Stationary Points, Concavity and Points of Inflection: The Second Derivative Test and Curve Sketching

In 30 seconds

A stationary point is where $ \frac{dy}{dx} = 0 $. It can be a local maximum, a local minimum or a stationary point of inflection.
The second derivative test: if $ f”(x) < 0 $ at a stationary point it is a maximum; if $ f”(x) > 0 $ it is a minimum; if $ f”(x) = 0 $ the test is inconclusive.
A point of inflection is where the curve changes concavity, from bending up to bending down or the reverse. There, $ f”(x) = 0 $ and changes sign.
In our case, a new app’s user curve has its inflection at month 10 with 2,000 users: growth is fastest there, at 300 new users a month, and slows after.
Exam point: $ f”(x) = 0 $ alone does not prove a point of inflection. You must show that $ f” $ changes sign either side.

The case: when does an app’s growth start to slow?

A start-up launches a revision app for a school district. The marketing team tracks total users each month. In the early months, growth speeds up as students tell their friends. Later, as most of the market has signed up, growth slows down and the total levels off. Plotted over time, the total looks like a stretched letter S.

The team fits a simple model to the first year of data. Let $ t $ be months since launch and $ N(t) $ the total number of users. For the first 20 months the model is

\[ N(t) = 30t^2 – t^3, \qquad 0 \le t \le 20 \]

At month 5 there are 625 users. At month 10 there are 2,000. At month 15 there are 3,375. By month 20 the total reaches 4,000 and stops rising. The team has three questions. When is growth fastest? When does the curve change from “accelerating” to “slowing down”? And where does the total peak? The first two questions are about the shape of the curve, not just its highest point. They need the second derivative. (The figures are illustrative.)

The answer matters for the business. The month when growth peaks is the point where marketing money buys the most new users. After it, every extra user is harder to win. Analysts call it the tipping point of an adoption curve.

The concept: what the first and second derivatives tell you

The first derivative, $ f'(x) $, is the gradient. It tells you whether the curve is going up ($ f’ > 0 $) or down ($ f’ < 0 $). Where $ f'(x) = 0 $ the tangent is horizontal. These points are called stationary points, because the function is momentarily neither rising nor falling.

The second derivative, $ f”(x) $, is the rate of change of the gradient. It tells you how the curve bends. If $ f” > 0 $, the gradient is increasing and the curve bends upwards like a cup. We call it concave up (some books say convex). If $ f” < 0 $, the gradient is decreasing and the curve bends downwards like a cap. We call it concave down (or concave).

Putting these together classifies a stationary point. At the bottom of a cup, the gradient is zero and the curve bends up: a minimum. At the top of a cap, the gradient is zero and the curve bends down: a maximum. Where the bending switches from cup to cap, or cap to cup, we have a point of inflection.

Key termPoint of inflection: a point on a curve where the concavity changes, so $ f”(x) $ changes sign. At a point of inflection the gradient $ f'(x) $ reaches a local maximum or minimum. If $ f'(x) = 0 $ there too, it is a stationary point of inflection; otherwise it is a non-stationary point of inflection.

\[ f'(a) = 0 \text{ and } \begin{cases} f”(a) < 0 & \Rightarrow \text{local maximum} \\ f”(a) > 0 & \Rightarrow \text{local minimum} \\ f”(a) = 0 & \Rightarrow \text{test inconclusive} \end{cases} \]

Here $ a $ is the $ x $-value of a stationary point, found by solving $ f'(x) = 0 $. $ f”(a) $ is the second derivative evaluated at that point. A negative value means the curve bends down, so the stationary point is a peak. A positive value means it bends up, so it is a trough. Zero means the second derivative alone cannot decide, and you must look at the sign of $ f'(x) $ on either side instead. This “first derivative test” always works ^[1].

Mathematics Made Simple explains stationary points, concavity and curve sketching with worked examples and graphs.Get the textbook · $12

Points of inflection and stationary points are different ideas, even though they can coincide. A stationary point is about the gradient being zero. A point of inflection is about the bending changing. The curve $ y = x^3 $ at the origin is both. The adoption curve at month 10 is only an inflection: the gradient there is 300, far from zero.

Why this method fits the case

The team’s first question, “when is growth fastest?”, is a question about the maximum of $ N'(t) $, the growth rate. To maximise $ N’ $, we differentiate it and set the result to zero: $ N”(t) = 0 $. So the moment of fastest growth is exactly the point of inflection of $ N $. The second question, “when does growth switch from accelerating to slowing?”, asks where $ N” $ changes sign, which is the same point. The third question, “where does the total peak?”, is an ordinary stationary point, $ N'(t) = 0 $, classified by the second derivative test.

So one function and two derivatives answer all three questions. This is also the backbone of curve sketching: find where the curve rises and falls, where it bends which way, and mark the special points. With those, you can draw an accurate sketch without plotting dozens of points ^[2].

When is this not the right approach?

When you only need the largest value on an interval. That is an optimisation problem. You still find stationary points, but you must also check the end points. Our separate post on optimisation problems covers that workflow.
When $ f”(a) = 0 $ at a stationary point. The second derivative test gives no answer. $ x^4 $, $ -x^4 $ and $ x^3 $ all have $ f'(0) = f”(0) = 0 $, yet they give a minimum, a maximum and an inflection. Use the first derivative test.
When the model only fits part of the range. Our cubic falls after month 20, which makes no sense for total users (people do not “un-join” in bulk). The analysis is valid only for $ 0 \le t \le 20 $. A logistic model fits whole adoption curves better.

Setting it up

Quantity	Formula	Meaning in the case
Total users	$ N(t) = 30t^2 – t^3 $	Users after $ t $ months
First derivative	$ N'(t) = 60t – 3t^2 $	New users per month (growth rate)
Second derivative	$ N”(t) = 60 – 6t $	Whether growth is speeding up (+) or slowing (−)
Stationary points	$ N'(t) = 3t(20 – t) = 0 $	$ t = 0 $ and $ t = 20 $
Candidate inflection	$ N”(t) = 0 $	$ t = 10 $

A sign table is the clearest way to organise the shape. Between the special points, pick any test value and check the signs.

Interval	Sign of $ N’ $	Sign of $ N” $	Shape
$ 0 < t < 10 $	+	+	Rising, concave up (growth accelerating)
$ 10 < t < 20 $	+	−	Rising, concave down (growth slowing)
$ t > 20 $	−	−	Falling (outside the model’s valid range)

Solving it

Worked example

Find and classify the stationary points of $ N(t) = 30t^2 – t^3 $, and find the point of inflection.

Step 1$ N'(t) = 60t – 3t^2 = 3t(20 – t) $. Set to zero: $ t = 0 $ or $ t = 20 $.

Step 2$ N”(t) = 60 – 6t $. At $ t = 0 $: $ N” = 60 > 0 $, a local minimum, $ N(0) = 0 $.

Step 3At $ t = 20 $: $ N” = 60 – 120 = -60 < 0 $, a local maximum, $ N(20) = 12000 – 8000 = 4000 $.

Step 4Inflection: $ 60 – 6t = 0 \Rightarrow t = 10 $. Check the sign change: $ N”(9) = 6 > 0 $ and $ N”(11) = -6 < 0 $. It changes, so this is a point of inflection.

Step 5$ N(10) = 3000 – 1000 = 2000 $ and $ N'(10) = 600 – 300 = 300 $. The gradient is not zero, so it is a non-stationary inflection.

Growth is fastest at month 10, with 2,000 users and 300 new users a month; the total peaks at 4,000 users at month 20.

Worked example

Sketch $ y = x^3 – 3x^2 – 9x + 5 $ by finding its stationary points and point of inflection.

Step 1$ y’ = 3x^2 – 6x – 9 = 3(x – 3)(x + 1) $. Stationary at $ x = -1 $ and $ x = 3 $.

Step 2$ y” = 6x – 6 $. At $ x = -1 $: $ y” = -12 < 0 $, maximum. $ y = -1 – 3 + 9 + 5 = 10 $.

Step 3At $ x = 3 $: $ y” = 12 > 0 $, minimum. $ y = 27 – 27 – 27 + 5 = -22 $.

Step 4Inflection: $ 6x – 6 = 0 \Rightarrow x = 1 $, $ y = 1 – 3 – 9 + 5 = -6 $. $ y” $ changes from − to + there.

Step 5$ y $-intercept: $ (0, 5) $. As $ x \to \infty $, $ y \to \infty $; as $ x \to -\infty $, $ y \to -\infty $.

Maximum $ (-1, 10) $, minimum $ (3, -22) $, inflection $ (1, -6) $, halfway between them, as always for a cubic.

Worked example

When the second derivative test fails. Classify the stationary point of $ y = x^4 $ and of $ y = x^3 $ at $ x = 0 $.

Step 1For $ y = x^4 $: $ y’ = 4x^3 = 0 $ at $ x = 0 $; $ y” = 12x^2 = 0 $ at $ x = 0 $. Inconclusive.

Step 2First derivative test: $ y'(-0.1) = -0.004 < 0 $ and $ y'(0.1) = 0.004 > 0 $. Falling then rising: a minimum.

Step 3For $ y = x^3 $: $ y’ = 3x^2 $, $ y” = 6x $, both zero at 0. Inconclusive again.

Step 4$ y'(-0.1) = 0.03 > 0 $ and $ y'(0.1) = 0.03 > 0 $. Rising on both sides, so no turn: a stationary point of inflection (and $ y” $ changes sign).

When $ f”(a) = 0 $, check the sign of $ f’ $ either side: same sign means inflection, a change of sign means a turning point.

Make stationary points and points of inflection automatic with the Practice Questions book: 1,565 questions, all fully worked.Get 1,565 questions · $9

What the answer tells us

The point of inflection at month 10 is the tipping point. Before it, each month brings more new users than the month before: 225 new users a month at month 5, rising to 300 at month 10. After it, each month brings fewer: back to 225 a month at month 15. The growth rate is symmetric around month 10 in this model, which is why months 5 and 15 match. Figure 1 shows the S-shape, with the tangent at month 10 drawn in. The tangent is steepest there; everywhere else it is flatter.

App adoption curve with its point of inflection users

Total users N(t) = 30t² − t³Tangent at inflection, gradient 300

Figure 1. Total users $ N(t) = 30t^2 – t^3 $. The curve is concave up before month 10 and concave down after it. The point of inflection $ (10, 2000) $ is where the tangent is steepest (300 users per month). The maximum is at $ (20, 4000) $. (Illustrative figures.)

Is the answer reasonable? At the inflection the curve has reached 2,000 users, exactly half of the eventual 4,000. That is a well-known pattern for symmetric S-curves: the fastest growth happens at the halfway mark. A quick numerical check also agrees: the actual increase from month 9 to month 10 is $ 2000 – 1701 = 299 $, and from month 10 to month 11 it is $ 2299 – 2000 = 299 $, both very close to the derivative of 300.

What does it mean for decisions? Before month 10, the team is in the accelerating phase, and word-of-mouth is doing a lot of the work. After month 10, growth naturally slows even if nothing goes wrong. A manager who sees the monthly increase fall from 300 new users at month 10 to 288 at month 12 should not panic: the shape was always going to do that. To beat the slowdown, the team must enlarge the market (new districts, new subjects) rather than push harder in a market that is half full.

How sensitive is the timing? For a general cubic $ N(t) = at^2 – t^3 $, the inflection is at $ t = \frac{a}{3} $ and the peak at $ t = \frac{2a}{3} $. So the inflection always comes halfway to the peak. If the market were larger, with $ a = 36 $, the inflection would move to month 12. Watching for the moment the monthly increase stops growing gives an early, data-based signal of where you are on the curve ^[4].

Common mistakeClaiming a point of inflection just because $ f”(x) = 0 $. For $ y = x^4 $, $ y”(0) = 0 $, but $ y” = 12x^2 $ is positive on both sides, so the curve is concave up throughout. The origin is a minimum, not an inflection. Always check that $ f” $ changes sign.

Common mistakeThinking a point of inflection must have a horizontal tangent. Most inflections are non-stationary, like month 10 in the case, where the gradient is 300. Only when $ f’ = 0 $ as well is it a stationary point of inflection.

Exam tipIn AP Calculus, justify every classification in words that link to the derivative: “$ f” $ changes from positive to negative at $ x = 10 $, so the graph of $ f $ has a point of inflection there” ^[2]. A bare “$ f”(10) = 0 $” earns no credit. In Cambridge 9709 and IB, “determine the nature” of a stationary point means show the second derivative’s sign, or a gradient sign check if it is zero ^[3].

For more on curve sketching, including asymptotes and how all the pieces fit together, see the Mathematics Made Simple textbook. To practise classifying stationary points until it is automatic, try the Mathematics Made Simple Practice Questions book.

Master curve sketching and differentiation with both Mathematics Made Simple books for $16.Both books · $16

Practice questions

For each question, find the stationary points first, then classify them.

EasyQ1. Find and classify the stationary point of $ y = x^2 – 6x + 11 $. Show worked answer

$ y’ = 2x – 6 = 0 \Rightarrow x = 3 $; $ y = 9 – 18 + 11 = 2 $.
$ y” = 2 > 0 $, so $ (3, 2) $ is a minimum.

EasyQ2. Find and classify the stationary points of $ y = x^3 – 12x $. Show worked answer

$ y’ = 3x^2 – 12 = 0 \Rightarrow x = \pm 2 $.
$ y” = 6x $. At $ x = 2 $: $ y” = 12 > 0 $, minimum, $ y = 8 – 24 = -16 $.
At $ x = -2 $: $ y” = -12 < 0 $, maximum, $ y = -8 + 24 = 16 $.

MediumQ3. For $ y = xe^{-x} $, find the maximum point and the point of inflection. Show worked answer

$ y’ = e^{-x} – xe^{-x} = (1 – x)e^{-x} = 0 \Rightarrow x = 1 $; $ y = e^{-1} \approx 0.368 $.
$ y” = -e^{-x} – (1 – x)e^{-x} = (x – 2)e^{-x} $. At $ x = 1 $: $ y” = -e^{-1} < 0 $, maximum.
$ y” = 0 $ at $ x = 2 $, and $ (x – 2) $ changes sign there, so inflection at $ (2, 2e^{-2}) \approx (2, 0.271) $.

MediumQ4. Find and classify the stationary points of $ y = x^4 – 4x^3 $, and find any non-stationary point of inflection. Show worked answer

$ y’ = 4x^3 – 12x^2 = 4x^2(x – 3) = 0 \Rightarrow x = 0, 3 $.
$ y” = 12x^2 – 24x $. At $ x = 3 $: $ y” = 36 > 0 $, minimum, $ y = 81 – 108 = -27 $.
At $ x = 0 $: $ y” = 0 $, inconclusive. $ y’ < 0 $ on both sides, so $ (0, 0) $ is a stationary point of inflection.
$ y” = 12x(x – 2) = 0 $ also at $ x = 2 $, with a sign change: non-stationary inflection at $ (2, -16) $.

ExamQ5. The curve $ y = 3x^4 – 16x^3 + 18x^2 $ is given. (a) Find the coordinates of the three stationary points. (b) Use the second derivative to determine their nature. (c) Find the $ x $-coordinates of the points of inflection, giving exact answers. [8] Show worked answer

(a) $ y’ = 12x^3 – 48x^2 + 36x = 12x(x – 1)(x – 3) $ [1]. Points $ (0, 0) $, $ (1, 5) $, $ (3, -27) $. [2]
(b) $ y” = 36x^2 – 96x + 36 $ [1]. $ y”(0) = 36 > 0 $ min; $ y”(1) = -24 < 0 $ max; $ y”(3) = 72 > 0 $ min. [2]
(c) $ 36x^2 – 96x + 36 = 0 \Rightarrow 3x^2 – 8x + 3 = 0 $ [1] $ \Rightarrow x = \frac{4 \pm \sqrt{7}}{3} $ (about 0.451 and 2.215); simple roots, so $ y” $ changes sign. [1]

Frequently asked questions

Is every point where $ f”(x) = 0 $ a point of inflection?

No. It is only a candidate. You must check that $ f” $ changes sign as $ x $ passes through it. $ y = x^4 $ has $ f”(0) = 0 $ but no inflection at 0.

What is the difference between concave and convex?

Books use different words. “Concave up” and “convex” mean the same thing: bending upwards, $ f” > 0 $. “Concave down” and “concave” mean bending downwards, $ f” < 0 $. Economics texts usually use convex and concave; AP uses concave up and concave down.

Why does the gradient have a maximum or minimum at an inflection?

Because $ f” $ is the derivative of $ f’ $. At an inflection, $ f” $ changes sign, so $ f’ $ changes from increasing to decreasing (a maximum of the gradient) or the reverse (a minimum). That is why the fastest growth in the case is at the inflection.

Can a curve have a point of inflection at a point where it is not defined?

No. A point of inflection must lie on the curve. $ y = \frac{1}{x} $ changes concavity at $ x = 0 $, but there is no point there, only an asymptote, so it has no point of inflection.

References

Strang, G. and Herman, E. (2016) Calculus Volume 1. Houston, TX: OpenStax. Available at: https://openstax.org/details/books/calculus-volume-1.
College Board (2020) AP Calculus AB and AP Calculus BC Course and Exam Description. New York: College Board.
Cambridge International (2024) Cambridge International AS & A Level Mathematics 9709: Syllabus for examination in 2026 and 2027. Cambridge: Cambridge University Press & Assessment. Available at: cambridgeinternational.org.
Sydsæter, K., Hammond, P., Strøm, A. and Carvajal, A. (2016) Essential Mathematics for Economic Analysis. 5th edn. Harlow: Pearson.

Stationary Points, Concavity and Points of Inflection: The Second Derivative Test and Curve Sketching

The case: when does an app’s growth start to slow?

The concept: what the first and second derivatives tell you

Why this method fits the case

Setting it up

Solving it

What the answer tells us

Practice questions

Frequently asked questions

Is every point where \( f”(x) = 0 \) a point of inflection?

What is the difference between concave and convex?

Why does the gradient have a maximum or minimum at an inflection?

Can a curve have a point of inflection at a point where it is not defined?

References

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