2.4.1 Maxima and minima

Peaks and valleys

A local maximum is a point of the graph higher than every point near it: a hilltop. A local minimum is lower than every point near it: a valley floor. "Near" means within some small distance on either side.

The global maximum of a function on an interval is its highest value there, and the global minimum its lowest. A local maximum need not be global: another hill may be taller.

An answer has two parts: where, an input $x$ in input units, and how much, the value $f(x)$ in output units. For $f(x) = x^3 - 3x$ there is a hilltop at $x = -1$ of height 2. Yet $f(2.5) = 8.125$ , higher than the hilltop.

Critical points

At a smooth hilltop the graph rises, then falls. The slope goes from positive to negative and passes through 0 on the way, so the tangent is flat for an instant. At a valley floor the slope goes from negative to positive.

A graph can also turn at a corner, where there is no tangent at all. $|x|$ has its lowest point at 0, where lesson 2.2.2 found no derivative.

Rule: Critical point

\begin{gathered} f'(c) = 0 \\ \text{or } f'(c) \text{ does not exist} \end{gathered}

A point $c$ in the domain of $f$ where either holds is a critical point. Every local maximum or minimum inside the domain is one. The reverse fails: a critical point can be neither, so test each one.

For $f(x) = x^3 - 3x$ , the power and sum rules (lesson 2.3.1) give $f'(x) = 3x^2 - 3$ . It factorises as $3(x - 1)(x + 1)$ , which is zero at $x = -1$ and $x = 1$ . The values there are $f(-1) = 2$ and $f(1) = -2$ .

A polynomial has a slope everywhere, so these two are its only critical points.

The first-derivative test

The slope can change sign only at a critical point. So one test input in each gap between critical points tells you the sign of $f'$ on the whole gap.

Rule: First-derivative test

\begin{gathered} f' \colon + \text{ then } - \;\Rightarrow\; \text{maximum} \\ f' \colon - \text{ then } + \;\Rightarrow\; \text{minimum} \end{gathered}

Rising then falling is a peak; falling then rising is a valley. If $f'$ has the same sign on both sides, the point is neither.

For $x^3 - 3x$ , test one input in each gap:

$x$	$-2$	$-1$	0	1	2
$f'(x)$	9	0	$-3$	0	9
graph	rising	flat	falling	flat	rising

So $x = -1$ is a local maximum and $x = 1$ a local minimum. The factors give the same signs without test inputs. In $3(x - 1)(x + 1)$ , both brackets are negative below $-1$ and both positive above 1, so the product is positive there; in between, they have opposite signs.

Slide the point across $x = -1$ below. The slope graph underneath crosses zero from above to below: rising turns into falling. At $x = 1$ it crosses from below to above.

Then switch to $|x|$ . At 0 the slope jumps from $-1$ to 1, and the sign still changes from minus to plus.

Explore

Slide the point along the curve. Its tangent shows the slope there, and the lower graph plots that slope for every x: the derivative.

The slope of y = x³ − 3x, which is 3x² − 3

Point at x = −1.8

Trace as you go

Point: (−1.8, −0.432)
Slope there: 6.72
The curve is: rising: the slope is positive, so the slope graph is above zero

Neither: a flat point on a rising curve

$x^3$ has slope $3x^2$ , which is zero at 0, so 0 is a critical point. But $3x^2$ is positive on both sides: 0.03 at $-0.1$ and at $0.1$ . The graph rises to 0, flattens for an instant, and keeps rising.

So 0 is neither a maximum nor a minimum. Just left of it the value is lower ( $-0.001$ at $-0.1$ ), and just right it is higher ( $0.001$ at $0.1$ ). Lesson 2.2.1 met this point: the flat tangent $y = 0$ crosses the curve there.

A corner

$|x|$ has slope $-1$ for $x < 0$ and 1 for $x > 0$ (lesson 2.2.2). Minus then plus: 0 is a minimum, with value 0. The test needs a slope on each side, not at the point itself.

x^3 - 3x

import sympy as sp x = sp.symbols("x") f = x**3 - 3 * x slope = sp.diff(f, x) critical = sp.solve(slope, x) print(slope, critical, [f.subs(x, c) for c in critical]) for t in [-2, 0, 2]: print(t, slope.subs(x, t))

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x^3

def slope(f, a, h=1e-6): return (f(a + h) - f(a - h)) / (2 * h) def cube(x): return x**3 for a in [-0.1, 0.1]: print(a, round(slope(cube, a), 6), round(slope(abs, a), 6))

⌘/Ctrl + Enter runs the cell

Largest and smallest on a closed interval

A closed interval $[a, b]$ includes its two ends. A continuous function on a closed interval always has a largest and a smallest value there. Each one sits either at a critical point inside the interval or at an end.

Rule: Extremes on a closed interval

\begin{gathered} \text{compare } f(a) \text{ and } f(b) \\ \text{with } f(c) \text{ at each} \\ \text{critical point } a < c < b \end{gathered}

The largest of these values is the global maximum on $[a, b]$ , and the smallest is the global minimum. Critical points outside the interval do not count.

For $x^3 - 3x$ on $[-2, 2.5]$ , the candidates are both ends and the two critical points:

$x$	$-2$ (end)	$-1$	1	2.5 (end)
$f(x)$	$-2$	2	$-2$	8.125

The largest value is 8.125, at the right end. The smallest is $-2$ , reached twice: at the left end and at the local minimum.

The cubic $x^3 - 3x$ on $[-2, 2.5]$ . The tangents are flat at the local maximum and minimum, but the largest value, 8.125, is at an end. The smallest, $-2$ , is reached at an end and at the local minimum.

A second example: $xe^{-x}$ on $[0, 4]$

Start with a table of $g(x) = xe^{-x}$ :

$x$	0	0.5	1	2	4
$g(x)$	0	0.3033	0.3679	0.2707	0.0733

The values climb, then fall, so the peak is near 1. Algebra pins it down. The product rule (lesson 2.3.2), with $-e^{-x}$ as the slope of $e^{-x}$ (lesson 2.3.4), gives

g'(x) = e^{-x} - xe^{-x} = (1 - x)e^{-x}

$e^{-x}$ is always positive, so $g'$ has the sign of $1 - x$ : plus before 1, minus after. So 1 is a maximum, with $g(1) = e^{-1} \approx 0.3679$ . Comparing with the ends, $g(0) = 0$ and $g(4) \approx 0.0733$ , the largest value is 0.3679 at 1 and the smallest is 0 at the left end.

Critical points outside, and no ends at all

For $x^2 - 4x$ on $[3, 6]$ , the slope $2x - 4$ is zero only at 2, outside the interval. So only the ends compete: the smallest value is $-3$ at 3, the largest 12 at 6.

Without ends there is no guarantee. On the whole number line $x^3 - 3x$ has no largest value: it is 970 at $x = 10$ and 999,700 at $x = 100$ .

f

import sympy as sp x = sp.symbols("x") def candidates(f, a, b): inside = [c for c in sp.solve(sp.diff(f, x), x) if a < c < b] return {c: round(float(f.subs(x, c)), 4) for c in [a, *inside, b]} values = candidates(x**3 - 3 * x, -2, sp.Rational(5, 2)) print(values, max(values, key=values.get)) values = candidates(x * sp.exp(-x), 0, 4) print(values, max(values, key=values.get), min(values, key=values.get)) print(candidates(x**2 - 4 * x, 3, 6))

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Lesson 2.4.2 tells a maximum from a minimum with the second derivative. When $f'(x) = 0$ has no formula, lesson 2.4.3 solves it with Newton's method.

Common mistakes

$f'(c) = 0$ does not guarantee a peak or a valley. $x^3$ is flat at 0 and still rising on both sides.
A critical point can have no derivative. $|x|$ has its minimum at 0, where $f'$ does not exist.
Read the sign on both sides. A maximum needs plus then minus; one side alone decides nothing.
Check the ends of a closed interval. For $x^3 - 3x$ on $[-2, 2.5]$ the largest value is at the end 2.5.
Drop critical points outside the interval. For $x^2 - 4x$ on $[3, 6]$ , the critical point 2 does not count.
Local is not global. The hilltop 2 of $x^3 - 3x$ is not its largest value on $[-2, 2.5]$ .
Report where and how much: the $x$ of the maximum and the value $f(x)$ there.

Maxima and minima

0 of 10 correct

1.
At what $x$ is $x^2 - 8x + 3$ smallest?
2.
What is the smallest value of $x^2 - 8x + 3$ ?
3.
$x^3 - 12x$ has two critical points. What is the larger one?
4.
What is the local maximum value of $x^3 - 12x$ ?
5.
$f'(x)$ is positive for $x < 3$ and negative for $x > 3$ . Is $x = 3$ a maximum (type 1) or a minimum (type 0)?
6.
$x^3 + 1$ has a critical point at 0. How many local maxima and minima does it have?
7.
At what $x$ is $|x - 2| + 5$ smallest?
8.
What is the largest value of $xe^{-x}$ on $[0, 4]$ ?
9.
What is the largest value of $x^3 - 3x$ on $[-2, 3]$ ?
10.
What is the smallest value of $x^2 - 4x$ on $[3, 6]$ ?

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'(c) = 0

Maxima and minima

Peaks and valleys

Critical points

Rule: Critical point

The first-derivative test

Rule: First-derivative test

Neither: a flat point on a rising curve

A corner

Largest and smallest on a closed interval

Rule: Extremes on a closed interval

A second example: $xe^{-x}$ on $[0, 4]$

Critical points outside, and no ends at all

Common mistakes

Maxima and minima

Summary card

Free further reading

Peaks and valleys

Critical points

Rule: Critical point

The first-derivative test

Rule: First-derivative test

Neither: a flat point on a rising curve

A corner

Largest and smallest on a closed interval

Rule: Extremes on a closed interval

A second example: xe−xxe^{-x}xe−x on [0,4][0, 4][0,4]

Critical points outside, and no ends at all

Common mistakes

Summary card

Free further reading

A second example: $xe^{-x}$ on $[0, 4]$