Concavity and Points of Inflection: Steps and Examples

What Are Concavity And Points Of Inflection?

Concavity and points of inflection describe how a graph bends, separate from whether it rises or falls. A function is concave up on an interval when its slope $f'(x)$ increases across that interval, and concave down when $f'(x)$ decreases. A point of inflection is a point on the graph where $f$ is continuous and the concavity changes from one type to the other.

The working test uses the second derivative. For a function $f$ that is twice differentiable on an open interval $I$:

$$f''(x) > 0 \text{ for all } x \text{ in } I ;\Rightarrow; f \text{ is concave up on } I$$

$$f''(x) < 0 \text{ for all } x \text{ in } I ;\Rightarrow; f \text{ is concave down on } I$$

Here $f'(x)$ is the slope of the tangent line at $x$, and $f''(x)$ is the rate at which that slope changes. A point $(c, f(c))$ is a point of inflection when two conditions hold together:

• $f$ is continuous at $c$, so the point actually sits on the graph;

• $f''(x)$ has one sign just to the left of $c$ and the opposite sign just to the right.

Concavity is a statement about the slope, so it sits one level above increasing and decreasing functions. A graph can be rising and concave down, or falling and concave up; the two descriptions are independent, and a full picture of a graph needs both. The OpenStax treatment of derivatives and the shape of a graph uses the same definitions.

How Does The Second Derivative Tell You The Concavity?

The second derivative measures how fast the slope is changing, so its sign tells you whether the slope is climbing or sliding. If $f''(x) > 0$ on an interval, then $f'(x)$ is increasing there, which is the definition of concave up: the tangent lines rotate counterclockwise as you move right and sit below the graph. Negative $f''(x)$ means the tangents rotate clockwise and sit above it.

One caution keeps the test honest. A positive second derivative guarantees concave up, but concave up does not force $f''(x)$ to stay strictly positive: $y = x^4$ is concave up everywhere, yet $f''(0) = 0$. An isolated zero of $f''(x)$ therefore tells you nothing by itself.

How Do You Find Points Of Inflection Step By Step?

To find points of inflection, compute $f''(x)$, list where it could change sign, and check the sign on each side of every candidate. The common question "how do I solve for the points of inflection" always reduces to five moves:

1. Find the domain of $f$. Points outside the domain can never be inflection points.

2. Compute and factor $f''(x)$. Factored form makes the sign chart almost automatic.

3. List the candidates. Write down every $x$ where $f''(x) = 0$. Actually, that list is incomplete on its own: also add every $x$ in the domain where $f''(x)$ does not exist. These are the candidates, and none of them is an inflection point yet.

4. Build a sign chart. The candidates split the domain into intervals. Pick one test value in each interval and record the sign of $f''(x)$ there.

5. Keep only real sign changes. A candidate where the sign flips, and where $f$ is continuous, is a point of inflection. Substitute it into $f$ (not into $f''$) to get its $y$-coordinate.

Here is the full procedure for $f(x) = x^4 - 4x^3$. The domain is all real numbers.

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

$$f''(x) = 12x^2 - 24x = 12x(x - 2)$$

Setting $f''(x) = 0$ gives the candidates $x = 0$ and $x = 2$. The sign chart:

The sign flips at both candidates, so both are inflection points. With $f(0) = 0$ and $f(2) = 16 - 32 = -16$, the points of inflection are $(0, 0)$ and $(2, -16)$.

Why Is $f''(c) = 0$ Not Enough For An Inflection Point?

A zero of the second derivative is only a candidate, because the sign of $f''(x)$ can touch zero and bounce back without flipping. The candidate list also works the other way: an inflection point can sit where $f''(x)$ does not exist at all, and a sign flip can happen at a point that is not on the graph. Each of the three cases below breaks a different shortcut.

When $f''(c) = 0$ But Nothing Flips

A forum question puts it exactly: why does $f(x) = x^4$ have $f''(0) = 0$ but no inflection point at $x = 0$? Because $f''(x) = 12x^2$ is positive on both sides of $0$. The curve is concave up to the left, concave up to the right, and merely very flat at the bottom. Such a point is sometimes called an undulation point: the second derivative vanishes, but the concavity never changes.

The general rule behind this, which you will meet again with Taylor polynomials, is that the first nonzero derivative beyond the first decides the matter: odd order gives an inflection point, even order does not. For $x^4$ that derivative is the fourth, which is even. The Wikipedia article on inflection points states the same sufficient condition.

When $f''$ Does Not Exist But The Concavity Still Flips

The cube root $f(x) = x^{1/3}$ has

$$f''(x) = -\frac{2}{9x^{5/3}},$$

which is undefined at $x = 0$. For $x < 0$ the denominator is negative, so $f''(x) > 0$ and the graph is concave up; for $x > 0$ it is concave down. The function is continuous at $0$, so $(0, 0)$ is a point of inflection, with a vertical tangent line there.

This case took real effort to accept the first time: an inflection point at a place where the second derivative does not even exist. It is the reason step 3 of the procedure insists on listing the undefined points too.

When The Concavity Flips But $f$ Does Not Exist

For $f(x) = \frac{1}{x}$, the second derivative $\frac{2}{x^3}$ is negative for $x < 0$ and positive for $x > 0$. The concavity changes across $x = 0$, but $f(0)$ does not exist, so there is no point on the graph to call an inflection point. A vertical asymptote can separate two concavities without creating one.

Can A Point Of Inflection Also Be A Critical Point?

Yes, a point of inflection can also be a critical point, and then it is called a stationary point of inflection: the tangent is horizontal and the concavity changes at the same place. The curve $y = x^3$ at the origin is the standard case, and in the sign-chart example above, $f(x) = x^4 - 4x^3$ has $f'(0) = 0$ as well as a concavity change at $0$. Its other inflection point, $(2, -16)$, has $f'(2) = -16$, so it is a non-stationary inflection point with a sloping tangent.

The two ideas answer different questions. Critical points come from $f'(x)$ and locate where the graph may turn around; inflection points come from $f''(x)$ and locate where the bending changes. A stationary inflection point is also one of the stationary points of the function, but it is not a maximum or minimum, because the graph keeps heading the same way through it.

This is also why the second derivative test goes silent when $f''(c) = 0$ at a critical point. The value zero cannot tell a minimum (like $x^4$) from a stationary inflection point (like $x^3$), so that test is inconclusive there.

How Do You Read Concavity From The Graph Of $f'$ Or $f''$?

On a graph of $f''(x)$, the inflection points of $f$ sit where the graph of $f''$ crosses the $x$-axis, not where it merely touches it. That answers a common homework question: "Where are the inflection points of f(x) if the graph shows the second derivative f''(x)?" Above the axis, $f$ is concave up; below the axis, concave down.

On a graph of $f'(x)$, look for rising and falling instead. Wherever $f'$ is increasing, $f$ is concave up, and wherever $f'$ is decreasing, $f$ is concave down. The inflection points of $f$ line up with the local maxima and minima of $f'$, because that is exactly where the slope stops growing and starts shrinking (or the reverse).

How Are Concavity And Points Of Inflection Used In Curve Sketching?

In curve sketching, concavity and points of inflection supply the bends between the turning points, so the sketch has the right shape and not only the right peaks. The critical points tell you where the graph turns; the sign chart of $f''(x)$ tells you how it curves on the way.

A forum request asked for a sketch of $y = x^4 - 4x^3 - 20x^2 + 150$ with critical points and inflection points. The concavity half of that job looks like this:

$$f'(x) = 4x^3 - 12x^2 - 40x = 4x(x^2 - 3x - 10)$$

$$f''(x) = 12x^2 - 24x - 40 = 4(3x^2 - 6x - 10)$$

The critical points are $x = -2$, $0$ and $5$. Solving $3x^2 - 6x - 10 = 0$ with the quadratic formula gives

$$x = 1 \pm \frac{\sqrt{39}}{3},$$

which is approximately $x = -1.0817$ and $x = 3.0817$. Since $f''(x)$ is an upward-opening parabola, it is positive outside these roots and negative between them. The curve is concave up, then concave down between the two inflection points, then concave up again, which places the local maximum at $(0, 150)$ inside the concave-down stretch and the minima at $(-2, 118)$ and $(5, -225)$ in the concave-up stretches.

Examples Of Concavity And Points Of Inflection

Example 1

Find the intervals of concavity and the point of inflection of $f(x) = x^3 - 6x^2 + 5$.

$$f'(x) = 3x^2 - 12x$$

$$f''(x) = 6x - 12 = 6(x - 2)$$

The only candidate is $x = 2$. Test values: $f''(1) = -6 < 0$ and $f''(3) = 6 > 0$, so the sign flips.

$$f(2) = 8 - 24 + 5 = -11$$

Final answer: concave down on $(-\infty, 2)$, concave up on $(2, \infty)$, point of inflection $(2, -11)$.

Example 2

Find all points of inflection of $f(x) = x^5 - 5x^4$.

The tempting route is to solve $f''(x) = 0$ and report every solution.

$$f'(x) = 5x^4 - 20x^3$$

$$f''(x) = 20x^3 - 60x^2 = 20x^2(x - 3)$$

Setting this to zero gives $x = 0$ and $x = 3$, so the tempting answer is "two inflection points, $(0, 0)$ and $(3, -162)$". Now check the signs. At $x = -1$, $f''(-1) = -80$; at $x = 1$, $f''(1) = -40$. The second derivative is negative on both sides of $0$, so the concavity does not change there and $(0, 0)$ is not an inflection point.

The factor $x^2$ is the giveaway: a squared factor never changes sign. At $x = 3$ the factor $(x - 3)$ does flip, with $f''(1) = -40$ and $f''(4) = 320$.

$$f(3) = 243 - 405 = -162$$

Final answer: one point of inflection, $(3, -162)$; the curve is concave down on $(-\infty, 3)$ and concave up on $(3, \infty)$.

Example 3

Describe the concavity of $f(x) = x + \frac{1}{x}$ and decide whether it has a point of inflection.

$$f'(x) = 1 - \frac{1}{x^2}$$

$$f''(x) = \frac{2}{x^3}$$

The second derivative is never zero. It is undefined at $x = 0$, but $0$ is outside the domain. Test values: $f''(-1) = -2 < 0$ and $f''(1) = 2 > 0$.

Final answer: concave down on $(-\infty, 0)$, concave up on $(0, \infty)$, and no point of inflection, because the concavity changes only across the vertical asymptote $x = 0$.

The first instinct here is to search for $f''(x) = 0$, find nothing, and declare the concavity the same everywhere; the sign flip at the asymptote is exactly what that search misses.

Example 4

Find the points of inflection of $f(x) = \sin x + \cos x$ on $[0, 2\pi]$.

$$f'(x) = \cos x - \sin x$$

$$f''(x) = -\sin x - \cos x$$

Setting $\sin x + \cos x = 0$ gives $\tan x = -1$, so $x = \frac{3\pi}{4}$ or $x = \frac{7\pi}{4}$ in the interval. Test values: $f''\left(\frac{\pi}{2}\right) = -1$, $f''(\pi) = 1$, and just past $\frac{7\pi}{4}$ the value is negative again (for example, about $-0.4179$ at $x = \frac{7\pi}{4} + 0.3$).

$$f\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0$$

$$f\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 0$$

Final answer: points of inflection at $\left(\frac{3\pi}{4}, 0\right)$ and $\left(\frac{7\pi}{4}, 0\right)$; concave down on $\left(0, \frac{3\pi}{4}\right)$, up on $\left(\frac{3\pi}{4}, \frac{7\pi}{4}\right)$, down on $\left(\frac{7\pi}{4}, 2\pi\right)$.

Example 5

Find the point of inflection of $f(x) = x e^{-x}$ and the slope of the tangent there.

$$f'(x) = e^{-x} - x e^{-x} = (1 - x)e^{-x}$$

$$f''(x) = -e^{-x} - (1 - x)e^{-x} = (x - 2)e^{-x}$$

Because $e^{-x} > 0$ for every $x$, the sign of $f''(x)$ is the sign of $x - 2$, which flips at $x = 2$.

$$f(2) = 2e^{-2} \approx 0.2707$$

$$f'(2) = -e^{-2} \approx -0.1353$$

Final answer: point of inflection $\left(2, 2e^{-2}\right) \approx (2, 0.2707)$, where the tangent slope is about $-0.1353$, the steepest descent the curve ever reaches.

Example 6

A population grows as $P(t) = \frac{1000}{1 + 9e^{-t/2}}$, with $t$ in years. When is the population growing fastest, and how large is it then?

The fastest growth happens where $P'(t)$ is largest, which is the point of inflection of $P$. Solving $P''(t) = 0$ for this model reduces to

$$9e^{-t/2} = 1$$

$$t = 2\ln 9 = \ln 81 \approx 4.3944$$

The sign of $P''(t)$ checks out: about $18.59$ at $t = 3$ and about $-20.36$ at $t = 6$. At the inflection point the denominator is $1 + 1 = 2$:

$$P(\ln 81) = \frac{1000}{2} = 500$$

$$P'(\ln 81) = 125$$

Final answer: the population grows fastest at $t \approx 4.3944$ years, when it reaches $500$ (half of its limit of $1000$) and is growing at $125$ per year.

Why Do Inflection Points Matter? "The Moment Growth Stops Speeding Up"

Inflection points exist because a first derivative tells you how fast something changes, but not when that change itself reverses. The point of inflection is the earliest signal that acceleration has become deceleration, before the quantity itself stops rising.

• It marks the peak rate, not the peak value. In Example 6 the population is still rising at $t = 4.3944$, but its growth rate peaks there and falls from then on.

• It separates two regimes. On one side, each extra unit of time adds more than the last; on the other, each adds less.

• It works when there is no turning point. A function with no maximum or minimum, like $x^3$ or an S-curve, can still have an inflection point, so concavity adds information the critical points cannot.

Where Do Points Of Inflection Show Up In Real Life?

Points of inflection appear wherever growth first speeds up and then slows down, which is the shape of most real processes with a limit.

• Population models. The logistic growth model, introduced by Pierre-François Verhulst in 1838, has its inflection point at exactly half the carrying capacity, as described in the logistic function reference. Example 6 is a small version of that model.

• Epidemic curves. The cumulative number of cases is the running total of daily new cases, so the day new cases peak is an inflection point of the cumulative curve.

• Economics. A total-output curve that first rises faster and then slower has its inflection point at the point of diminishing marginal returns, the moment each extra worker or machine starts adding less than the one before.

• Technology adoption. The S-shaped adoption curve of a new product bends from concave up to concave down at its inflection point, when new users join fastest.

Where Do Students Trip Up On Points Of Inflection?

Most mistakes on this topic come from treating a candidate as a confirmed answer.

Mistake 1: Treating Every Zero Of $f''$ As An Inflection Point

Where it slips in: Any function whose second derivative has a squared factor, such as $f''(x) = 20x^2(x - 3)$.

Don't do this: Solve $f''(x) = 0$ and report every solution as an inflection point.

The correct way: Build the sign chart and keep only candidates where the sign actually flips.

The rusher does this most: the equation $f''(x) = 0$ produces neat solutions, and stopping there feels finished. Students who draw the sign chart before writing any conclusion stop announcing phantom inflection points, because the chart forces the one question that matters: did the sign actually change?

Mistake 2: Skipping Points Where $f''$ Is Undefined

Where it slips in: Functions with fractional powers, such as $x^{1/3}$ or $(x + 1)^{1/3}$.

Don't do this: Look only for zeros of $f''(x)$ and conclude there is no inflection point.

The correct way: Add every point in the domain where $f''(x)$ is undefined to the candidate list, then test the signs around it.

Mistake 3: Calling A Point Outside The Domain An Inflection Point

Where it slips in: Rational functions with vertical asymptotes, such as $\frac{1}{x}$ or $x + \frac{1}{x}$.

Don't do this: Report $x = 0$ as an inflection point because the concavity changes across it.

The correct way: Check that $f(c)$ exists and that $f$ is continuous at $c$, a condition the continuity of a function page explains in full.

Mistake 4: Reading Concavity From The Sign Of $f'$

Where it slips in: Graph-reading questions, and any problem that also asks about increasing and decreasing intervals.

Don't do this: Assume a rising graph is concave up and a falling graph is concave down.

The correct way: Use the sign of $f''(x)$ for concavity and the sign of $f'(x)$ for direction, keeping the first derivative test for turning points only.

The memorizer is most exposed here, because "positive means up" holds for both derivatives and the two rules blur. Label each sign chart with the derivative it came from before reading anything off it.

Practice Problems On Concavity And Points Of Inflection

1. Find the point of inflection of $f(x) = x^3 + 3x^2 - 9x + 1$.

2. Find the intervals of concavity and the points of inflection of $f(x) = x^4 - 6x^2$.

3. Decide whether $f(x) = (x - 2)^4 + 1$ has a point of inflection, and give its concavity.

4. Find $a$ and $b$ so that $f(x) = ax^3 + bx^2$ has a point of inflection at $(1, 2)$.

5. Find the points of inflection of $f(x) = \ln(x^2 + 1)$.

6. Find the point of inflection of $f(x) = (x + 1)^{1/3}$ and say what is unusual about $f''(x)$ there.

Answer to Problem 1: $f''(x) = 6(x + 1)$ changes sign at $x = -1$, so the point of inflection is $(-1, 12)$.

Answer to Problem 2: $f''(x) = 12(x - 1)(x + 1)$; concave up on $(-\infty, -1)$ and $(1, \infty)$, concave down on $(-1, 1)$; points of inflection $(-1, -5)$ and $(1, -5)$.

Answer to Problem 3: $f''(x) = 12(x - 2)^2 \geq 0$, so the function is concave up everywhere and has no point of inflection, even though $f''(2) = 0$.

Answer to Problem 4: $f''(1) = 6a + 2b = 0$ and $f(1) = a + b = 2$ give $a = -1$, $b = 3$, so $f(x) = -x^3 + 3x^2$.

Answer to Problem 5: $f''(x) = \frac{2(1 - x^2)}{(x^2 + 1)^2}$ changes sign at $x = \pm 1$, so the points of inflection are $(-1, \ln 2)$ and $(1, \ln 2)$, with $\ln 2 \approx 0.6931$.

Answer to Problem 6: $f''(x) = -\frac{2}{9(x + 1)^{5/3}}$ is undefined at $x = -1$ but changes sign there, and $f$ is continuous, so $(-1, 0)$ is a point of inflection with a vertical tangent.

Key Takeaways

• Concavity and points of inflection come from the sign of $f''(x)$: positive means concave up, negative means concave down.

• A point of inflection needs a genuine sign change of $f''(x)$ at a point where $f$ is continuous.

• A zero of $f''(x)$ is only a candidate, as $x^4$ shows, and points where $f''(x)$ is undefined are candidates too.

• A concavity change across a vertical asymptote does not create an inflection point.

• A stationary point of inflection has a horizontal tangent but is not a maximum or minimum.

Your Next Move

Work through Practice Problems 1 to 6 with a sign chart for every candidate; if Problem 3 or Problem 6 trips you up, return to the section "Why Is f''(c) = 0 Not Enough For An Inflection Point?" and compare it with $x^4$ and the cube root. Want a live Bhanzu trainer to check your sign charts with you? Book a free demo class.

Read More

• Maxima and minima of a function

• Graphing functions with calculus

• Applications of derivatives

• Tangent line equations