What is the difference between an inflection point and a critical point?

A critical point is where f'(x) = 0 or f' is undefined; it marks where the function has a local maximum or minimum. An inflection point is where f''(x) = 0 (or f'' is undefined) and the second derivative changes sign; it marks where the curve changes concavity. They are found using different derivatives and address different aspects of the function's shape.

Can a function have an inflection point where f''(x) is undefined?

Yes. An inflection point can occur where f'' is undefined if f'' changes sign at that point. For example, f(x) = x^{1/3} has an inflection point at x = 0 even though f''(0) is undefined, because f''(x) changes from negative to positive as x passes through 0.

If f''(x) = 0 at a point, is that point always an inflection point?

No. You must also check that f'' changes sign at that point. If f'' is positive on both sides or negative on both sides, then there is no sign change and the point is not an inflection point, even though f''(x) = 0. For instance, if f''(x) = (x - 1)^2, then f''(1) = 0, but f''(x) > 0 everywhere else, so there is no inflection point at x = 1.

How do inflection points relate to the graph of the derivative f'(x)?

Inflection points of f occur where the graph of f' has a local maximum or minimum. When f''(x) > 0, the graph of f' is increasing (curving upward), and when f''(x) < 0, the graph of f' is decreasing (curving downward). An inflection point of f is the moment when f' transitions from increasing to decreasing or vice versa.

Concavity and Inflection Points

Pinpoint where a function's curve changes from bending upward to downward, revealing the transition point between concave up and concave down.

f''(x) = 0 \text{ and sign changes}

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What each symbol means

What Concavity and Inflection Points takes

Symbol	Meaning
$f$	$f$ is the original function being studied; it is not the same as $f'$ or $f''$ , and confusing these will lead to taking derivatives of the wrong function.
$x$	$x$ is the independent variable (the input to $f(x)$ ), and it is the variable we solve for when we set $f''(x) = 0$ to find inflection points.

When to use it

Use inflection points when you need to find where a function's concavity changes, or where the rate of change of the function transitions from accelerating to decelerating.

Level

Usually taught in: Calculus I

Worked examples

1. Find inflection points of a cubic function

Problem

Find the inflection points of

f(x) = x^3 - 3x^2 + 2x

$f'(x) = 3x^2 - 6x + 2$
Take the first derivative.
$f''(x) = 6x - 6$
Take the second derivative to identify where concavity might change.
$6x - 6 = 0$
Set $f''(x)$ equal to zero to find candidate inflection points.
$x = 1$
Solve for $x$ .
$f''(0) = 6(0) - 6 = -6 < 0$
Check the sign of $f''$ to the left of $x = 1$ ; the function is concave down here.
$f''(2) = 6(2) - 6 = 6 > 0$
Check the sign of $f''$ to the right of $x = 1$ ; the function is concave up here.

Answer: $x = 1$

Since $f''$ changes from negative to positive at $x = 1$ , the curve changes from bending downward to bending upward. Both conditions are satisfied: $f''(1) = 0$ and the sign changes, so $x = 1$ is an inflection point.

2. Quartic function where the second derivative touches zero but does not change sign

Problem

Find all inflection points of

f(x) = x^4 - 4x^3 + 6x^2

$f'(x) = 4x^3 - 12x^2 + 12x$
Take the first derivative.
$f''(x) = 12x^2 - 24x + 12$
Take the second derivative.
$f''(x) = 12(x^2 - 2x + 1) = 12(x - 1)^2$
Factor out 12 and recognize the perfect square trinomial.
$12(x - 1)^2 = 0$
Set $f''(x)$ equal to zero to find candidates.
$x = 1$
Solve for $x$ .
$f''(0) = 12(0 - 1)^2 = 12 > 0$
Check the sign of $f''$ to the left of $x = 1$ .
$f''(2) = 12(2 - 1)^2 = 12 > 0$
Check the sign of $f''$ to the right of $x = 1$ ; the second derivative stays positive.

Answer: $\text{No inflection points}$

Even though $f''(x) = 0$ at $x = 1$ , there is no inflection point there because the second derivative never changes sign—it remains non-negative everywhere. An inflection point requires both $f''(x) = 0$ and a sign change, so this point fails the second condition.

3. Coffee shop revenue with two inflection points

Problem

A coffee shop's daily revenue is

R(t) = -t^4 + 8t^3 - 18t^2 + 16t + 50

(in hundreds of dollars), where

t

is the time in hours after opening. Find the times when the rate of revenue growth changes from accelerating to decelerating.

$R'(t) = -4t^3 + 24t^2 - 36t + 16$
Take the first derivative to find the rate of change of revenue.
$R''(t) = -12t^2 + 48t - 36$
Take the second derivative to identify where the concavity (acceleration of revenue growth) changes.
$R''(t) = -12(t^2 - 4t + 3) = -12(t - 1)(t - 3)$
Factor out $-12$ and then factor the quadratic.
$-12(t - 1)(t - 3) = 0$
Set $R''(t)$ equal to zero to find candidate inflection points.
$t = 1 \text{ or } t = 3$
Solve for the two values of $t$ .
$R''(0) = -12(0 - 1)(0 - 3) = -12(-1)(-3) = -36 < 0$
For $t < 1$ , the second derivative is negative, so revenue growth is decelerating.
$R''(2) = -12(2 - 1)(2 - 3) = -12(1)(-1) = 12 > 0$
For $1 < t < 3$ , the second derivative is positive, so revenue growth is accelerating.
$R''(4) = -12(4 - 1)(4 - 3) = -12(3)(1) = -36 < 0$
For $t > 3$ , the second derivative is negative again, so revenue growth is decelerating.

Answer: $t = 1 \text{ and } t = 3$

The second derivative changes sign at both $t = 1$ (from negative to positive) and $t = 3$ (from positive to negative). This means the rate at which revenue grows accelerates between hour 1 and hour 3, then decelerates after hour 3. Both times are inflection points of the revenue function.

Common mistakes

Where Concavity and Inflection Points usually goes wrong

The mistake	Why it is wrong	The fix
I found that $f''(x) = 0$ at $x = 2$ , so $x = 2$ is an inflection point.	Finding where $f''(x) = 0$ is necessary but not sufficient; you must also verify that the second derivative changes sign at that point.	Check the sign of $f''$ on both sides of the candidate point. Only if the sign changes from positive to negative (or vice versa) is it a true inflection point.
When I found $f''(x) = 0$ at $x = 1$ , I checked that $f''(0) = 2 > 0$ and $f''(2) = 3 > 0$ , so I concluded that $x = 1$ is an inflection point.	If $f''$ has the same sign on both sides of the point where $f''(x) = 0$ , then the concavity does not change, and the point is not an inflection point.	Remember that both conditions are required: $f''(x) = 0$ and a sign change in $f''$ . If there is no sign change, the point is not an inflection point, even if $f''(x) = 0$ .
I found the inflection points by solving $f'(x) = 0$ , then checking where $f''$ changes sign.	Solving $f'(x) = 0$ gives critical points (local extrema), not inflection points; inflection points are found by solving $f''(x) = 0$ .	Set the second derivative equal to zero, $f''(x) = 0$ , to find candidate inflection points. Use the First Derivative Test or Second Derivative Test only for critical points where $f'(x) = 0$ .

Tips and when to use something else

Inflection points and critical points are different concepts: critical points (where $f'(x) = 0$ or $f'$ is undefined) mark local maxima and minima, while inflection points (where $f''(x) = 0$ with a sign change) mark where concavity changes. Use the First Derivative Test or Second Derivative Test for critical points, not inflection points.
Always verify a sign change when you find a point where $f''(x) = 0$ . A perfect square factor like $(x - a)^2$ in $f''(x)$ is a red flag—check both sides to confirm the sign actually changes.
If $f''(x)$ is always positive or always negative (for example, $f''(x) = x^2 + 1$ ), the function has no inflection points because concavity never changes.
Inflection points can also occur where $f''(x)$ is undefined (such as a corner or cusp), provided $f''$ changes sign at that point. Do not assume that $f''(x) = 0$ is the only place an inflection point can occur.

Related formulas

Critical PointsCritical points mark where a function's derivative is zero or undefined, revealing local extrema and where the graph changes direction.Second Derivative TestThe Second Derivative Test determines whether a critical point is a local minimum or maximum using the concavity at that point.Higher Order DerivativesHigher order derivatives describe how the rate of change is changing; use them to find concavity and acceleration in functions.

Frequently asked questions

What is the difference between an inflection point and a critical point?: A critical point is where $f'(x) = 0$ or $f'$ is undefined; it marks where the function has a local maximum or minimum. An inflection point is where $f''(x) = 0$ (or $f''$ is undefined) and the second derivative changes sign; it marks where the curve changes concavity. They are found using different derivatives and address different aspects of the function's shape.
Can a function have an inflection point where $f''(x)$ is undefined?: Yes. An inflection point can occur where $f''$ is undefined if $f''$ changes sign at that point. For example, $f(x) = x^{1/3}$ has an inflection point at $x = 0$ even though $f''(0)$ is undefined, because $f''(x)$ changes from negative to positive as $x$ passes through $0$ .
If $f''(x) = 0$ at a point, is that point always an inflection point?: No. You must also check that $f''$ changes sign at that point. If $f''$ is positive on both sides or negative on both sides, then there is no sign change and the point is not an inflection point, even though $f''(x) = 0$ . For instance, if $f''(x) = (x - 1)^2$ , then $f''(1) = 0$ , but $f''(x) > 0$ everywhere else, so there is no inflection point at $x = 1$ .
How do inflection points relate to the graph of the derivative $f'(x)$ ?: Inflection points of $f$ occur where the graph of $f'$ has a local maximum or minimum. When $f''(x) > 0$ , the graph of $f'$ is increasing (curving upward), and when $f''(x) < 0$ , the graph of $f'$ is decreasing (curving downward). An inflection point of $f$ is the moment when $f'$ transitions from increasing to decreasing or vice versa.

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Reviewed 2026-09-18