← Skill tree MathScape MATH161

Concavity and Inflection Points

Reference: Stewart §3.3 • Chapter: 3 • Section: 3

Navigation: Wiki Home > Skills > Concavity and Inflection Points

Beyond Rising and Falling: How the Curve Bends

Two functions can both be increasing, yet look completely different. One might curve upward like a rocket accelerating skyward; the other might curve downward like a ball thrown upward that's slowing down. The first derivative tells us direction; the second derivative tells us shape.

Concavity describes how a curve bends. Understanding concavity is essential for accurate graph sketching and has deep physical meaning: in motion problems, concavity relates to whether you're speeding up or slowing down.

Prerequisite Map

Quick Reference

Property	Value
Concept	Graph Shape Analysis
Chapter	Chapter 4, Section 3
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

Definitions of Concavity

Concave Upward (CU): The graph lies above all of its tangent lines on an interval.

The curve "holds water" like a bowl
The slope $f'(x)$ is increasing
Sometimes called "concave up" or "convex"

Concave Downward (CD): The graph lies below all of its tangent lines on an interval.

The curve "spills water" like an umbrella
The slope $f'(x)$ is decreasing
Sometimes called "concave down"

Visual Comparison

Concave Upward (CU)           Concave Downward (CD)
    f'' > 0                        f'' < 0

        ╱                          ╲
       ╱                            ╲
      ╱  curve above tangent         ╲  curve below tangent
     ╱────────                  ──────╲
    ╱                                  ╲

   "holds water"               "spills water"
   slope increasing            slope decreasing

The Concavity Test

$$\boxed{\begin{aligned} &\text{If } f''(x) > 0 \text{ on an interval } I, \text{ then } f \text{ is } \textbf{concave upward} \text{ on } I. \\ &\text{If } f''(x) < 0 \text{ on an interval } I, \text{ then } f \text{ is } \textbf{concave downward} \text{ on } I. \end{aligned}}$$

Why This Works

The key insight is that $f''(x)$ is the derivative of $f'(x)$.

If $f'' > 0$, then $f'$ is increasing (the slope gets steeper)
An increasing slope means the curve bends upward

This is exactly like the I/D Test, but applied to $f'$ instead of $f$.

Inflection Points

Definition: A point $P$ on a curve $y = f(x)$ is an inflection point if $f$ is continuous there and the concavity changes at $P$.

At an inflection point:

The graph crosses its tangent line
The curve changes from "holding water" to "spilling water" (or vice versa)
$f''(x)$ changes sign

Finding Inflection Points

Find where $f''(x) = 0$ or $f''(x)$ is undefined
Check that $f''(x)$ actually changes sign at each candidate
Verify that $f$ is continuous at that point

Warning: $f''(c) = 0$ does NOT guarantee an inflection point. You must verify a sign change.

Situation	Example	Inflection point?
$f''$ changes from $+$ to $-$	$f(x) = -x^3$ at $x = 0$	Yes
$f''$ changes from $-$ to $+$	$f(x) = x^3$ at $x = 0$	Yes
$f'' = 0$ but no sign change	$f(x) = x^4$ at $x = 0$	No

Physical Interpretation

In motion, if $s(t)$ is position:

$s'(t) = v(t)$ is velocity
$s''(t) = a(t)$ is acceleration

$s'(t)$	$s''(t)$	Concavity	Motion
$+$	$+$	CU	Moving right, speeding up
$+$	$-$	CD	Moving right, slowing down
$-$	$-$	CU	Moving left, speeding up
$-$	$+$	CD	Moving left, slowing down

An inflection point in position corresponds to the moment when acceleration changes direction.

Practice Problems

Level 1 Identifying Concavity from a Graph

A function $f$ has the following properties on $[-2, 4]$:

$f''(x) > 0$ on $(-2, 1)$
$f''(1) = 0$
$f''(x) < 0$ on $(1, 4)$

Where is $f$ concave upward?
Where is $f$ concave downward?
Find the x-coordinate of any inflection points.

Thought Process

Directly apply the Concavity Test: $f'' > 0$ means concave up, $f'' < 0$ means concave down. An inflection point occurs where $f''$ changes sign.

Show Answer

(a) $f$ is concave upward where $f'' > 0$: on $(-2, 1)$

(b) $f$ is concave downward where $f'' < 0$: on $(1, 4)$

(c) At $x = 1$, $f''$ changes from positive to negative, so $x = 1$ is an inflection point.

Level 2 Finding Concavity Intervals

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

Thought Process

Compute $f'(x)$, then $f''(x)$
Find where $f''(x) = 0$
Test the sign of $f''$ in each interval
Identify where concavity changes

Show Answer

Step 1: Find $f''(x)$. $$f'(x) = 3x^2 - 12x + 9$$ $$f''(x) = 6x - 12 = 6(x - 2)$$

Step 2: Find where $f''(x) = 0$. $$6(x - 2) = 0 \Rightarrow x = 2$$

Step 3: Sign chart for $f''(x)$.

Interval	$f''(x) = 6(x-2)$	Concavity
$(-\infty, 2)$	$-$	Concave down
$(2, \infty)$	$+$	Concave up

Step 4: Check for inflection point. At $x = 2$: $f''$ changes from $-$ to $+$ → inflection point

The y-coordinate: $f(2) = 8 - 24 + 18 + 1 = 3$

Answer:

Concave down on $(-\infty, 2)$
Concave up on $(2, \infty)$
Inflection point at $(2, 3)$

Level 3 Polynomial with Multiple Inflection Points

Find the intervals of concavity and all inflection points of $f(x) = x^4 - 4x^3 + 6$.

Thought Process

The second derivative of a degree-4 polynomial is degree-2 (quadratic), so there could be 0, 1, or 2 places where $f'' = 0$. Factor $f''$ and check for sign changes at each root.

Show Answer

Step 1: Find $f''(x)$. $$f'(x) = 4x^3 - 12x^2$$ $$f''(x) = 12x^2 - 24x = 12x(x - 2)$$

Step 2: Find where $f''(x) = 0$. $$12x(x - 2) = 0 \Rightarrow x = 0 \text{ or } x = 2$$

Step 3: Sign chart for $f''(x) = 12x(x-2)$.

Interval	$12x$	$(x-2)$	$f''(x)$	Concavity
$(-\infty, 0)$	$-$	$-$	$+$	CU
$(0, 2)$	$+$	$-$	$-$	CD
$(2, \infty)$	$+$	$+$	$+$	CU

Step 4: Identify inflection points.

At $x = 0$: $f''$ changes from $+$ to $-$ → inflection point
At $x = 2$: $f''$ changes from $-$ to $+$ → inflection point

Step 5: Find y-coordinates. $$f(0) = 0 - 0 + 6 = 6$$ $$f(2) = 16 - 32 + 6 = -10$$

Answer:

Concave up on $(-\infty, 0) \cup (2, \infty)$
Concave down on $(0, 2)$
Inflection points at $(0, 6)$ and $(2, -10)$

Level 4 When $f'' = 0$ Is Not an Inflection Point

Let $f(x) = x^4$.

Show that $f''(0) = 0$.
Prove that $(0, 0)$ is NOT an inflection point.
What type of point is $(0, 0)$ on the graph?

Thought Process

Just because $f''(c) = 0$ doesn't mean there's an inflection point. We need to check whether $f''$ actually changes sign at $c$. If it doesn't change sign, there's no inflection point.

Show Answer

(a) Showing $f''(0) = 0$: $$f(x) = x^4$$ $$f'(x) = 4x^3$$ $$f''(x) = 12x^2$$ $$f''(0) = 12(0)^2 = 0 \checkmark$$

(b) Proving $(0, 0)$ is NOT an inflection point:

For an inflection point, $f''$ must change sign. Check:

For $x < 0$: $f''(x) = 12x^2 > 0$ (positive)
For $x > 0$: $f''(x) = 12x^2 > 0$ (positive)

Since $f''(x) > 0$ on both sides of $x = 0$, there is no sign change.

Therefore, $(0, 0)$ is not an inflection point.

(c) What is $(0, 0)$?

From the First Derivative Test:

$f'(x) = 4x^3 < 0$ for $x < 0$
$f'(x) = 4x^3 > 0$ for $x > 0$

$f'$ changes from $-$ to $+$ at $x = 0$, so $(0, 0)$ is a local minimum.

The graph is concave up everywhere, and $(0, 0)$ is the bottom of the "bowl."

Level 5 Graph Sketching from Derivative Information

Sketch a possible graph of a function $f$ satisfying all of the following conditions:

$f(0) = 0$, $f(2) = 3$, $f(4) = 6$
$f'(x) > 0$ for $0 < x < 4$
$f'(x) < 0$ for $x < 0$ and $x > 4$
$f''(x) > 0$ for $x < 2$
$f''(x) < 0$ for $x > 2$

Identify the local extrema and inflection points.

Thought Process

Organize the information:

Three points we must pass through
$f' > 0$ on $(0,4)$ means increasing there; $f' < 0$ elsewhere means decreasing
$f'' > 0$ for $x < 2$ means concave up; $f'' < 0$ for $x > 2$ means concave down
The switch in $f'$ sign at $x = 0$ and $x = 4$ tells us about extrema
The switch in $f''$ sign at $x = 2$ tells us about inflection points

Show Answer

Analysis:

Property	Condition	Conclusion
$f' < 0$ for $x < 0$	Decreasing	approaching a minimum
$f' > 0$ for $0 < x < 4$	Increasing	rising from minimum
$f' < 0$ for $x > 4$	Decreasing	falling from maximum
$f'' > 0$ for $x < 2$	Concave up	curving upward
$f'' < 0$ for $x > 2$	Concave down	curving downward

Extrema:

At $x = 0$: $f'$ changes from $-$ to $+$ → Local minimum at $(0, 0)$
At $x = 4$: $f'$ changes from $+$ to $-$ → Local maximum at $(4, 6)$

Inflection point:

At $x = 2$: $f''$ changes from $+$ to $-$ → Inflection point at $(2, 3)$

Sketch:

    y
    │                  ●(4,6) local max
  6 │                 ╱ ╲
    │                ╱   ╲
    │               ╱     ╲
    │              ╱       ╲
  3 │        ●(2,3) inflection
    │       ╱
    │      ╱
    │     ╱
    │    ╱
  0 │───●(0,0)───────────────── x
    │  ╱    local min
    │ ╱
    │╱

The curve:

Falls toward $(0,0)$ from the left (decreasing, concave up)
Rises from $(0,0)$ to $(4,6)$ (increasing)
Bends upward until $x = 2$, then bends downward
Falls after $(4,6)$ (decreasing, concave down)

Conceptual Questions (CCI-Style)

Conceptual Interpreting Concavity

A population $P(t)$ is growing. The government announces that "the rate of growth is slowing down." In terms of derivatives, which of the following must be true?

(A) $P'(t) < 0$ (B) $P''(t) < 0$ (C) $P'(t) > 0$ and $P''(t) > 0$ (D) $P'(t) > 0$ and $P''(t) < 0$

Thought Process

"Growing" means $P$ is increasing, so $P' > 0$. "Rate of growth is slowing" means the rate of change $P'$ is decreasing, which means $P'' < 0$.

Show Answer

Answer: (D)

"Population is growing" → $P(t)$ is increasing → $P'(t) > 0$
"Rate of growth is slowing" → $P'(t)$ is decreasing → $P''(t) < 0$

Together: $P'(t) > 0$ and $P''(t) < 0$

This is the concave-down portion of a growth curve: the population is still increasing but the growth is decelerating.

Conceptual Inflection Point Misconception

True or False: If $f''(c) = 0$, then $c$ is an inflection point of $f$.

Thought Process

This is a common misconception. The condition $f''(c) = 0$ is necessary for an inflection point (at points where $f''$ is defined), but it's not sufficient. We need $f''$ to actually change sign.

Show Answer

False

Counterexample: $f(x) = x^4$

We have $f''(x) = 12x^2$, so $f''(0) = 0$.

But $f''(x) = 12x^2 > 0$ for all $x \neq 0$, so $f''$ does not change sign at $x = 0$.

Therefore, $x = 0$ is not an inflection point (it's actually a local minimum).

The correct statement: If $f''(c) = 0$ and $f''$ changes sign at $c$, then $c$ is an inflection point.

Mastery Checklist

[ ] I can define concave upward and concave downward
[ ] I can apply the Concavity Test using $f''(x)$
[ ] I can find intervals of concavity by analyzing the sign of $f''$
[ ] I understand that $f''(c) = 0$ does not guarantee an inflection point
[ ] I can identify inflection points by checking for sign changes in $f''$
[ ] I can interpret concavity in physical contexts (acceleration, rates)

Mental Model

The Bending Pipe:

Imagine bending a flexible pipe. When you bend it into a U-shape (opening upward), that's concave up, $f'' > 0$. When you flip it to an arch (opening downward), that's concave down, $f'' < 0$.

An inflection point is where you're actively reversing the bend: the moment the pipe transitions from one curve direction to the other. At that instant, the pipe is momentarily straight (tangent line), and crosses from one side of that line to the other.

Connections

Looking back:

The Concavity Test is the I/D Test applied to $f'$ instead of $f$
Computing $f''$ requires fluent differentiation

Looking ahead:

The Second Derivative Test uses concavity to classify extrema
Curve sketching combines concavity with monotonicity

Physical interpretation:

In kinematics, $s''(t) = a(t)$ is acceleration
Positive acceleration doesn't mean moving right; it means velocity is increasing

Previous	Up	Next
First Derivative Test	Skills Index	Second Derivative Test

Last updated: 2026-01-22