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Function Inflection Points

Prerequisite: calculus — this chapter uses the second derivative

f''(x)

to study concavity and inflection points. If you have not yet studied derivatives, consider learning “the concept of the derivative” first, or skip the “second/third sufficient conditions” sections for now.

Basic Idea

An inflection point is a special point on a function’s graph where the concavity changes. It divides the graph into intervals of different concavity.

Definition

Definition of an Inflection Point

Let $f(x)$ be continuous on a neighborhood of $x_0$ . If there exists $\delta > 0$ such that:

On $(x_0 - \delta, x_0)$ the graph of $f(x)$ is concave and on $(x_0, x_0 + \delta)$ it is convex
Or on $(x_0 - \delta, x_0)$ it is convex and on $(x_0, x_0 + \delta)$ it is concave

then $(x_0, f(x_0))$ is called an inflection point of $f(x)$ .

几何解释

An inflection point is where the graph changes its bending direction.

符号说明

Symbol	Type	Reading/Explanation	Meaning in This Article
$\delta$	Greek letter	Delta	A sufficiently small positive number

How to Find Inflection Points

Necessary Condition

定理1

If $f(x)$ is twice differentiable at $x_0$ and $(x_0, f(x_0))$ is an inflection point, then $f''(x_0) = 0$ .

Note: $f''(x_0) = 0$ is necessary but not sufficient for an inflection point.

Sufficient Conditions

First Sufficient Condition

定理2

Let $f(x)$ be continuous on a neighborhood of $x_0$ and twice differentiable on a punctured neighborhood, with $f''(x_0) = 0$ or $f''(x_0)$ undefined:

If $f''(x) < 0$ for $x < x_0$ and $f''(x) > 0$ for $x > x_0$ , then $(x_0, f(x_0))$ is an inflection point
If $f''(x) > 0$ for $x < x_0$ and $f''(x) < 0$ for $x > x_0$ , then $(x_0, f(x_0))$ is an inflection point

Second Sufficient Condition

定理3

Let $f(x)$ be three times differentiable at $x_0$ with $f''(x_0) = 0$ :

If $f'''(x_0) \neq 0$ , then $(x_0, f(x_0))$ is an inflection point
If $f'''(x_0) = 0$ , further analysis is needed

Steps to Locate Inflection Points

Basic Steps

Find the second derivative: compute $f''(x)$
Find critical points: solve $f''(x) = 0$ or find where $f''(x)$ is undefined
Analyze the sign: examine the sign of $f''(x)$ on both sides of each critical point
Determine inflection points: if the sign changes, the point is an inflection point

Caveats

A point where the second derivative vanishes is not necessarily an inflection point
A point where $f''(x)$ is undefined may be an inflection point
Check that the function is continuous at the point

Worked Examples

Example 1: Find the inflection points of $f(x) = x^3 - 3x^2 + 2$ .

Solution:

$f'(x) = 3x^2 - 6x$
$f''(x) = 6x - 6 = 6(x-1)$
Setting $f''(x) = 0$ gives $x = 1$
For $x < 1$ , $f''(x) < 0$ , so the function is concave
For $x > 1$ , $f''(x) > 0$ , so the function is convex
Therefore $(1, f(1)) = (1, 0)$ is an inflection point

Example 2: Find the inflection points of $f(x) = x^4$ .

Solution:

$f'(x) = 4x^3$
$f''(x) = 12x^2$
Setting $f''(x) = 0$ gives $x = 0$
For $x < 0$ , $f''(x) > 0$ , so the function is convex
For $x > 0$ , $f''(x) > 0$ , so the function is convex
The sign does not change, so $x = 0$ is not an inflection point

This example shows that a zero second derivative does not always give an inflection point—you must check whether the sign changes.

Relationship Between Inflection Points and Extrema

Important Properties

An inflection point is not necessarily an extremum point: the function value at an inflection point need not be a local maximum or minimum
An extremum point is not necessarily an inflection point: extrema and concavity changes are different questions; check each separately
Geometric character: when a tangent exists and the usual smoothness conditions hold, the curve near an inflection point changes its bending direction relative to the tangent

Example

For $f(x) = x^3$ :

$f'(x) = 3x^2$ , so $f'(0) = 0$ and $x = 0$ is a stationary point
$f''(x) = 6x$ , so $f''(0) = 0$ , but the sign changes
Hence $(0,0)$ is a stationary point (also called a saddle point in the one-variable setting) and an inflection point, but not an extremum point

Exercises

Exercise 1

Find the inflection points of $f(x) = x^3 - 6x^2 + 9x + 1$ .

Reference Answer(3 个标签)

inflection pointsecond derivativepolynomial

Solution Approach: Differentiate twice, find critical points, and analyze sign changes.

Detailed Steps:

$f'(x) = 3x^2 - 12x + 9$
$f''(x) = 6x - 12 = 6(x-2)$
Setting $f''(x) = 0$ gives $x = 2$
For $x < 2$ , $f''(x) < 0$ , so the function is concave
For $x > 2$ , $f''(x) > 0$ , so the function is convex
$f(2) = 8 - 24 + 18 + 1 = 3$

Answer: The inflection point is $(2, 3)$ .

Exercise 2

Find the inflection points of $f(x) = \frac{x^3}{x^2 + 1}$ .

Reference Answer(3 个标签)

inflection pointsecond derivativerational function

Solution Approach: Differentiate twice, find critical points, and analyze sign changes.

Detailed Steps:

$f'(x) = \frac{3x^2(x^2 + 1) - x^3 \cdot 2x}{(x^2 + 1)^2} = \frac{x^2(x^2 + 3)}{(x^2 + 1)^2}$
$f''(x) = \frac{(2x(x^2 + 3) + x^2 \cdot 2x)(x^2 + 1)^2 - x^2(x^2 + 3) \cdot 2(x^2 + 1) \cdot 2x}{(x^2 + 1)^4}$
Simplifying: $f''(x) = \frac{2x(3 - x^2)}{(x^2 + 1)^3}$
Setting $f''(x) = 0$ gives $x = 0, \pm \sqrt{3}$
Analyzing the sign changes:
- $x = 0$ : sign changes from negative to positive, so it is an inflection point
- $x = \pm \sqrt{3}$ : sign changes from positive to negative, so they are inflection points

Answer: The inflection points are $(0, 0)$ , $(\sqrt{3}, \frac{3\sqrt{3}}{4})$ , $(-\sqrt{3}, -\frac{3\sqrt{3}}{4})$ .

Exercise 3

Find the inflection points of $f(x) = \sin x$ .

Reference Answer(3 个标签)

inflection pointsecond derivativetrigonometric function

Solution Approach: Differentiate twice, find critical points, and analyze sign changes.

Detailed Steps:

$f'(x) = \cos x$
$f''(x) = -\sin x$
Setting $f''(x) = 0$ gives $\sin x = 0$ , i.e. $x = k\pi$ ( $k$ an integer)
Analyzing sign changes:
- As $x$ moves from $(2k-1)\pi$ to $2k\pi$ , $f''(x)$ changes from positive to negative
- As $x$ moves from $2k\pi$ to $(2k+1)\pi$ , $f''(x)$ changes from negative to positive
Therefore every $x = k\pi$ is an inflection point

Answer: The inflection points are $(k\pi, 0)$ , where $k$ is an integer.

Exercise 4

Suppose $\lim\limits_{x\to1} \frac{f(x)}{\ln x}=1$ and $f(x)$ is twice differentiable at $x=1$ . Determine whether $x=1$ is an inflection point of $f(x)$ .

Reference Answer(2 个标签)

inflection pointlimit

Solution Approach: First use the limit to determine $f(1)$ and $f'(1)$ , then check whether the hypotheses are enough to fix $f''(1)$ and its sign on both sides.

Detailed Steps:

From $\lim\limits_{x\to1} \frac{f(x)}{\ln x}=1$ , we have $f(x)\sim \ln x$ as $x\to1$
As $x\to1$ , $\ln x\to0$ , so $\lim\limits_{x\to1} f(x)=0$ , i.e. $f(1)=0$
By the definition of the derivative: $f'(1)=\lim_{x\to1}\frac{f(x)-f(1)}{x-1} =\lim_{x\to1}\frac{f(x)}{\ln x}\cdot\frac{\ln x}{x-1}=1$
But the hypotheses are not enough to determine the second derivative. Take $f_c(x)=\ln x+c(x-1)^2\quad (x>0),\qquad f_c(1)=0,$ which satisfies the hypotheses for every constant $c$ , with $f_c''(1)=2c-1$ .
For example, when $c=0$ , $f_c''(x)=-1/x^2<0$ , so $x=1$ is not an inflection point; when $c=\frac12$ , $f_c''(x)=1-1/x^2$ changes sign around $x=1$ , so $x=1$ is an inflection point.

Answer: From the given hypotheses we cannot determine whether $x=1$ is an inflection point.

Exercise 5

Adapted from Question 3 of the 2023 National Postgraduate Entrance Exam (Math I)

Let $y=f(x)$ be defined by the parametric equations $\begin{cases}x=2t+|t|\\y=|t|\sin t\end{cases}$ . Determine whether $x=0$ is an inflection point of $f(x)$ .

Reference Answer(3 个标签)

inflection pointparametric equationsecond derivative

Solution Approach: Use the parametric differentiation formulas and analyze the sign change of the second derivative.

Detailed Steps:

For $t\geq0$ : $x=3t$ , $y=t\sin t$ , so $y=\frac{x}{3}\sin\frac{x}{3}$
For $t<0$ : $x=t$ , $y=-t\sin t$ , so $y=-x\sin x$
Hence $f(x)=\begin{cases}\frac{x}{3}\sin\frac{x}{3}, & x\geq0\\-x\sin x, & x<0\end{cases}$
First derivative: $f'(x)=\begin{cases}\frac{1}{3}\sin\frac{x}{3}+\frac{x}{9}\cos\frac{x}{3}, & x\geq0\\-\sin x-x\cos x, & x<0\end{cases}$
Second derivative: $f''(x)=\begin{cases}\frac{2}{9}\cos\frac{x}{3}-\frac{x}{27}\sin\frac{x}{3}, & x\geq0\\-2\cos x+x\sin x, & x<0\end{cases}$
At $x=0$ $x = 0$ :
- Left derivative: $f''_-(0)=-2\cos 0+0\cdot\sin 0=-2$
- Right derivative: $f''_+(0)=\frac{2}{9}\cos 0-0\cdot\sin 0=\frac{2}{9}$
Since the left and right second derivatives differ, $f''(0)$ does not exist
Sign change:
- For $x<0$ , $f''(x)<0$ (concave)
- For $x>0$ , $f''(x)>0$ (convex)

Answer: $x=0$ is an inflection point of $f(x)$ .

Summary

Symbols Used in This Article

Symbol	Type	Reading/Explanation	Meaning in This Article
$f(x)$	Math symbol	f of x	A function of variable $x$
$x_0$	Math symbol	x zero	The abscissa of the inflection point
$\delta$	Greek letter	Delta	A sufficiently small positive number defining a neighborhood
$f'(x)$	Math symbol	f prime of x	The first derivative of the function
$f''(x)$	Math symbol	f double prime of x	The second derivative of the function
$f'''(x)$	Math symbol	f triple prime of x	The third derivative of the function

English–Chinese Glossary

English term	Chinese term	Phonetic	Explanation
inflection point	拐点	/ɪnˈflekʃən pɔɪnt/	A point where the graph changes its concavity
concavity/convexity	凹凸性	/ˌkɒnkəˈvɪti/ /ˌkɒnvekˈsɪti/	The bending direction of a function’s graph
concave function	凹函数	/kɒnˈkeɪv ˈfʌŋkʃən/	A function whose graph bends downward
convex function	凸函数	/ˈkɒnveks ˈfʌŋkʃən/	A function whose graph bends upward
second derivative	二阶导数	/ˈsekənd dɪˈrɪvətɪv/	The derivative of the first derivative
third derivative	三阶导数	/θɜːd dɪˈrɪvətɪv/	The derivative of the second derivative
critical point	临界点	/ˈkrɪtɪkəl pɔɪnt/	A point where the second derivative is zero or undefined
neighborhood	邻域	/ˈneɪbəhʊd/	An open interval containing a point
deleted neighborhood	去心邻域	/dɪˈliːtɪd ˈneɪbəhʊd/	A neighborhood with the center point removed
saddle point	鞍点	/ˈsædl pɔɪnt/	A stationary point that is neither a maximum nor a minimum

Convexity and Concavity of Functions

Concave and convex functions, second-derivative tests, and applications such as Jensen's inequality, with worked examples.

Special Function Forms

Composite, inverse, piecewise, implicit, and parametric functions—definitions, quick methods, and practice.

Function Inflection Points

Basic Idea

蓝色曲线：f(x) = x³

绿色曲线：f(x) = x³ - 3x² + 2

Definition

How to Find Inflection Points

Necessary Condition

Sufficient Conditions

First Sufficient Condition

Second Sufficient Condition

Steps to Locate Inflection Points

Basic Steps

Caveats

Worked Examples

Relationship Between Inflection Points and Extrema

Important Properties

Example

Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Summary

Symbols Used in This Article

English–Chinese Glossary