Extrema and concavity

A look at first and second derivative tests.

Higher derivatives
Purpose
Stationary points
Extrema
Concavity
First derivative test
Second derivative test
Global extrema
Points of inflexion
Tips

Higher derivatives

Derivative functions can also be differentiated. The second-derivative of $f$ is

\frac{\d ^2}{\d x^2} f(x) = \frac{\d ^2f}{\d x^2} = \frac{\d }{\d x} f^\prime(x) = f^{\prime\prime}(x)

At HL, there is also the notation

\frac{\d ^n}{\d x^n} f(x) = f^{(n)}(x)

for the $n$ th derivative.

Purpose

In this course, the purposes of identifying extrema are primarily

Curve sketching
Identifying intervals of increasing and decreasing
Optimization problems
Kinetics problems

The purpose of identifying concavity is mainly curve sketching-related.

So extrema and concavity are not only goals, but also tools.

Stationary points

A stationary point is where $f^\prime(x) = 0$ . It is either a local extremum or a horizontal point of inflexion. There are of course other extrema or points of inflexion outside of stationary points.

IB excludes stationary points from intervals of increasing or decreasing.

Extrema

If the function is not defined over all reals, then these endpoints are automatically local extrema, provided that the function is defined on these endpoints.

Otherwise, $x = x_0$ is a critical point, or candidate for a local maximum or local minimum if

$f^\prime(x_0) = 0$ , OR
$f(x_0)$ is defined, but $f^\prime(x_0)$ is undefined (not differentiable).

Both tests are applicable when $f^\prime(x_0) = 0$ . When $f^\prime(x_0)$ is undefined you should evaluate function at $x_0$ and nearby values. It is rare for question to include when $f^\prime(x_0)$ is undefined.

Passing either the first or second derivative test means an extremum on $x = x_0$ . Failing either test (as long as it is applicable) means there is no extremum on $x = x_0$ .

The global maximum (minimum) is the greatest (least) of the local maxima (minima).

Concavity

Informally, a portion of the graph looking like $\cup$ is concave up, while ones looking like $\cap$ is concave down. The point where a graph changes concavity is known as the point of inflexion.

First derivative test

With critical point $x = x_0$ , pick $a$ and $b$ such that $a < x_0 < b$ . There cannot be other critical points in $[a, b]$ . Try to make the algebra easy, if applying first derivative test by hand.

case	result
${f^\prime(a) > 0 > f^\prime(b)}$	local maximum; $\cap$ concave down; or ⋏ concave up
${f^\prime(a) < 0 < f^\prime(b)}$	local minimum; $\cup$ concave up; or ⋎ concave down
no sign change, $f^\prime(x_0) = 0$	horizontal (stationary) point of inflexion (eg $(0,0)$ on $x \mapsto x^3$ )
no sign change, $f^\prime(x_0)$ is undefined	not extremum or POI

Second derivative test

The second derivative test requires the first derivative to be defined (and continuous).

Find $f^{\prime\prime}(x)$ , and evaluate $f^{\prime\prime}(x_0)$ .

case	result
${f^{\prime\prime}(x_0) < 0}$	local maximum; $\cap$ concave down
${f^{\prime\prime}(x_0) > 0}$	local minimum; $\cup$ concave up
${f^{\prime\prime}(x_0) = 0}$	horizontal (stationary) point of inflexion (eg $(0, 0)$ on $x \mapsto x^3$ )

Global extrema

Global max and min can be determined by comparing all the $y$ -values of the endpoints and critical points, using $f(x)$ .

Example: Consider

f(x) = \sin^3x,\;\; -\frac\pi2 \leq x < \frac{3\pi}2

find the $x$ -coordinates of all extrema, and classify each extremum as a minimum or a maximum.

\begin{align*} 0 &= f^\prime(x) \\ &= 3 \left(\sin^2x\right)\cdot\cos x \end{align*}

Either

\sin^2 x = 0 \implies x = 0, \pi

\cos x = 0 \implies x = -\frac{\pi}{2}, \frac{\pi}{2}

There are no points in the interval where the derivative is undefined.

The endpoints of the domain, if any, should be considered. However, $-\frac\pi2$ is already a critical point from the first derivative being zero, and $\frac{3\pi}{2}$ is outside the domain so it is ignored.

As the critical points have derivatives being zero, we can proceed with either the first derivative test or the second derivative test. Both are shown, but either alone works.

Method 1: First derivative test

Note that $\sin^2 x$ does not change the sign of $f^\prime(x)$ , so we only care about the sign of $\cos x$ .

$\cos x$ is positive over $-\frac\pi2 < x < \frac\pi2$ , and negative over $\frac\pi2 < x < \frac{3\pi}{2}$ . For some $a < x_0 < b$ which contains only one critical point, namely $x_0$ ,

$x_0$	$f^\prime(a)$	$f^\prime(b)$
$-\frac\pi2$	don’t care*	$+$
$0$	$+$	$+$
$\frac\pi2$	$+$	$-$
$\pi$	$-$	$-$

If the first derivative changes signs from positive to negative, it is a local max. So a max at $x = \frac\pi2 \qed$

If the derivative does not change signs, it is a horizontal point of inflexion, such as at $x = 0, \pi$ .

At the endpoint of the domain, we just care about the the interval that is defined. So since the function increases after $x = -\frac\pi2$ , it is a minimum $\qed$

Method 2: Second derivative test

\begin{align*} f^{\prime\prime}(x) &= 3\cos x(2\sin x \cos x) - 3\sin^3 x \\ &= 3 \sin x (2\cos^2 x - \sin^2 x) \end{align*}

As the second derivative is zero at $x = 0, \pi$ , they are horizontal points of inflexion and not extrema.

f^{\prime\prime}\left(-\frac\pi2\right) = 3(-1)(0 - 1) = 3 > 0

f^{\prime\prime}\left(\frac\pi2\right) = 3(1)(0 - 1) = -3 < 0

$f$ is concave up at $x = -\frac\pi2$ so it is a min; $f$ is concave down at $x = \frac\pi2$ so it is a max $\qed$

The choice between first and second derivative tests depends on which method is easier for the specific function, and whether the question specifies a method. It is imperative that you are familiar with both methods.

Points of inflexion

A point of inflexion at $x = x_1$ requires

$f^\prime(x_1)$ is defined
$f^{\prime\prime}(x_1) = 0$
no sign change around $f^\prime(x_1)$ or sign change around $f^{\prime\prime}(x_1)$ .

Note that a POI is only a horizontal POI if the first derivative is also $0$ . A POI need not satisfy either the first or second derivative tests.

Example: Find all points of inflexion in

f(x) = \frac{\cos x + 2}{\e^x},\,\, -\pi \leq x \leq 2\pi

\begin{align*} 0 &= f^{\prime\prime}(x) \\ &= \frac{\d}{\d x}\frac{\e^x(-\sin x) - (\cos x + 2)\e^x}{\e^{2x}} \\ &= \frac{\d}{\d x} \frac{- \sin x - \cos x - 2}{\e^x} \\ &= \frac{\e^x(-\cos x + \sin x) + (\sin x + \cos x + 2)\e^x}{\e^{2x}}\\ &= \frac{2\sin x + 2}{\e^x} \\ \\ 0 &= 2(\sin x + 1) \\ -1 &= \sin x \\ x &= -\frac\pi2, \frac{3\pi}2 \end{align*}

From our knowledge of trigonometric functions and their transformations, we have

2\sin x + 2 \geq 0

Hence there is no sign change in the second derivative of either values, so $f$ does not have any points of inflexion $\qed$

Tips

Does the question care about all extrema or just ones on an open or closed interval?
Does question state that a max or min exist, or do you have to prove it’s a max or min?
Did you check the endpoints? Do you need to check them?
Does question want $x$ or $y$ or both?
Is it easier to find the second derivative or evaluate at values near the critical point?