derivative. It measures the rate of change of the slope. In simple terms, it tells you whether the curve is bending upward or downward.
Concave up
A function is concave up on an interval where f''(x) > 0. The curve bends upward like a bowl. Think of it as: if you drew a tangent line at any point in this interval, the curve would sit above the tangent.
Concave down
A function is concave down on an interval where f''(x) < 0. The curve bends downward like an arch. The curve sits below the tangent line.
Inflection point
A point where concavity changes from up to down or down to up. At an inflection point, f''(x) = 0 or f''(x) is undefined, and the sign of f'' changes.
Second derivative test
A method for classifying a critical point x = c (where f'(c) = 0): if f''(c) > 0, then c is a relative minimum. If f''(c) < 0, then c is a relative maximum. If f''(c) = 0, the test is inconclusive and you fall back to the first derivative test.
Objective function
The function you are trying to maximise or minimise in an optimisation problem. You build it from the problem's geometry or business scenario, then reduce it to one variable using a constraint equation.
Average cost function
The total cost divided by the number of units: C-bar(x) = C(x)/x. Minimising average cost is a common optimisation problem in business calculus.
Differentiate the first derivative again. The same rules apply (power, product, quotient, chain).
Worked example: f(u) = u / (u² + 5)
f'(u) = [(1)(u² + 5) − u(2u)] / (u² + 5)² = (5 − u²) / (u² + 5)²
f''(u) requires the quotient rule on f'(u). The result is f''(u) = (2u³ − 30u) / (u² + 5)³ = 2u(u² − 15) / (u² + 5)³.
Worked example: g(t) = e^(3t)
g'(t) = 3e^(3t)
g''(t) = 9e^(3t)
Worked example: d²/dx² [ln(x) + √x]
First derivative: 1/x + (1/2)x^(−1/2)
Second derivative: −1/x² − (1/4)x^(−3/2)
Where f''(x) > 0, the graph is concave up (slopes are increasing).
Where f''(x) < 0, the graph is concave down (slopes are decreasing).
Inflection points occur where f''(x) changes sign.
For graph-reading problems (like problem 31), you identify concavity visually: concave up sections look like the bottom of a cup, concave down sections look like the top of a hill.
Given a critical point x = c where f'(c) = 0:
f''(c) > 0: relative minimum (concave up, so the curve is bowl-shaped at c)
f''(c) < 0: relative maximum (concave down, so the curve is arch-shaped at c)
f''(c) = 0: inconclusive, revert to the first derivative test
The general procedure:
Read the problem and identify what quantity to maximise or minimise.
Write the objective function (the formula for that quantity).
If the objective function has two variables, use the constraint equation to eliminate one.
Differentiate, set equal to zero, and solve.
Verify it is a max or min (using the second derivative test or the context of the problem).
Worked example (problem 32): P(x) = −0.02x² + 400x − 13,000
P'(x) = −0.04x + 400. Set to zero: x = 10,000.
P''(x) = −0.04 < 0, confirming a maximum.
Cannon should produce 10,000 cameras per month.
Worked example (problem 16): C(q) = 2q³ − 33q² + 60q + 1500
C'(q) = 6q² − 66q + 60 = 6(q² − 11q + 10) = 6(q − 1)(q − 10). Critical points: q = 1 and q = 10.
C''(q) = 12q − 66. C''(1) = −54 < 0 (relative max). C''(10) = 54 > 0 (relative min).
The cost is minimised at q = 10 (that is, 1,000 units). Minimum cost: C(10) = 2(1000) − 33(100) + 60(10) + 1500 = 2000 − 3300 + 600 + 1500 = 800 (in hundred thousands, so $80,000,000).
Worked example (problem 33, average cost): C(x) = 0.5(0.09x² + 190) = 0.045x² + 95
Average cost: C-bar(x) = C(x)/x = 0.045x + 95/x
C-bar'(x) = 0.045 − 95/x². Set to zero: 95/x² = 0.045, x² = 95/0.045 ≈ 2111.1, x ≈ 46.
Approximately 46 units minimises average cost.
Second derivative test summary:
Condition | Conclusion |
|---|---|
f'(c) = 0 and f''(c) > 0 | Relative minimum at x = c |
f'(c) = 0 and f''(c) < 0 | Relative maximum at x = c |
f'(c) = 0 and f''(c) = 0 | Test inconclusive |
Profit, revenue, cost relationships:
P(x) = R(x) − C(x)
Marginal cost = C'(x), marginal revenue = R'(x), marginal profit = P'(x)
Average cost = C(x) / x
To minimise average cost: set the derivative of C(x)/x equal to zero and solve.
Optimisation is the bread and butter of business calculus. Companies use these methods to determine optimal production quantities, pricing strategies, and resource allocation. The average cost minimisation problem models the point at which economies of scale are balanced against rising marginal costs. In engineering, the same framework applies to minimising material waste or maximising structural strength.
Students often confuse concave up with increasing. A function can be concave up and still decreasing (it is just decreasing less steeply). Concavity describes the bending, not the direction.
When the second derivative test gives f''(c) = 0, students sometimes conclude "no extremum." The correct conclusion is "test inconclusive" and you need the first derivative test to decide.
In optimisation problems, students sometimes forget to check that their answer is actually a maximum (or minimum) rather than the other. Use the second derivative test or domain reasoning.
When minimising average cost, students sometimes minimise total cost instead. These are different questions with different answers.
⚠️ Problems 21 through 23 are straightforward second-derivative computation. Know the quotient rule and chain rule cold.
⚠️ Problem 31 (graph reading) tests whether you can identify increasing, decreasing, concave up, and concave down intervals from a picture.
⚠️ Problems 16 and 32 are applied optimisation. Write the objective function, differentiate, solve, and verify. Show your work.
⚠️ Problem 33 asks you to minimise average cost, not total cost. Read carefully.
⚠️ Problem 15 (population model with eᵗ) requires you to interpret the critical point in context, so include a sentence about what the numbers mean.
True or False: If f''(x) > 0, the function is increasing.
False. f''(x) > 0 means concave up, not increasing. Increasing is determined by f'(x) > 0.
Fill in the blank: If f'(c) = 0 and f''(c) < 0, then x = c is a relative ______.
Maximum.
True or False: The second derivative test can always classify a critical point.
False. If f''(c) = 0, the test is inconclusive.
Fill in the blank: Average cost is minimised by setting the derivative of ______ equal to zero.
C(x)/x (the average cost function).
True or False: A point where f''(x) = 0 is always an inflection point.
False. f'' must actually change sign at that point for it to be an inflection point.
Q: The population of a town is modelled by P(t) = eᵗ − 10t + 15 for t years since 1960. Find and interpret any critical points.
A: P'(t) = eᵗ − 10. Set to zero: eᵗ = 10, so t = ln 10 ≈ 2.303 years after 1960. P''(t) = eᵗ, and P''(ln 10) = 10 > 0, so this is a local minimum. The population bottomed out at about t ≈ 2.3 (roughly mid-1962) with P(ln 10) = 10 − 10 ln 10 + 15 ≈ 25 − 23.03 ≈ 1.97 thousand.
Q: Find the second derivative of g(t) = e^(3t).
A: g'(t) = 3e^(3t). g''(t) = 9e^(3t).
Q: A company's profit is P(x) = −0.02x² + 400x − 13,000. How many units maximise profit?
A: P'(x) = −0.04x + 400 = 0 gives x = 10,000. P''(x) = −0.04 < 0 confirms a maximum. Produce 10,000 units.
Q: C(x) = 0.5(0.09x² + 190). Find the production level that minimises average cost.
A: Average cost = (0.045x² + 95)/x = 0.045x + 95/x. Differentiate: 0.045 − 95/x² = 0, x² = 95/0.045 ≈ 2111, x ≈ 46 units.
Concavity and the second derivative test build directly on the critical-point material from the first set of notes. Optimisation problems often involve setting up an objective function that uses implicit differentiation or related rates (covered in the next set of notes). The profit, cost, and revenue relationships here also appear in the elasticity of demand extra-credit topic.
Second derivative, concavity, concave up, concave down, inflection point, second derivative test, optimisation, optimization, objective function, constraint equation, average cost, marginal cost, marginal revenue, profit maximisation, MAT 210, brief calculus, ASU, exam 3