Use second derivatives and interpret concavity

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Calculus

Using the rate of change of a function’s gradient to understand its shape

The first derivative describes how a function changes; the second derivative describes how its gradient changes. This gives a practical way to analyse a graph’s shape and interpret changing rates in a context. You will use familiar differentiation rules, sign analysis, and a graphing calculator as a check. The method applies on intervals where the function and its second derivative are defined.

What you will learn

Find a second derivative by differentiating a function twice.
Use the sign of the second derivative to describe concavity on an interval.
Identify a possible point of inflection and check whether concavity changes there.
Connect algebraic results with a graph, a numerical check, and a contextual interpretation.

1. Prerequisite: gradients and derivatives

A derivative gives the gradient of a function at each input. If a function is written as

y=f(x)

, its first derivative is written as

f'(x)

or

\frac{dy}{dx}

. To find the second derivative, differentiate the first derivative again:

f''(x)=\frac{d}{dx}\bigl(f'(x)\bigr)

. It is also written as

\frac{d^2y}{dx^2}

.

For example, the power rule says that the derivative of

x^n

is

nx^{n-1}

. Applying it twice to a polynomial is usually enough to find its second derivative. Constants differentiate to zero, and each term can be differentiated separately.

Think of a moving object whose position is

s(t)

. Its first derivative gives its rate of change of position, while its second derivative tells how that rate is changing. This same idea applies to other contexts, provided the variables and units are stated.

f''(x)=\frac{d}{dx}\bigl(f'(x)\bigr)

Differentiate the original function to find $f'(x)$ , then differentiate again to find $f''(x)$ .
The second derivative describes change in gradient, not the function’s height.

2. Concavity from the sign of the second derivative

On an interval where

f''(x)>0

, the gradient

f'(x)

is increasing. The graph bends upwards and is called concave up on that interval. A useful visual cue is that the graph lies below its tangent lines nearby, though the sign test is the reliable method.

On an interval where

f''(x)<0

, the gradient is decreasing. The graph bends downwards and is called concave down. The function itself may still be increasing or decreasing: concavity describes how its gradient changes, not whether its output rises or falls.

To analyse concavity, find

f''(x)

, solve

f''(x)=0

where possible, and also note any points where f'' is undefined or the function is not defined. These values divide the domain into intervals. Test the sign of f'' in each interval. A sign chart or a few test values can make the reasoning clear.

A point of inflection is a point on the graph where concavity changes. A zero of f'' is only a candidate: check the sign of f'' on both sides. If the signs do not change, there is no change in concavity there. The point must also be on the graph, so the function must be defined at the input.

f''(x)>0\Rightarrow\text{concave up},\qquad f''(x)<0\Rightarrow\text{concave down}

$f''(x)>0$ means concave up; $f''(x)<0$ means concave down.
A candidate point of inflection requires a change in concavity, not simply $f''(x)=0$ .

3. Representations and a purposeful technology check

Algebraically, the sign of f'' gives the concavity on each interval. Graphically, concave-up sections bend like a bowl, and concave-down sections bend like an upside-down bowl. The first derivative links these views: when f' increases, the graph is concave up; when f' decreases, it is concave down.

Numerically, compare values of f' at nearby inputs. For instance, if the gradient rises from one nearby input to the next, that supports concave-up behaviour in that region. Numerical samples are evidence, not a replacement for an interval sign analysis.

A graphing calculator can help you see the likely intervals and locate a possible change in concavity. Enter the original function, graph it on a useful window, and compare the view with your calculated second derivative. If your calculator can graph f'', use it to inspect where its sign changes. Zoom in near a suspected point and check the function’s value there. The algebra remains the justification: graphs can hide features if the viewing window or scale is unsuitable.

In a context, interpret the sign using the units. If

f

measures distance in metres and

x

is time in seconds, then f'' has units of metres per second squared. A positive value means the rate of change of distance is increasing; it does not, by itself, say that distance is decreasing or increasing.

Use technology to inspect and check, not to replace the derivative and sign analysis.
State the relevant interval, and include units when the variables represent quantities.

4. A reliable written method

For an exam-style response, show the derivative calculation, identify the interval boundaries, and state the sign of the second derivative on each interval. If asked for a point of inflection, give its coordinates and explain that the sign changes there. Keep the domain in view: a point excluded from the function cannot be an inflection point.

For contextual questions, first define the variables and their units. Then explain what increasing or decreasing gradient means in that situation. Avoid using the everyday meaning of words such as “positive” or “accelerating” without connecting them to the model’s variables.

Show enough sign analysis to support each concavity statement.
A complete inflection-point answer gives the graph point and confirms a concavity change.

Worked example

Polynomial: concavity intervals and an inflection point

For

f(x)=x^3-6x^2+9x+2

, find the intervals of concavity and any point of inflection.

Differentiate twice
Apply the power rule term by term. The constant disappears in the first derivative, and differentiating once more gives the second derivative.
$f'(x)=3x^2-12x+9,\qquad f''(x)=6x-12$
Find the boundary
The second derivative is zero when $6x-12=0$ , so the sign can change at $x=2$ . The polynomial is defined for every real input.
$6x-12=0\quad\Rightarrow\quad x=2$
Test each interval
For $x<2$ , choose $x=0$ : $f''(0)=-12<0$ , so the graph is concave down. For $x>2$ , choose $x=3$ : $f''(3)=6>0$ , so it is concave up. The sign changes at $2$ .
$(-\infty,2):f''<0,\qquad (2,\infty):f''>0$
Find the graph point
Substitute $x=2$ into the original function. Since the concavity changes there, the point is an inflection point.
$f(2)=8-24+18+2=4$

Answer: The graph is concave down for

x<2

and concave up for

x>2

. Its point of inflection is

(2,4)

.

Check: A graph of the cubic should show a change from downward bending to upward bending near

(2,4)

. The sign change in f'' confirms this independently of the viewing window.

Worked example

A zero of the second derivative that is not an inflection point

For

g(x)=x^4

, determine its concavity and decide whether it has a point of inflection at

x=0

.

Find the second derivative
Differentiate the power twice. The result is defined for every real input.
$g'(x)=4x^3,\qquad g''(x)=12x^2$
Check the signs
For every $x\ne0$ , $12x^2$ is positive. For example, it is positive at both $x=-1$ and $x=1$ . Thus the graph is concave up on either side of zero.
$g''(-1)=12>0,\qquad g''(1)=12>0$
Decide about inflection
Although $g''(0)=0$ , the sign does not change as the input passes through zero. Therefore the graph does not change concavity there.
$g''(0)=0$

Answer: The function is concave up on both sides of

0

. It has no point of inflection at

(0,0)

.

Check: The graph has a bowl shape through the origin. This agrees with the sign analysis and shows why a zero of the second derivative alone is not sufficient.

Worked example

Interpreting a changing rate

A model gives a cyclist’s distance from a start point, in metres, after

t

seconds:

s(t)=t^3-6t^2+12t

for

0\le t\le5

. Describe the concavity of the distance graph and interpret it.

Differentiate with respect to time
The first derivative is the rate of change of distance, in metres per second. Differentiate again to find how that rate changes.
$s'(t)=3t^2-12t+12,\qquad s''(t)=6t-12$
Locate the sign change
The second derivative is zero at $t=2$ . On the allowed interval, it is negative before $2$ and positive after $2$ .
$0\le t<2:s''(t)<0,\qquad 2<t\le5:s''(t)>0$
Interpret the result
Before $2$ seconds, the distance graph is concave down, so its gradient is decreasing. After $2$ seconds, it is concave up, so its gradient is increasing. At $t=2$ , the concavity changes. The second derivative’s units are metres per second squared.
$s(2)=8-24+24=8$

Answer: The distance graph is concave down for

0\le t<2

and concave up for

2<t\le5

. The point of inflection is

(2,8)

, meaning the model’s distance rate changes from decreasing to increasing at that time.

Check: A graphing calculator can display

s(t)

on

0\le t\le5

and

s''(t)

on the same domain. Confirm that s'' crosses zero at

t=2

; the derivative calculation establishes the intervals and units.

Common mistakes and how to avoid them

Calling a graph concave up whenever the function is increasing.

Correction: Increasing or decreasing concerns the sign of f'. Concavity concerns whether f' is increasing or decreasing, determined by the sign of f''.

Treating every solution of

f''(x)=0

as an inflection point.

Correction: Check the signs of f'' on both sides and confirm that the function is defined at the point. Concavity must change.

Giving a concavity statement without specifying where it applies.

Correction: Use intervals separated by zeros or undefined values of f'', and respect the function’s domain.

Reporting only what a calculator graph appears to show.

Correction: Use the graph as a visual check, then support the conclusion with the second derivative and its sign.

Lesson summary

Find the second derivative by differentiating the function twice.
Positive second derivative means concave up; negative second derivative means concave down.
Check signs on intervals separated by zeros or undefined values of the second derivative.
A point of inflection requires a change of concavity and must lie on the graph.
Interpret units and changing rates carefully in applications.

Check your understanding

Question 1

If

h''(x)<0

throughout an interval, what is true on that interval?

The graph is concave down.
The function must be decreasing.
The graph must cross the horizontal axis.
The function has a point of inflection at every input.

Show answer and explanation

The graph is concave down.

A negative second derivative means the gradient is decreasing, so the graph is concave down. It does not determine whether the function itself is increasing or decreasing.

Question 2

For

p(x)=x^3

, what can be concluded about

x=0

?

It is a point of inflection because p'' changes sign there.
It is not a point of inflection because $p''(0)=0$ .
The graph is concave down on both sides of zero.
The graph is concave up on both sides of zero.

Show answer and explanation

It is a point of inflection because p'' changes sign there.

Here

p''(x)=6x

, which is negative for

x<0

and positive for

x>0

. Concavity changes at zero, and the graph point is

(0,0)

.

Key terms

First derivative: The derivative $f'(x)$ gives the gradient of the function at each input.
Second derivative: The derivative of $f'(x)$ , written $f''(x)$ , describes how the gradient changes.
Concave up: A graph shape associated with a positive second derivative on an interval.
Concave down: A graph shape associated with a negative second derivative on an interval.
Point of inflection: A point on a graph where the concavity changes.

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