MATH 140 maximum and minimum values: revision sheet

Q: What is the difference between an absolute and a local maximum?

An absolute maximum is the largest value the function takes on its whole domain or on the interval in the question. A local maximum is only the largest value compared with nearby points, on some small open interval around it. So a local maximum can be beaten elsewhere, and in the textbook used at McGill an endpoint is never a local extremum, even when it is the absolute one.

Q: Is every critical number a maximum or a minimum?

No. A critical number is only a candidate. The function x cubed has a horizontal tangent at zero and keeps increasing through it, so there is no maximum or minimum there. Fermat's theorem says that an extremum inside the domain happens at a critical number, never the reverse. To decide, compare values: on a closed interval, the closed interval method does it for you.

Q: Why do I have to check the endpoints in the closed interval method?

Because Fermat's theorem only speaks about points inside the interval. At an endpoint the derivative does not have to be zero, and the largest or smallest value is very often there. For x squared on the interval from minus one to three, the only critical number is zero, yet the absolute maximum, nine, is at the endpoint three.

Q: Can a function have no maximum on an interval?

Yes, when a hypothesis of the Extreme Value Theorem fails. On an open interval, x squared between minus one and two gets as close to four as you like without reaching it, so it has no maximum. A jump or an asymptote inside a closed interval can have the same effect. When the function is continuous on a closed interval, both extrema always exist.

The thread of the chapter

Fermat NOMINATES, the table DECIDES: a critical number, where $f'(c) = 0$ or $f'(c)$ does not exist with $c$ in the domain, is only a candidate, and the absolute extrema of a function continuous on $[a, b]$ are found by comparing the values at the critical numbers INSIDE the interval AND at the two endpoints.

Topics covered

Absolute maximum and minimum
Local maximum and minimum
Critical number
Extreme Value Theorem
Fermat's theorem
Closed interval method
Continuity on a closed interval
Corners and cusps
Converse of a theorem
Contrapositive

The essentials

Absolute or local: two questions, two definitions

• Absolute maximum on $D$ : $f(c) \ge f(x)$ for ALL $x$ in $D$ . It depends on the domain: change the interval, and the answer can change.
• Local maximum: $f(c) \ge f(x)$ for all $x$ in some OPEN interval around $c$ . With Stewart's definition, an endpoint is never a local extremum.
• An absolute extremum reached at an INTERIOR point is also a local one. The converse is false: a local maximum can be lower than a local minimum ( $x + \frac{1}{x}$ : local max $-2$ , local min $2$ ).
• Every answer has two parts, the VALUE $f(c)$ and the PLACE $c$ , and a value reached twice is reported at both places.

The local maximum

5

x = 2

is beaten by the endpoint value

9

x = 6

, which is absolute but not local; the absolute minimum

0

is also at an endpoint.

Markers read the words: absolute maximum VALUE $9$ , attained AT $x = 6$ . Writing max at $9$ when $9$ is the value costs the location mark.

Candidates: critical numbers, and what Fermat does not say

• Fermat's theorem: if $f$ has a local extremum at $c$ AND $f'(c)$ exists, then $f'(c) = 0$ .
• Critical number: $c$ in the DOMAIN of $f$ with $f'(c) = 0$ or $f'(c)$ undefined. Corners ( $|x|$ ) and cusps ( $x^{2/3}$ ) are critical; an asymptote ( $\frac{1}{x - 2}$ at $2$ ) is not.
• The converse is false: $f'(0) = 0$ for $x^3$ , with no extremum. A critical number is a CANDIDATE, never a verdict.
• Contrapositive, the useful direction: if $f'(c)$ exists and $f'(c) \ne 0$ , there is no local extremum at $c$ .
• Extreme Value Theorem: $f$ continuous on a CLOSED interval $[a, b]$ attains an absolute maximum and an absolute minimum there. Sufficient, not necessary.

Left, an extremum with

f' = 0

; middle, an extremum where

f'

does not exist; right,

f' = 0

with no extremum. Only the first is what students expect.

Write $f'$ as ONE factored fraction before reading it: zeros of the numerator give $f' = 0$ , zeros of the denominator give $f'$ undefined, and the domain of $f$ filters both.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Does the Extreme Value Theorem guarantee the extrema?

Read a line as: this function on this interval, and what the EVT promises. A red cell is not a no: it means the theorem is silent, and the extrema must be studied by hand.

Function	Interval	EVT verdict
$x^2$	$[-1, 2]$	max and min exist
Example: Candidates $f(-1) = 1$ , $f(0) = 0$ , $f(2) = 4$ : max $4$ at $x = 2$ , min $0$ at $x = 0$ .
$\frac{1}{x}$	$[1, 3]$	max and min exist
Example: $0 \notin [1, 3]$ , so continuous; no critical number, max $1$ at $x = 1$ , min $\frac{1}{3}$ at $x = 3$ .
$x^2$	$(-1, 2)$	not closed no guarantee
Example: Values fill $[0, 4)$ : min $0$ at $x = 0$ , NO max, since $4$ is approached and never reached. Same form, other result: $\cos x$ on $(-\pi, 3\pi)$ is also on an open interval, yet has max $1$ at $x = 0$ and min $-1$ at $x = \pi$ . What to do: Study the values near the open end: is the limit there reached inside the interval, or only approached?
$x + \frac{4}{x}$	$[-1, 4]$	not continuous no guarantee
Example: $0 \in [-1, 4]$ : $r(0.1) = 40.1$ and $r(-0.1) = -40.1$ , no max and no min at all. Same form, other result: $0$ on $[0, 1)$ and $1$ on $[1, 2]$ jumps at $1$ , yet has max $1$ and min $0$ . What to do: Find the discontinuity inside $[a, b]$ and study the values on each side of it before quoting any candidate.

The two rows in blue are the only ones where the closed interval method may be used as a recipe. Everywhere else, the first line of the answer is the hypothesis that fails.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Calling an endpoint a local extremum

1 mark per endpoint misclassified

What not to write

“ $f(0) = 7$ is a local maximum of $f$ on $[0, 10]$ , since $f$ decreases right after $0$ .”

What to write

“ $f(0) = 7$ is the absolute maximum of $f$ on $[0, 10]$ . It is not a local maximum: no open interval around $0$ lies in the domain.”

f(0) = 7

is the highest value on

[0, 10]

, so it is the absolute maximum; there is no domain on its left (shaded), so it cannot be compared on both sides and is not a local one.

Why: Stewart's local extremum needs $f(c)$ compared with values on BOTH sides of $c$ . The endpoints are handled by the absolute question, which is why the closed interval method lists them separately.

2. Solving only f'(x) = 0 and missing a cusp

the whole question when the cusp carries the answer

What not to write

“ $k(x) = x - 3x^{2/3}$ on $[-8, 8]$ : $k'(x) = 0$ gives $x = 8$ , so the candidates are $-8$ and $8$ , and the maximum is $k(8) = -4$ .”

What to write

“ $k'(x) = \frac{x^{1/3} - 2}{x^{1/3}}$ is undefined at $0$ , where $k(0) = 0$ exists, so $0$ is critical. Values $-20$ , $0$ , $-4$ : the maximum is $0$ at $x = 0$ .”

k(x) = x - 3x^{2/3}

has a cusp at

(0, 0)

, where

k'

does not exist: that critical number gives the maximum

0

, higher than

k(8) = -4

and

k(-8) = -20

Why: The equation $f'(x) = 0$ only sees the numerator. The denominator of $f'$ gives the other half of the critical numbers, the ones where the graph has a corner, a cusp or a vertical tangent.

3. Counting a point outside the domain as a critical number

1 mark, and a wrong candidate carried into every later question

What not to write

“ $f(x) = \frac{1}{x - 2}$ : $f'(2)$ does not exist, so $2$ is a critical number.”

What to write

“ $f'(x) = -\frac{1}{(x - 2)^2}$ is never $0$ , and $2$ is not in the domain of $f$ : $f$ has no critical number.”

\frac{1}{x - 2}

has no point at all above

x = 2

, only an asymptote: there is no value there to compare, so

2

is not a critical number.

Why: A critical number is a place where the function has a VALUE to compare. At a vertical asymptote there is none. Check the domain of $f$ before listing anything.

4. Keeping a critical number that lies outside the interval

2 marks, the wrong maximum

What not to write

“ $g(x) = x^3 - 12x$ on $[0, 3]$ : $g'(x) = 0$ at $x = \pm 2$ , $g(-2) = 16$ , so the absolute maximum is $16$ .”

What to write

“ $g'(x) = 3(x - 2)(x + 2)$ ; $-2 \notin (0, 3)$ , rejected. Values $g(0) = 0$ , $g(2) = -16$ , $g(3) = -9$ : max $0$ at $x = 0$ , min $-16$ at $x = 2$ .”

The peak at

x = -2

is real but lies outside

[0, 3]

(dashed part); on the solid part the highest point is the endpoint

(0, 0)

Why: The method ranks the values TAKEN on $[a, b]$ . A value taken outside the interval is not a value of the restricted function, however large it is.

5. Forgetting the endpoints

half the question

What not to write

“ $f(x) = x^2$ on $[-1, 3]$ : the only critical number is $0$ , so the absolute maximum and minimum are both at $x = 0$ .”

What to write

“Candidates: $f(-1) = 1$ , $f(0) = 0$ , $f(3) = 9$ . Absolute maximum $9$ at $x = 3$ , absolute minimum $0$ at $x = 0$ .”

[-1, 3]

the critical number

0

gives the minimum, but the maximum

9

is at the endpoint

x = 3

, where

f'(3) = 6 \ne 0

Why: Fermat's theorem only speaks about interior points. At an endpoint the derivative need not vanish, $f'(3) = 6$ here, and the EVT's extremum is very often there.

6. Taking a horizontal tangent for an extremum

1 to 2 marks

What not to write

“ $u(x) = x^4 - 4x^3$ , $u'(0) = 0$ , so $u$ has a local extremum at $0$ .”

What to write

“ $u(x) = x^3(x - 4)$ is positive just left of $0$ and negative just right of $0$ , while $u(0) = 0$ : no local extremum at $0$ .”

u(x) = x^4 - 4x^3

has a horizontal tangent at

0

but keeps going down through it: no extremum there. The only local extremum is the minimum at

x = 3

Why: $f'(c) = 0$ is the converse of Fermat, and it is false. In this chapter, decide with values: by the definition, as here, or with the closed interval method on a small interval around $c$ .

7. Applying the method across a discontinuity

the whole question

What not to write

“ $r(x) = x + \frac{4}{x}$ on $[-1, 4]$ : $r(-1) = -5$ , $r(2) = 4$ , $r(4) = 5$ , so the minimum is $-5$ and the maximum is $5$ .”

What to write

“ $r$ is not continuous at $0 \in [-1, 4]$ , so the EVT does not apply. Near $0$ the values are unbounded both ways: no absolute maximum, no absolute minimum.”

[-1, 4]

the function

x + \frac{4}{x}

is cut at

0

and escapes to

-\infty

and

+\infty

on either side: the values

-5

and

5

at the ends are neither the minimum nor the maximum.

Why: The list of candidates is complete ONLY because the EVT guarantees that the extrema exist. Without continuity on $[a, b]$ that guarantee is gone, and so is the method.

8. Answering a maximum that is approached but never reached

1 to 2 marks

What not to write

“ $f(x) = x^2$ on $(-1, 2)$ has absolute maximum $4$ , at $x = 2$ .”

What to write

“For $-1 < x < 2$ , $0 \le x^2 < 4$ : the values approach $4$ but never reach it. No absolute maximum; absolute minimum $0$ at $x = 0$ .”

(-1, 2)

the curve climbs toward the empty dot

(2, 4)

without reaching it: every value is below

4

, and no point of the domain gives

4

. No absolute maximum.

Why: A maximum is a value TAKEN at a point of the domain. On an open interval the endpoint is not in the domain, so its value is only a limit, and a limit is not an extremum.

Which method to choose

Which tool, by what the question asks

Read the verb and the interval of the question before differentiating anything

If find the absolute extrema of $f$ on $[a, b]$ , with $f$ continuous → closed interval method: critical numbers inside, endpoints, compare the values

Example: $x^3 - 6x^2 + 9x + 2$ on $[0, 4]$ : max $6$ at $x = 1$ and $x = 4$ , min $2$ at $x = 0$ and $x = 3$
If show that $f$ HAS a maximum and a minimum → Extreme Value Theorem: state continuity on $[a, b]$ with its reason

Example: $\frac{1}{x}$ on $[1, 3]$ is continuous since $0 \notin [1, 3]$
If show that $f$ has NO local extremum → contrapositive of Fermat: $f'$ exists everywhere and is never $0$

Example: $x^5 + x^3 + 2x - 7$ : $f'(x) \ge 2 > 0$
If is there an extremum at a given critical number $c$ ? → compare values near $c$ by an inequality, or apply the method on a small $[c - h, c + h]$

Example: $x + \frac{1}{x}$ on $\left[\frac{1}{2}, 2\right]$ : min $2$ at the interior point $1$ , so a local min
If the interval is open, or the whole real line → the EVT is silent: study the ends, or trap $f$ outside a closed interval

Example: $x^4 - 4x \ge 8$ for $x \ge 2$ and $> 0$ for $x \le -2$ , so min $-3$ at $x = 1$ on $\mathbb{R}$
If $f$ is piecewise or contains $|\cdot|$ → split into cases; the junction points are candidates where $f'$ may not exist

Example: $|x^2 - 2x - 3|$ on $[0, 4]$ : the corner $x = 3$ is the minimum $0$

The first derivative test and the second derivative test belong to section 4.3: in a question on 4.1, decide with values. When no branch applies, go back to the definitions of absolute and local extremum.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

The closed interval method, written for full marks

When to use it: Any question asking for the absolute maximum and minimum values of $f$ on $[a, b]$

1 State the hypothesis with its reason: $f$ is continuous on $[a, b]$ (polynomial, or quotient whose denominator does not vanish on $[a, b]$ , or composition of continuous functions).
2 Compute $f'$ and write it as one factored fraction. List where $f' = 0$ and where $f'$ is undefined while $f$ is defined.
3 Keep only the critical numbers in the OPEN interval $(a, b)$ , writing rejected next to each one outside, with its reason.
4 Evaluate $f$ exactly at every kept critical number and at $a$ and $b$ , in a small table.
5 Compare with justified inequalities, then give each extremum as a value AND every place where it is attained.

Concluding sentence

“ $f$ is continuous on the closed interval $[0, 4]$ , so by the Extreme Value Theorem it attains an absolute maximum and minimum. Comparing $f(0) = 2$ , $f(1) = 6$ , $f(3) = 2$ and $f(4) = 6$ : the absolute maximum value is $6$ , attained at $x = 1$ and $x = 4$ , and the absolute minimum value is $2$ , attained at $x = 0$ and $x = 3$ .”

The trap: Skipping the continuity sentence: without it the method is a recipe applied on faith, and on $[-1, 4]$ with $x + \frac{4}{x}$ it produces a false answer.

Marking: Typically 1 mark for continuity, 3 for the critical numbers (including the undefined ones and the rejections), 3 for the exact values at every candidate, 3 for the comparison and the conclusion with locations.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

One more point, between min and max

Evaluate f at a point you did not list, a midpoint for instance. Its value must lie between your minimum and your maximum; if not, a candidate is missing.

For $h(x) = |x^2 - 2x - 3|$ on $[0, 4]$ with min claimed $3$ : $h(3) = 0 < 3$ exposes the forgotten corner.
Every kept candidate is in the interval

Read the list once more against the interval: each critical number must lie strictly between a and b, and both endpoints must appear.

$x^3 - 12x$ on $[0, 3]$ : $-2$ must be marked rejected, and $0$ and $3$ must be in the table.
The sign of f agrees with the answer

If f is visibly nonnegative, the minimum cannot be negative; if it is a sum of a square and a constant, that constant bounds it.

$x^2e^{-x} \ge 0$ , so a minimum below $0$ is impossible; $2\sin x + \cos 2x = \frac{3}{2} - 2\left(\sin x - \frac{1}{2}\right)^2 \le \frac{3}{2}$ .
Each critical number really cancels f'

Substitute every critical number back into f' before evaluating f: a sign error in the factorization shows at once.

$f'(x) = 6x^2 - 6x - 12$ at $x = 2$ : $24 - 12 - 12 = 0$ ; at $x = -2$ it gives $24$ , so $-2$ was wrong.

The typical problem, taken apart

A trigonometric function, five candidates, a tie and an endpoint

Find the absolute maximum and minimum values of $f(x) = 2\sin x + \cos 2x$ on $\left[0, \frac{3\pi}{2}\right]$ , and where they are attained.

No calculator. Every step must be justified as on a MATH 140 final.

The graph of

f(x) = 2\sin x + \cos 2x

on the closed interval

\left[0, \frac{3\pi}{2}\right]

, where

f

is continuous: the Extreme Value Theorem applies.

Step 1

$f$ is a sum of compositions of continuous functions, so it is continuous on the closed interval $\left[0, \frac{3\pi}{2}\right]$ . By the Extreme Value Theorem, the absolute maximum and minimum exist.

Why

This sentence is what makes the finite list of candidates COMPLETE. It is short, and it is a mark.

Step 2

By the chain rule (inner function $2x$ ): $f'(x) = 2\cos x - 2\sin 2x = 2\cos x - 4\sin x\cos x = 2\cos x(1 - 2\sin x)$ . It exists everywhere.

Why

The identity $\sin 2x = 2\sin x\cos x$ turns $f'$ into a PRODUCT, and only a product can be read factor by factor. Leaving $2\cos x - 2\sin 2x = 0$ unfactored is where solutions get lost.

Step 3

$\cos x = 0$ in $\left(0, \frac{3\pi}{2}\right)$ : $x = \frac{\pi}{2}$ only ( $\frac{3\pi}{2}$ is an endpoint). $\sin x = \frac{1}{2}$ : $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ . Critical numbers: $\frac{\pi}{6}$ , $\frac{\pi}{2}$ , $\frac{5\pi}{6}$ .

Why

Each factor is solved on the OPEN interval, and each equation for the sine has two solutions per period. Missing $\frac{5\pi}{6}$ here would lose one of the two locations of the maximum.

Step 4

Values: $f(0) = 0 + 1 = 1$ ; $f\left(\frac{\pi}{6}\right) = 1 + \cos\frac{\pi}{3} = \frac{3}{2}$ ; $f\left(\frac{\pi}{2}\right) = 2 + \cos\pi = 1$ ; $f\left(\frac{5\pi}{6}\right) = 1 + \cos\frac{5\pi}{3} = \frac{3}{2}$ ; $f\left(\frac{3\pi}{2}\right) = -2 + \cos 3\pi = -3$ .

Why

Five exact values, no decimals: the unit circle gives each one. The endpoint $\frac{3\pi}{2}$ is evaluated although $f'$ vanishes there too, because endpoints are ALWAYS on the list.

Step 5

Maximum $\frac{3}{2}$ , at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ . Minimum $-3$ , at $x = \frac{3\pi}{2}$ . Check: with $s = \sin x$ , $f = 2s + 1 - 2s^2 = \frac{3}{2} - 2\left(s - \frac{1}{2}\right)^2 \le \frac{3}{2}$ , with equality exactly when $\sin x = \frac{1}{2}$ .

Why

The algebraic rewriting confirms the maximum independently of the derivative. The critical number $\frac{\pi}{2}$ turned out to be neither the max nor the min on the interval: a candidate, dismissed by the table.

The conclusion, written out

“ $f$ is continuous on $\left[0, \frac{3\pi}{2}\right]$ . Comparing its values at the critical numbers $\frac{\pi}{6}$ , $\frac{\pi}{2}$ , $\frac{5\pi}{6}$ and at the endpoints, the absolute maximum value is $\frac{3}{2}$ , attained at $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$ , and the absolute minimum value is $-3$ , attained at $x = \frac{3\pi}{2}$ .”

The classic mistake on this problem: Solving only $\cos x = 0$ or only $\sin x = \frac{1}{2}$ ; or dropping the endpoint $\frac{3\pi}{2}$ because it is a zero of $\cos x$ , and answering min $1$ .

Learn by heart

• Absolute: compared with the whole domain. Local: compared with an open interval around $c$ , never at an endpoint.
• EVT: continuous on a CLOSED interval $[a, b]$ implies an absolute max and min exist. Not the converse.
• Fermat: local extremum and $f'(c)$ exists imply $f'(c) = 0$ . Not the converse: $x^3$ .
• Critical number: $c$ in the domain, $f'(c) = 0$ OR $f'(c)$ undefined.
• Closed interval method: continuity, critical numbers in $(a, b)$ , endpoints, compare exact values.
• Report every extremum as a value AND all its locations; write rejected next to any candidate outside $(a, b)$ .

Frequently asked questions

What is the difference between an absolute and a local maximum?

An absolute maximum is the largest value the function takes on its whole domain or on the interval in the question. A local maximum is only the largest value compared with nearby points, on some small open interval around it. So a local maximum can be beaten elsewhere, and in the textbook used at McGill an endpoint is never a local extremum, even when it is the absolute one.

Is every critical number a maximum or a minimum?

No. A critical number is only a candidate. The function x cubed has a horizontal tangent at zero and keeps increasing through it, so there is no maximum or minimum there. Fermat's theorem says that an extremum inside the domain happens at a critical number, never the reverse. To decide, compare values: on a closed interval, the closed interval method does it for you.

Why do I have to check the endpoints in the closed interval method?

Because Fermat's theorem only speaks about points inside the interval. At an endpoint the derivative does not have to be zero, and the largest or smallest value is very often there. For x squared on the interval from minus one to three, the only critical number is zero, yet the absolute maximum, nine, is at the endpoint three.

Can a function have no maximum on an interval?

Yes, when a hypothesis of the Extreme Value Theorem fails. On an open interval, x squared between minus one and two gets as close to four as you like without reaching it, so it has no maximum. A jump or an asymptote inside a closed interval can have the same effect. When the function is continuous on a closed interval, both extrema always exist.

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Corrected exercises: Maximum and minimum values, MATH 140 at McGill

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Revision sheet: maximum and minimum values (MATH 140)

The essentials

Absolute or local: two questions, two definitions

Candidates: critical numbers, and what Fermat does not say

The rules in table form

Does the Extreme Value Theorem guarantee the extrema?

The mistakes that cost marks

1. Calling an endpoint a local extremum

2. Solving only f'(x) = 0 and missing a cusp

3. Counting a point outside the domain as a critical number

4. Keeping a critical number that lies outside the interval

5. Forgetting the endpoints

6. Taking a horizontal tangent for an extremum

7. Applying the method across a discontinuity

8. Answering a maximum that is approached but never reached

Which method to choose

Which tool, by what the question asks

How the answer is expected to be written

The closed interval method, written for full marks

Check before you hand in

The typical problem, taken apart

A trigonometric function, five candidates, a tie and an endpoint

Learn by heart

Frequently asked questions

What is the difference between an absolute and a local maximum?

Is every critical number a maximum or a minimum?

Why do I have to check the endpoints in the closed interval method?

Can a function have no maximum on an interval?

Corrected exercises: Maximum and minimum values, MATH 140 at McGill

See also

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