Determining Concavity of Functions over Their Domains

AP Calculus · Analytical Applications of Differentiation

Determining Concavity

Which way the curve bends, decided by $f''$ alone — and independent of whether it rises or falls.

key idea$f''\gt 0\Rightarrow\text{concave up};\quad f''\lt 0\Rightarrow\text{concave down}$

Reading the graph

Concavity describes which way a curve bends, and it is decided by the sign of $f''$ alone. It has nothing to do with whether the function is rising or falling — those are separate questions with separate answers.

concave up — holds water

concave down — sheds water

inflection at $x=0$

Concave up bends upward like a cup; concave down bends downward like a cap. An inflection point is where the bending switches — and at such a point $f''$ must actually change sign.

sign chart — $f''(x)=6x$ for $f(x)=x^{3}$

	$x\lt 0$	$x=0$	$x\gt 0$
$f''=6x$	−	0	+
verdict	concave down	inflection	concave up

$f'$	$f''$	Shape of the curve	In words
$+$	$+$	rising, bending up	increasing at an increasing rate
$+$	$-$	rising, bending down	increasing at a decreasing rate
$-$	$+$	falling, bending up	decreasing at an increasing rate
$-$	$-$	falling, bending down	decreasing at a decreasing rate

All four combinations occur. The right-hand column is the phrasing AP questions use — parse it as direction from $f'$ and rate change from $f''$.

Worked examples

EX 1

Find the concavity intervals

$f(x)=x^{4}-6x^{2}$

Find where $f(x)=x^{4}-6x^{2}$ is concave up and concave down.

Differentiate twice

$f'(x)=4x^{3}-12x$ and $f''(x)=12x^{2}-12=12\left(x^{2}-1\right)$.

Find where $f''=0$

$x=\pm1$. These are the only places concavity can switch.

Test each interval

$f''(-2)=36\gt 0$; $f''(0)=-12\lt 0$; $f''(2)=36\gt 0$.

Translate

Concave up on $(-\infty,-1)$ and $(1,\infty)$; concave down on $(-1,1)$.

Identify the inflection points

$f''$ changes sign at both $x=-1$ and $x=1$, so both are inflection points.

EX 2

Zero second derivative, no inflection

$f(x)=x^{4}$

Does $f(x)=x^{4}$ have an inflection point at $x=0$?

Differentiate twice

$f'(x)=4x^{3}$ and $f''(x)=12x^{2}$.

Check $f''(0)$

$f''(0)=0$, so $x=0$ is a candidate.

Test the sign on each side

$f''(-1)=12\gt 0$ and $f''(1)=12\gt 0$.

Apply the criterion

$f''$ does not change sign — it is positive on both sides.

Conclude

No inflection point. The curve is concave up throughout; it merely flattens momentarily. $f''=0$ is necessary but not sufficient.

EX 3

Parse the exam phrasing

“increasing at a decreasing rate”

A quantity is described as “increasing at a decreasing rate”. What are the signs of $f'$ and $f''$?

Split the phrase

“Increasing” describes the quantity; “at a decreasing rate” describes its rate of change.

First clause

The quantity rises, so $f'\gt 0$.

Second clause

The rate is falling, so $f'$ is decreasing, which means $f''\lt 0$.

Name the shape

Rising and concave down — row 2 of the table.

The common error

Reading “decreasing” as applying to the quantity itself. It applies to the rate. The quantity is still growing.

EX 4

Concavity where $f''$ is undefined

$f(x)=x^{1/3}$

Discuss the concavity of $f(x)=x^{1/3}$.

Differentiate twice

$f'(x)=\tfrac13 x^{-2/3}$ and $f''(x)=-\tfrac29 x^{-5/3}$.

Find where $f''$ is zero or undefined

Never zero; undefined at $x=0$.

Test each side

For $x\gt 0$: $x^{-5/3}\gt 0$, so $f''\lt 0$ — concave down. For $x\lt 0$: $x^{-5/3}\lt 0$, so $f''\gt 0$ — concave up.

Conclude

Concave up on $(-\infty,0)$, concave down on $(0,\infty)$.

Is $x=0$ an inflection point?

The concavity does change there and $f(0)=0$ is defined, so yes — even though $f''$ does not exist at that point.

Watch out

Concavity is independent of direction

A falling curve can be concave up. Read $f'$ for direction and $f''$ for bending, and never let one answer the other's question.

$f''=0$ does not guarantee an inflection

The sign must actually change. $x^{4}$ at the origin is the standard counterexample.

Inflection points can occur where $f''$ is undefined

As in Example 4. Candidates are where $f''=0$ or $f''$ fails to exist, provided $f$ itself is defined.

Parse the two-clause phrasing carefully

“Increasing at a decreasing rate” means $f'\gt 0$ and $f''\lt 0$. The second adjective never describes the quantity.

Course map. Topic 5.6 · LO FUN-4.A — justify conclusions about the behaviour of a function based on the behaviour of its derivatives (EK FUN-4.A.4, FUN-4.A.5).