2.4.2 Concavity and the second derivative

The slope of the slope

The derivative $f'$ is a function, so it has a derivative of its own. That is the second derivative, written $f''(x)$ and read "f double prime of x". It says how fast the slope is changing.

For $f(x) = x^2$ the slope is $2x$ . At $x = -1, 0, 1, 2$ it is $-2, 0, 2, 4$ : each step of 1 in $x$ adds 2 to the slope. So $f''(x) = 2$ everywhere.

Rule: Second derivative

\begin{gathered} f''(x) = \frac{d}{dx}\, f'(x) \\[4pt] \text{also written } \frac{d^2y}{dx^2} \end{gathered}

Differentiate twice. Its units are output units per input unit, per input unit: the change in the slope per unit of input.

For $x^3 - 3x$ , the power rule (lesson 2.3.1) gives $f'(x) = 3x^2 - 3$ and then $f''(x) = 6x$ . For $e^x$ , every derivative is $e^x$ again (lesson 2.3.4).

A stone dropped from rest falls $s(t) = 4.9t^2$ metres in $t$ seconds. Its speed is $s'(t) = 9.8t$ metres per second. Its second derivative, $s''(t) = 9.8$ metres per second per second (m/s²), is its acceleration.

Concave up and concave down

The sign of $f''$ says which way the graph bends.

$f'' > 0$ : the slope rises from left to right, and the graph bends up like a cup. It is concave up, also called convex.
$f'' < 0$ : the slope falls, and the graph bends down like a cap. It is concave down, or simply concave.

Take $\ln x$ . Its slope $\frac{1}{x}$ is 2 at $x = 0.5$ , 1 at 1, 0.5 at 2 and 0.25 at 4. The slope keeps falling, so $\ln x$ bends down.

The algebra agrees: $\frac{1}{x} = x^{-1}$ has slope $-x^{-2} = -\frac{1}{x^2}$ , negative for every $x > 0$ .

Rule: Concavity

\begin{gathered} f'' > 0: \ \text{concave up (convex)} \\ f'' < 0: \ \text{concave down (concave)} \end{gathered}

A concave-up curve lies above its tangent lines; a concave-down curve lies below them. Near the point of contact, the gap grows like the square of the distance, as in lesson 2.2.1.

The tangent to $e^x$ at 0 is $y = 1 + x$ . The curve is above it everywhere except at 0: by 0.3679 at $x = -1$ , 0.1487 at 0.5 and 0.7183 at 1. The tangent to $\ln x$ at 1 is $y = x - 1$ , and the curve is below it: by 0.1931 at 0.5, 0.3069 at 2 and 0.9014 at 3.

A concave-up curve, $e^x$ , sits above its tangent at 0; a concave-down curve, $\ln x$ , sits below its tangent at 1. The vertical gaps are the distances between curve and tangent.

Direction and bend are separate

The sign of $f'$ says whether the graph rises or falls; the sign of $f''$ says how it bends. All four pairs happen.

	concave up, $f'' > 0$	concave down, $f'' < 0$
rising, $f' > 0$	$e^x$	$\ln x$
falling, $f' < 0$	$e^{-x}$	$-x^2$ for $x > 0$

$e^{-x}$ falls, yet bends up: its slope $-e^{-x}$ is negative but rising towards 0, and $f''(x) = e^{-x} > 0$ .

\ln

import sympy as sp x = sp.symbols("x") for f in [x**2, x**3 - 3 * x, sp.exp(x), sp.exp(-x), sp.log(x), sp.sqrt(x)]: print(f, "|", sp.diff(f, x), "|", sp.diff(f, x, 2))

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e^x - (1 + x)

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Estimating $f''$ from three values

When $f$ is only a program, estimate $f''$ as lesson 2.2.2 estimated $f'$ . The slope of the chord just right of $x$ , minus the slope just left of it, divided by $h$ , gives the second central difference.

Rule: Second central difference

f''(x) \approx \frac{f(x + h) - 2f(x) + f(x - h)}{h^2}

Three values of $f$ , a gap $h$ apart. Divide by $h^2$ , not $h$ . Its error shrinks like $h^2$ .

gap $h$	0.1	0.01	0.001
$e^x$ at 0 (true 1)	1.000834	1.000008	1.0
$\ln x$ at 2 (true $-0.25$ )	$-0.250313$	$-0.250003$	$-0.25$

Rounding bites sooner than for a first difference, because the error is divided by $h^2$ . At $h = 10^{-6}$ , the estimate for $e^x$ at 0 is 0.999978, worse than at $h = 0.001$ . Case study 2.4.5 uses this estimate to measure a bond's convexity.

e^x

import math def second(f, a, h): return (f(a + h) - 2 * f(a) + f(a - h)) / h**2 for h in [0.1, 0.01, 0.001]: print(h, round(second(math.exp, 0, h), 6), round(second(math.log, 2, h), 6)) print(second(math.exp, 0, 1e-6))

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Inflection points

Where $f''$ changes sign, the graph switches from bending one way to bending the other. That point is an inflection point.

Rule: Inflection point

f'' \text{ changes sign at } c

The curve is concave down on one side of $c$ and concave up on the other. There the tangent crosses the curve.

For $x^3 - 3x$ , $f''(x) = 6x$ is negative for $x < 0$ and positive for $x > 0$ . So 0 is an inflection point. The tangent there is $y = -3x$ , and the curve crosses it.

The cubic $x^3 - 3x$ bends down left of 0 and up right of 0. The maximum sits in the cap, the minimum in the cup, and the tangent at the inflection point crosses the curve.

For $x^3 - 6x^2 + 4$ , the second derivative is $6x - 12 = 6(x - 2)$ . It is $-6$ at 1 and 6 at 3, so the sign changes at $x = 2$ , an inflection point with value $-12$ .

$f'' = 0$ is not enough

$x^4$ has $f''(x) = 12x^2$ , which is 0 at 0. But it is 0.12 at both $-0.1$ and $0.1$ : no sign change. $x^4$ is concave up on both sides, so 0 is not an inflection point.

The second-derivative test

At a critical point where $f'(c) = 0$ , the bend tells you which kind it is. A cup has its lowest point at the bottom; a cap has its highest point at the top.

Rule: Second-derivative test

\begin{gathered} f'(c) = 0,\ f''(c) > 0 \;\Rightarrow\; \text{minimum} \\ f'(c) = 0,\ f''(c) < 0 \;\Rightarrow\; \text{maximum} \end{gathered}

If $f''(c) = 0$ , the test says nothing: go back to the first-derivative test of lesson 2.4.1.

For $x^3 - 3x$ : $f''(-1) = -6 < 0$ , a maximum, and $f''(1) = 6 > 0$ , a minimum. For $xe^{-x}$ from lesson 2.4.1, $f''(x) = (x - 2)e^{-x}$ , so $f''(1) = -e^{-1} \approx -0.3679 < 0$ : the critical point at 1 is a maximum. Both agree with the sign test of lesson 2.4.1, with one evaluation each.

When the test is silent

$x^4$ , $-x^4$ and $x^3$ all have $f'(0) = 0$ and $f''(0) = 0$ . They give three different answers, so $f''(0) = 0$ alone decides nothing.

function	slope just left of 0	slope just right of 0	at 0
$x^4$	$-0.004$	0.004	minimum
$-x^4$	0.004	$-0.004$	maximum
$x^3$	0.03	0.03	neither

The slopes are at $\pm 0.1$ . The test also needs $f'(c) = 0$ : at a corner such as $|x|$ at 0 there is no $f'$ or $f''$ , and only the sign test works.

f''

import sympy as sp x = sp.symbols("x") for f in [x**3 - 3 * x, x * sp.exp(-x)]: slope, bend = sp.diff(f, x), sp.diff(f, x, 2) for c in sp.solve(slope, x): print(f, "| critical", c, "| f''", bend.subs(x, c)) g = x**3 - 6 * x**2 + 4 print(sp.diff(g, x, 2), sp.solve(sp.diff(g, x, 2), x))

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-0.1

def slope(f, a, h=1e-6): return (f(a + h) - f(a - h)) / (2 * h) for name, f in [("x^4", lambda t: t**4), ("-x^4", lambda t: -t**4), ("x^3", lambda t: t**3)]: print(name, round(slope(f, -0.1), 6), round(slope(f, 0.1), 6))

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Lesson 2.4.3 uses $f''$ to find a minimum with Newton's method, and module 2.7 builds the best quadratic copy of a curve from it. Case study 2.4.5 uses convexity for bonds and for attitudes to risk.

Common mistakes

"Concave up" and "convex" mean the same thing, and so do "concave down" and "concave". Finance usually says convex and concave.
The bend is not the direction. $e^{-x}$ falls but is concave up; $\ln x$ rises but is concave down.
$f''(c) = 0$ is not a verdict. $x^4$ , $-x^4$ and $x^3$ all have $f''(0) = 0$ and give a minimum, a maximum and neither.
An inflection point needs a sign change, not just $f'' = 0$ . $x^4$ has none.
The second-derivative test needs $f'(c) = 0$ . At a corner, use the sign test.
The second central difference divides by $h^2$ , and a tiny $h$ drowns it in rounding.

Concavity and the second derivative

0 of 12 correct

1.
Find the second derivative of $x^4$ at $x = 1$
2.
Find the second derivative of $5x^3 - 2x$
3.
Find the second derivative of $e^{2x}$ at $x = 0$
4.
Find the second derivative of $\ln x$ at $x = 2$
5.
A car is $s(t) = 3t^2$ metres from the start after $t$ seconds. What is its acceleration $s''(t)$ , in m/s²?
6.
Is $e^{-x}$ concave up (type 1) or concave down (type 0)?
7.
Estimate $f''(3)$ for $f(x) = x^2$ by the second central difference with $h = 0.1$
8.
Where is the inflection point of $x^3 - 6x^2 + 4$ ?
9.
Where is the inflection point of $x^3 + 3x^2$ ?
10.
$f'(3) = 0$ and $f''(3) = -5$ . Is $x = 3$ a maximum (type 1) or a minimum (type 0)?
11.
$2x^3 - 9x^2 + 12x$ has critical points at 1 and 2. At which one is there a local minimum?
12.
How many inflection points does $x^4$ have?

How to type answers

Powers with ^: x^7, 2^(1/3), a^-2.
Multiply with * or by writing side by side: 3x^2, 2(x+1).
ln(x) is the natural log, log(x, 2) is log base 2, e is Euler's number, sqrt(x) is a square root.
sin(x), cos(x) and tan(x) take radians; write pi for π and sin(x)^2 for sin²x.
Your first answer to each question is saved: it counts towards your score and brings the question back for review.
Exact answers like ln(20)/ln(3) are best. Decimals need about 4 significant figures.

f''(x)

Concavity and the second derivative

The slope of the slope

Rule: Second derivative

Concave up and concave down

Rule: Concavity

Direction and bend are separate

Estimating $f''$ from three values

Rule: Second central difference

Inflection points

Rule: Inflection point

$f'' = 0$ is not enough

The second-derivative test

Rule: Second-derivative test

When the test is silent

Common mistakes

Concavity and the second derivative

Summary card

Free further reading

The slope of the slope

Rule: Second derivative

Concave up and concave down

Rule: Concavity

Direction and bend are separate

Estimating f′′f''f′′ from three values

Rule: Second central difference

Inflection points

Rule: Inflection point

f′′=0f'' = 0f′′=0 is not enough

The second-derivative test

Rule: Second-derivative test

When the test is silent

Common mistakes

Summary card

Free further reading

Estimating $f''$ from three values

$f'' = 0$ is not enough