Concavity, convexity, and inflection

Draw a smooth curve and it is always doing one of two things: bending upward like the inside of a bowl, or bending downward like the top of a dome. The second derivative $f''(x)$ is the instrument that tells you which — its sign, not its size, is the whole story. Where the bending direction actually flips, the curve passes through a point of inflexion, and the curve crosses its own tangent line there instead of hugging it. This sits in Unit 1 of MA10001, right after the mean value theorems and Taylor expansions, and it asks something narrower than full curve sketching: given $f''(x)$ , find where the curve is convex, where it is concave, and where genuine inflexion occurs. The one habit this topic drills hardest is refusing to trust $f''(c)=0$ on its own — a sign change must always be checked, and Taylor's theorem is exactly the tool that explains why.

The core insight

The sign of $f''(x)$ fixes the bending direction of a curve, and a point of inflexion is where that sign genuinely changes, which Taylor's theorem shows is guaranteed once $f''(c)=0$ and $f'''(c)\neq 0$ .

The core idea

Let $f$ be twice differentiable on an interval $I$ .

$f$ is convex (concave up) on $I$ if $f''(x) > 0$ throughout $I$ : the curve lies above every one of its own tangent lines there, the way $y = x^2$ sits above every tangent drawn to it.
$f$ is concave (concave down) on $I$ if $f''(x) < 0$ throughout $I$ : the curve lies below every tangent, the way $y = -x^2$ does.

A point $c$ is a point of inflexion if $f''$ changes sign as $x$ crosses $c$ — equivalently, the tangent line at $c$ crosses the curve instead of staying on one side of it. Two conditions must both hold before you can call $c$ an inflexion point:

f''(c) = 0 \quad \text{(or } f'' \text{ undefined at } c\text{)}, \qquad \text{and} \qquad f'' \text{ changes sign across } c.

where: $f''(c)=0$ (or non-existence of $f''(c)$ ) only flags a candidate; the sign change is the part that actually certifies inflexion. This distinction is the entire syllabus content of the topic, and the classic counterexample is $f(x) = x^4$ at $x=0$ : $f''(x) = 12x^2 \geq 0$ everywhere, so $f''(0)=0$ but there is no sign change, and no inflexion. The figure below shows the two regimes and a genuine crossing between them.

Figure

A cubic curve that is concave on the left, has a marked inflection point in the middle, and is convex on the right, with the tangent line at the inflection point crossing the curve

Procedure. Compute $f''(x)$ . Solve $f''(x)=0$ and note any points where $f''$ is undefined — these are candidates. Test the sign of $f''$ immediately on either side of each candidate (or use the third-derivative shortcut below). If the sign genuinely changes, $(c, f(c))$ is a confirmed point of inflexion.

Why it works

Derivation AWhy f''(c)=0 with f'''(c) nonzero forces a sign change

The goal is to explain, using the Taylor expansion already built earlier in this unit, why checking $f'''(c) \neq 0$ at a candidate point where $f''(c)=0$ is enough to guarantee a genuine sign change of $f''$ across $c$ — turning the sign test from a numerical check into a one-line certificate, and explaining exactly why $f(x)=x^4$ fails to produce an inflexion at its own candidate point.

Step 1. Expand $f''$ itself in a first-order Taylor approximation about the candidate $c$ , treating $f''$ as the function being expanded:

f''(x) = f''(c) + f'''(c)\,(x - c) + o(x-c) \quad \text{as } x \to c.

This is legitimate because $f$ is assumed to have enough derivatives near $c$ for the expansion to hold, which the curriculum already assumes once $f'''$ exists.

Step 2. Substitute the candidate condition $f''(c) = 0$ . The expansion collapses to

f''(x) \approx f'''(c)\,(x - c) \qquad \text{for } x \text{ close to } c.

Near $c$ , the sign of $f''(x)$ is therefore controlled entirely by the sign of the product $f'''(c)\,(x-c)$ , since the leading term dominates any smaller correction.

Step 3. Suppose $f'''(c) \neq 0$ , so it has a fixed sign — say positive (the negative case is identical with signs flipped). Then for $x$ slightly less than $c$ , $(x-c)<0$ , so $f''(x) \approx f'''(c)(x-c) < 0$ ; for $x$ slightly more than $c$ , $(x-c)>0$ , so $f''(x) > 0$ . The sign of $f''$ has flipped across $c$ purely because $(x-c)$ flips sign while $f'''(c)$ does not — this is a genuine, guaranteed sign change, so $c$ is a point of inflexion.

Step 4. Read off the failure case. If instead $f'''(c)=0$ too, the linear term vanishes and the argument gives no information — $f''(x)$ near $c$ is then controlled by a higher-order term whose sign need not flip. This is exactly what happens for $f(x)=x^4$ : $f''(x)=12x^2$ , so $f'''(x)=24x$ and $f'''(0)=0$ ; the shortcut is silent, and indeed the direct sign test shows $f''\geq 0$ on both sides, no inflexion. The third-derivative test is therefore a sufficient shortcut when it succeeds, never a substitute for the sign test when it is inconclusive.

Worked example

Locate all points of inflexion of $f(x) = x^3 - 6x^2 + 9x + 1$ , confirm each with the third-derivative shortcut, and find the tangent line at the inflexion point.

Step 1. Differentiate. $f'(x) = 3x^2 - 12x + 9$ , $f''(x) = 6x - 12$ , and $f'''(x) = 6$ .

Step 2. Find candidates. Set $f''(x)=0$ : $6x - 12 = 0 \implies x = 2$ , the only candidate.

Step 3. Apply the shortcut instead of a full sign test. $f'''(2) = 6 \neq 0$ , so by the derivation above the sign change is guaranteed — $x=2$ is certainly a point of inflexion, with no need to test points on either side.

Step 4. Locate the point and its tangent. $f(2) = 8 - 24 + 18 + 1 = 3$ , so the inflexion point is $(2,3)$ . The tangent slope there is $f'(2) = 3(4) - 12(2) + 9 = -3$ , so the tangent line is $y - 3 = -3(x-2)$ , i.e. $y = -3x + 9$ . Since $f'''(2)=6>0$ , $f''$ goes from negative to positive at $x=2$ , meaning the curve sits below this tangent just before $x=2$ (concave) and above it just after (convex) — the tangent line genuinely crosses the curve, the visual signature of inflexion.

The point of inflexion is $(2, 3)$ , with tangent $y = -3x + 9$ , concave down on $(-\infty, 2)$ and concave up on $(2, \infty)$ .

Competitive stretch

Try the shortcut on $f(x) = x^5$ at the candidate $x=0$ : $f''(x) = 20x^3$ , so $f''(0)=0$ , but $f'''(x) = 60x^2$ gives $f'''(0) = 0$ as well — the shortcut is inconclusive. Falling back to the direct sign test, $f''(x)=20x^3$ is negative for $x<0$ and positive for $x>0$ , a genuine change, so $x=0$ is an inflexion point here even though the shortcut gave no answer. This is the exam-relevant lesson: when $f'''(c) = 0$ too, you must return to sign-testing $f''$ directly — the third-derivative test can confirm inflexion but can never rule it out.

Common traps

Declaring inflexion the moment $f''(c)=0$ , without checking sign change or the third-derivative shortcut — fails for $f(x)=x^4$ at $x=0$ .
Using the third-derivative shortcut when $f'''(c)=0$ too, and wrongly concluding "no inflexion" — the shortcut is silent here, not negative; a direct sign test is required, as in $f(x)=x^5$ .
Confusing concavity (sign of $f''$ , the bending direction) with monotonicity (sign of $f'$ , whether the curve rises or falls) — these are independent properties of the same curve.
Forgetting to check points where $f''(x)$ is undefined, not just where $f''(x)=0$ , as additional candidate inflexion points.

Ask the tutor: if you want, ask the AI tutor to generate a rational or trigonometric function so you can practice combining the sign test with points where $f''$ is undefined.

Why do we even need the third-derivative shortcut if the sign test always works anyway

The sign test always works, but it forces you to compute $f''$ at two extra test points and compare signs, which is slower and more error-prone under exam time pressure. The shortcut lets you skip that entirely whenever $f'''(c)\neq 0$ , since Taylor's theorem guarantees the sign change without any further arithmetic. It is a speed tool layered on top of the fully rigorous method, not a replacement for it — you always fall back to the sign test the moment the shortcut is inconclusive.

Is a point of inflexion the same thing as a stationary point

No, and mixing these up is a common slip. A stationary point is where $f'(x)=0$ — the tangent is horizontal. A point of inflexion is where $f''(x)$ changes sign — the tangent, horizontal or not, crosses the curve. Some points, like $x=0$ for $f(x)=x^3$ , are both stationary and inflexion points at once, but most inflexion points have a tilted tangent, as in the worked example above where the tangent at $(2,3)$ has slope $-3$ , not zero.

Does convex here mean the same thing as convex functions in optimization courses

Yes, this is the same core idea, just introduced through the sign of $f''$ on an interval rather than through a global inequality. A twice-differentiable function with $f''(x)\geq 0$ on its whole domain is convex in the fuller sense used in optimization, where the chord between any two points lies on or above the curve. In this unit you only need the local, interval-by-interval version: $f''>0$ marks convex stretches, $f''<0$ marks concave stretches, and inflexion points are where the two stretches meet.