In short

A curve is concave up when it bends like a cup (the tangent line lies below the curve) and concave down when it bends like a cap (the tangent line lies above the curve). The second derivative tells you which: f''(x) > 0 means concave up, f''(x) < 0 means concave down. A point of inflection is where the curve switches from one to the other.

Take f(x) = x^3. Plot it and stare at the curve for a moment. On the right side of the origin, the curve swoops upward, bending like the inside of a bowl. On the left side, it swoops downward, bending like the top of a hill. Somewhere in the middle — right at the origin — the bending changes direction.

You already know the first derivative: it tells you whether the function is increasing or decreasing. But two functions can both be increasing and yet look completely different. Compare f(x) = x^2 and g(x) = \sqrt{x} on the interval (0, \infty). Both are increasing. But x^2 curves upward — it gets steeper and steeper — while \sqrt{x} curves downward — it gets flatter and flatter. The first derivative cannot distinguish between these two shapes. It only says "going up." It does not say how the going-up is itself changing.

That is what the second derivative is for. It measures the rate at which the slope changes, and that rate is exactly the bending of the curve.

Concave up and concave down

Hold a physical ruler against a curve. If the curve bends away from the ruler on the upper side — like water pooling in a bowl — the curve is concave up at that point. If it bends away on the lower side — like rain running off a dome — the curve is concave down.

There is a cleaner way to say this using the tangent line.

The parabola $y = x^2$ is concave up everywhere. Every tangent line sits below the curve — the curve bends away from its tangent upward, like the inside of a bowl.
The curve $y = \sqrt{x}$ is concave down for all $x > 0$. Every tangent line sits above the curve — the curve sags below its tangent, like the outside of a dome.

Think of it physically. If you are driving along a road shaped like y = x^2, your steering wheel is turned to the left (curving upward). If you are driving along a road shaped like y = -x^2, your steering wheel is turned to the right (curving downward). The second derivative measures how hard you are turning.

The second derivative test for concavity

The connection between concavity and the second derivative is direct.

Concavity test

Let f be a function that is twice differentiable on an interval I.

  • If f''(x) > 0 for all x in I, then f is concave up on I.
  • If f''(x) < 0 for all x in I, then f is concave down on I.

Why this works. The first derivative f'(x) gives the slope of the curve at each point. The second derivative f''(x) gives the rate at which that slope is changing.

If f''(x) > 0, then f'(x) is increasing — the slope is getting larger as you move to the right. That means the curve is getting steeper in the positive direction, which is exactly what "bending upward" looks like.

If f''(x) < 0, then f'(x) is decreasing — the slope is getting smaller as you move to the right. The curve is getting less steep (or more steeply negative), which is "bending downward."

Check this against the examples above:

Points of inflection

When the concavity changes — from up to down, or from down to up — the point where the switch happens is called a point of inflection.

Point of inflection

A point c in the domain of f is a point of inflection if f is continuous at c and the concavity of f changes at c. That is, f is concave up on one side of c and concave down on the other side (or vice versa).

For f(x) = x^3, the point of inflection is at x = 0. On the left of 0, f''(x) = 6x < 0 (concave down). On the right, f''(x) = 6x > 0 (concave up). The concavity flips at x = 0.

The curve $y = x^3$. The concavity changes at the origin: concave down for $x < 0$, concave up for $x > 0$. The tangent line at the origin actually crosses the curve there — this is a hallmark of an inflection point.

Notice something striking in the figure: at the inflection point, the tangent line crosses the curve. This is not a coincidence. At a regular point where the curve is concave up, the tangent sits below the curve. Where the curve is concave down, the tangent sits above. At the inflection point, the tangent switches from one side to the other — so it must cross through the curve.

How to find inflection points

Here is the procedure:

Step 1. Compute f''(x).

Step 2. Find all values of x where f''(x) = 0 or where f''(x) is undefined. These are candidates for inflection points.

Step 3. Check whether f''(x) actually changes sign at each candidate. If it does, you have an inflection point. If it does not — if f'' has the same sign on both sides — then the candidate is not an inflection point.

Step 3 is essential. The condition f''(c) = 0 is necessary but not sufficient for an inflection point. The function f(x) = x^4 has f''(x) = 12x^2, so f''(0) = 0. But f''(x) \geq 0 everywhere — no sign change. The origin is not an inflection point for x^4; the curve is concave up on both sides. It just happens to have a momentarily "flat" second derivative at x = 0.

A richer example: f(x) = x^4 - 4x^3

Compute the derivatives:

f'(x) = 4x^3 - 12x^2
f''(x) = 12x^2 - 24x = 12x(x - 2)

Set f''(x) = 0: the candidates are x = 0 and x = 2.

Check sign changes:

The sign changes at both x = 0 and x = 2. Both are inflection points.

At x = 0: f(0) = 0. At x = 2: f(2) = 16 - 32 = -16. The inflection points are (0, 0) and (2, -16).

Worked examples

Example 1: Concavity analysis of $f(x) = x^3 - 6x^2 + 9x + 1$

Step 1. Compute the first and second derivatives.

f'(x) = 3x^2 - 12x + 9
f''(x) = 6x - 12

Why: you need f'' to determine concavity, and f' to find critical points for context.

Step 2. Find where f''(x) = 0.

6x - 12 = 0 \implies x = 2

Why: this is the candidate for an inflection point — the only place where the concavity could change.

Step 3. Determine the sign of f''(x) on each side of x = 2.

  • For x < 2 (try x = 0): f''(0) = -12 < 0. Concave down.
  • For x > 2 (try x = 3): f''(3) = 6 > 0. Concave up.

Why: the sign changes from negative to positive, so x = 2 is a genuine inflection point.

Step 4. Find the coordinates of the inflection point.

f(2) = 8 - 24 + 18 + 1 = 3

The inflection point is at (2, 3).

Why: you need both coordinates to locate the point on the curve.

Step 5. Identify the critical points for a complete picture.

f'(x) = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

Critical points at x = 1 and x = 3. Since f''(1) = -6 < 0, x = 1 is a local maximum: f(1) = 5. Since f''(3) = 6 > 0, x = 3 is a local minimum: f(3) = 1.

Why: the second derivative test classifies each critical point. Negative second derivative means the curve bends downward there, so it is a peak. Positive means it bends upward, so it is a valley.

Result: f is concave down on (-\infty, 2) and concave up on (2, \infty), with an inflection point at (2, 3).

The curve $y = x^3 - 6x^2 + 9x + 1$. To the left of $x = 2$, the curve bends downward (concave down). To the right, it bends upward (concave up). The inflection point at $(2, 3)$ is where the bending reverses — notice how the curve has an S-shaped transition there, right between the local maximum and the local minimum.

The inflection point sits exactly halfway between the local maximum and local minimum, at x = 2. This is not a coincidence for cubics — for any cubic ax^3 + bx^2 + cx + d, the inflection point is always at x = -b/(3a), which is always the midpoint of the two critical points.

Example 2: Finding inflection points of $f(x) = \sin x$ on $[0, 2\pi]$

Step 1. Compute derivatives.

f'(x) = \cos x, \qquad f''(x) = -\sin x

Why: the second derivative of sine is negative sine — a fact worth memorising, since it says the sine curve always bends toward the axis.

Step 2. Find where f''(x) = 0 on [0, 2\pi].

-\sin x = 0 \implies \sin x = 0 \implies x = 0, \; \pi, \; 2\pi

Why: these are the candidates for inflection points within the given interval.

Step 3. Check sign changes.

  • On (0, \pi): \sin x > 0, so f''(x) = -\sin x < 0. Concave down.
  • On (\pi, 2\pi): \sin x < 0, so f''(x) = -\sin x > 0. Concave up.

Why: the sign of f'' changes at x = \pi. At x = 0 and x = 2\pi, the sign also changes (from positive on the left of 0 and from positive to negative on the right of 2\pi), but since 0 and 2\pi are endpoints of our interval, x = \pi is the interior inflection point.

Step 4. The inflection point on (0, 2\pi) is at x = \pi, where f(\pi) = 0.

The inflection point is (\pi, 0).

Why: the sine curve crosses zero at x = \pi, and that is exactly where the bending reverses — from the arch of the first hump to the trough of the second.

Result: \sin x is concave down on (0, \pi) and concave up on (\pi, 2\pi), with an inflection point at (\pi, 0).

One full period of $y = \sin x$. The first arch (from $0$ to $\pi$) bends downward — concave down. The second arch (from $\pi$ to $2\pi$) bends upward — concave up. The inflection point at $(\pi, 0)$ is where the curvature reverses. The tangent line at the inflection point crosses through the curve.

The sine curve has an inflection point at every integer multiple of \pi. Each time the function crosses zero, the bending reverses. So the concavity of sine alternates: concave down on (0, \pi), concave up on (\pi, 2\pi), concave down on (2\pi, 3\pi), and so on, forever.

Common confusions

Using concavity in curve sketching

Concavity information, combined with what you know about increasing/decreasing behaviour from the first derivative, gives you a much more accurate sketch of a function. Here is the key insight: at a local maximum where f''(c) < 0, the curve looks like the top of a hill (concave down). At a local minimum where f''(c) > 0, the curve looks like the bottom of a valley (concave up).

Between extrema, inflection points mark where the curve transitions from one shape to the other. A well-drawn curve sketch should show:

This is the foundation of the full curve-sketching procedure you will see in the article on Curve Sketching.

Going deeper

If you came here to understand what concavity and inflection points are, you have it — you can stop here. What follows is the rigorous version, some subtle edge cases, and the connection to the second derivative test for extrema.

The rigorous definition of concavity

The intuitive definition — "the tangent line lies below the curve" — can be turned into a precise algebraic statement.

A function f is concave up on an interval I if, for any two points a, b \in I and any t \in (0, 1):

f\bigl(ta + (1-t)b\bigr) \leq t\,f(a) + (1-t)\,f(b)

The left side is the value of f at a weighted average of a and b. The right side is the same weighted average of f(a) and f(b) — which is a point on the straight line (the chord) connecting (a, f(a)) to (b, f(b)).

So the inequality says: the curve lies below the chord between any two of its points. This is the same as saying all tangent lines lie below the curve, but it does not require differentiability. It works for functions with corners too.

A function is concave down (sometimes called concave) if the inequality is reversed: f\bigl(ta + (1-t)b\bigr) \geq t\,f(a) + (1-t)\,f(b).

The second derivative test for local extrema

There is a useful connection between concavity and the classification of critical points.

If f'(c) = 0 and f''(c) > 0, then c is a local minimum. The reasoning: f''(c) > 0 means the curve is concave up at c, bending like a valley. A critical point (horizontal tangent) at the bottom of a valley is a minimum.

If f'(c) = 0 and f''(c) < 0, then c is a local maximum. The curve is concave down at c, bending like a hilltop.

If f'(c) = 0 and f''(c) = 0, the test is inconclusive. You need more information. For example:

All three cases have f'(0) = 0 and f''(0) = 0, but the behaviour is different. When the second derivative test is inconclusive, fall back to the first derivative test: check whether f' changes sign at c.

Higher-order test for inflection

When f''(c) = 0, how do you know whether c is a genuine inflection point without checking sign changes? If f has continuous higher derivatives, there is a systematic approach.

Find the first n > 2 such that f^{(n)}(c) \neq 0:

For f(x) = x^4: f''(0) = 0, f'''(0) = 0, f^{(4)}(0) = 24 \neq 0. The first non-zero derivative is the fourth (n = 4, even), so x = 0 is not an inflection point.

For f(x) = x^5: f''(0) = 0, f'''(0) = 0, f^{(4)}(0) = 0, f^{(5)}(0) = 120 \neq 0. The first non-zero derivative is the fifth (n = 5, odd), so x = 0 is an inflection point.

Where this leads next

Concavity is one of the key tools in the complete analysis of a function's graph. The next articles build on it directly: