MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: extreme values of functions (MATH 203)

Q: What is a critical point in Thomas' Calculus?

It is an interior point of the domain of the function where the derivative is zero or does not exist. Endpoints of the domain or of the interval are never critical points in Thomas, even when the derivative is undefined there, but they are always evaluated in the closed interval method. A point outside the domain is not a critical point either.

Q: Can an endpoint be a local maximum or minimum in MATH 203?

Yes, with the definitions of Thomas used at Concordia. A local extremum compares the value at a point with the values of the function at nearby points of its domain, and at an endpoint those points lie on one side only. So the square root function on the interval from zero to four has a local minimum at zero. Some other textbooks forbid this, so name the convention on your copy.

Q: Why do I miss critical points when the derivative has fractional exponents?

Because a negative exponent is a denominator in disguise. If you set a sum like x to the minus one fifth plus x to the four fifths equal to zero and multiply through, the point where the denominator vanishes disappears. Factor out the lowest power first and write the derivative as one fraction: its numerator gives the zeros, its denominator gives the points where the derivative is undefined.

Q: My calculator gives an error for a negative number to the power two thirds. What do I do?

The number exists, the calculator simply refuses fractional powers of negative numbers. Take the cube root first and square it, or square first and take the cube root: negative eight to the power two thirds equals four. Never drop a candidate from the table because of that error message, it may be the endpoint that carries the maximum.

This sheet is not a summary of section 4.1 of Thomas' Calculus: you already have the course notes. It answers one question only, what makes students lose marks on extreme values of functions in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The chapter looks mechanical, differentiate, set to zero, plug in, and that is where the marks go: in the algebra that rewrites $f'$ , in the candidates the equation $f'(x) = 0$ never produces, in the endpoints, and in the calculator keystrokes. Thomas's two conventions are used throughout: a critical point is an interior point, and an endpoint can carry a local extremum.

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The thread of the chapter

The candidates hide in the algebra of $f'$ : only $f'$ rewritten as ONE factored fraction shows both the interior points where $f' = 0$ and those where $f'$ is undefined, and the absolute extrema on $[a, b]$ are then decided by the values of $f$ at those points AND at the two endpoints, which the calculator only evaluates.

This chapter is part of MATH 203, Calculus I (Concordia)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Absolute extrema
Local extrema
Extreme Value Theorem
First Derivative Theorem for Local Extrema
Critical point
Closed interval method
Endpoint extrema
Fractional exponents
Continuity on a closed interval
Radian mode

The essentials

Absolute or local, with Thomas's 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 $x$ of the domain in some open interval containing $c$ . At an endpoint only one side is in the domain, so in Thomas an ENDPOINT can be a local extremum.
• Consequence in Thomas: every absolute extremum is also a local one. The converse is false: a local maximum can be beaten elsewhere.
• Every answer has two parts, the VALUE $f(c)$ and the PLACE $c$ , and a value reached twice is reported at both places.

The left endpoint is a local minimum without being absolute; the maximum

3

is reached twice, inside at

x = -1

and at the endpoint

x = 2

Other textbooks refuse local extrema at endpoints. In MATH 203 follow Thomas, and write the word endpoint next to the answer so the marker sees the convention.

Candidates: critical points, and what Theorem 2 does not say

• Theorem 2: local extremum at an INTERIOR point $c$ and $f'(c)$ defined imply $f'(c) = 0$ .
• Critical point: interior point of the domain with $f'(c) = 0$ or $f'(c)$ undefined. Cusps ( $x^{2/3}$ ), corners, vertical tangents ( $x^{1/3}$ ) are critical; an endpoint is not; a point outside the domain is not.
• Theorem 2 has no converse: $f'(c) = 0$ or $f'(c)$ undefined makes $c$ a CANDIDATE, never a verdict.
• Theorem 1 (EVT): $f$ continuous on a CLOSED interval $[a, b]$ attains an absolute maximum and minimum there. It gives existence, not location, not uniqueness.
• Closed interval method: the candidates are the critical points in $(a, b)$ plus $a$ and $b$ ; the values of $f$ decide.

Left,

f'

undefined and a minimum; middle,

f'

undefined and no extremum; right,

f' = 0

and a maximum. Undefined is no more a verdict than zero.

Write $f'$ as ONE factored fraction before reading it: numerator zeros give $f' = 0$ , denominator zeros 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.

Critical point, candidate, or neither?

Read a line as: this point, what happens there, and its status in Thomas. A red cell is not a critical point: it says what to do with the point instead.

Point	What happens there	Critical point?
$x = \frac{4}{3}$ for $(x + 2)\sqrt{3 - x}$	interior, $f' = 0$	yes, a candidate
Example: $h'(x) = \frac{4 - 3x}{2\sqrt{3 - x}}$ vanishes at $\frac{4}{3}$ , inside the domain $(-\infty, 3]$ .
$x = 0$ for $x^{4/5}(x - 9)$	interior, $f'$ undefined, $f(0) = 0$	yes, a candidate
Example: $f'(x) = \frac{9(x - 4)}{5x^{1/5}}$ : the denominator vanishes at $0$ , where $f$ is defined.
$x = 5$ for $x^2\sqrt{5 - x}$ on $[-1, 5]$	endpoint, $f'$ undefined	not critical endpoint, still a candidate
Example: $g(5) = 0$ is half of the absolute minimum on $[-1, 5]$ , although $5$ is not a critical point. Same form, other result: On $[0, 3]$ , $2x^3 - 9x^2 + 12x - 1$ has its maximum $8$ at the endpoint $3$ , where $f'(3) = 12 \ne 0$ . What to do: Put every endpoint in the table of values anyway: the method evaluates critical points AND endpoints.
$x = 0$ for $x^2 + \frac{16}{x}$	outside the domain	not critical not a candidate at all
Example: $g'(x) = \frac{2(x^3 - 8)}{x^2}$ is undefined at $0$ , but $g(0)$ does not exist: no value to compare. Same form, other result: On $[1, 4]$ the function is continuous and the method gives min $12$ at $x = 2$ ; on $[-1, 4]$ there is no max and no min. What to do: Discard the point; if it lies inside $[a, b]$ , the EVT no longer applies and the values near it must be studied.

The two blue rows are the critical points. The red rows are the two ways a student either loses a candidate (the endpoint) or invents one (the point outside the domain).

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. Dividing by a factor and losing a critical point

1 to 2 marks, and the whole question if the lost point carries the extremum

What not to write

“ $k'(x) = 2\sin x\cos x - \sin x = 0$ , so $2\cos x = 1$ after dividing by $\sin x$ : the critical points in $(0, 2\pi)$ are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ .”

What to write

“ $k'(x) = \sin x(2\cos x - 1) = 0$ : $\sin x = 0$ gives $x = \pi$ , $\cos x = \frac{1}{2}$ gives $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ . Three critical points.”

Why: Dividing by an expression assumes it is not zero, and its zeros are exactly the solutions thrown away. Factor it out and set EACH factor to zero.

2. Leaving a negative exponent in f' and missing the undefined point

the whole question when the cusp carries the minimum

What not to write

“ $f(x) = x^{4/5}(x - 9)$ : $\frac{9}{5}x^{4/5} - \frac{36}{5}x^{-1/5} = 0$ , multiply by $x^{1/5}$ : $x = 4$ is the only critical point.”

What to write

“Factor the lowest power: $f'(x) = \frac{9}{5}x^{-1/5}(x - 4) = \frac{9(x - 4)}{5x^{1/5}}$ . Critical points: $4$ ( $f' = 0$ ) and $0$ ( $f'$ undefined, $f(0) = 0$ ).”

Why: A negative exponent is a DENOMINATOR in disguise. Multiplying through by it hides the one point where it vanishes, which is a critical point whenever $f$ itself is defined there.

3. Typing a negative number without brackets

2 to 4 marks: both extrema wrong

What not to write

“ $p(x) = x^4 - 2x^2$ on $[-2, 1]$ : $p(-2) = -24$ on my calculator, so the absolute minimum is $-24$ at $x = -2$ .”

What to write

“ $p(-2) = (-2)^4 - 2(-2)^2 = 16 - 8 = 8$ . Values $8$ , $-1$ , $0$ , $-1$ : max $8$ at $x = -2$ , min $-1$ at $x = -1$ and $x = 1$ .”

The lowest points are

(-1, -1)

and

(1, -1)

: the curve never goes below

-1

, so a value of

-24

x = -2

is impossible; the true point is

(-2, 8)

Why: On a calculator, $-2^4$ means $-(2^4) = -16$ : the power is computed before the sign. Type the brackets around every negative input, every time.

4. Taking a calculator ERROR for an undefined value

1 mark per dropped candidate, the whole question if it was the extremum

What not to write

“ $h(x) = (x^2 - 4)^{2/3}$ : my calculator says ERROR for $h(0) = (-4)^{2/3}$ , so $0$ is not in the domain and I drop it.”

What to write

“ $(-4)^{2/3} = \left((-4)^2\right)^{1/3} = 16^{1/3} \approx 2.5198$ . $0$ stays in the table.”

Why: Many scientific calculators compute $a^{p/q}$ with logarithms and refuse $a < 0$ . An odd root of a negative number exists: square first, or take the cube root first.

5. Evaluating in degree mode

the conclusion, 3 marks

What not to write

“ $k(x) = \arctan x - \frac{x}{2}$ on $[0, 3]$ : $k(3) = 71.5651 - 1.5 = 70.0651$ , so the maximum is at $x = 3$ .”

What to write

“In radians, $k(3) = \arctan 3 - 1.5 \approx -0.2510$ . Values $0$ , $0.2854$ , $-0.2510$ : max $\frac{\pi}{4} - \frac{1}{2}$ at $x = 1$ .”

Why: Every derivative formula of the course, $(\arctan x)' = \frac{1}{1 + x^2}$ and $(\sin x)' = \cos x$ first, assumes radians. Check the mode on $\sin\frac{\pi}{6} = 0.5$ before the first value.

6. Making a table of f' instead of f

the whole conclusion

What not to write

“ $f(x) = x^3 - 3x$ on $[0, 2]$ : $f'(0) = -3$ , $f'(1) = 0$ , $f'(2) = 9$ , so the absolute maximum is $9$ .”

What to write

“ $f(0) = 0$ , $f(1) = -2$ , $f(2) = 2$ : absolute maximum $2$ at $x = 2$ , absolute minimum $-2$ at $x = 1$ .”

Why: At a critical point $f'$ is $0$ or undefined BY DEFINITION, so a table of $f'$ says nothing. The question is about the values of the function itself.

7. Keeping a critical point outside the interval

1 mark, the location

What not to write

“ $g(x) = x^4 - 8x^2 + 3$ on $[-1, 3]$ : critical points $-2$ , $0$ , $2$ ; the minimum $-13$ is at $x = -2$ and $x = 2$ .”

What to write

“ $g'(x) = 4x(x - 2)(x + 2)$ ; $-2 \notin (-1, 3)$ , rejected. Values $-4$ , $3$ , $-13$ , $12$ : min $-13$ at $x = 2$ only, max $12$ at $x = 3$ .”

Why: The method ranks values TAKEN on $[a, b]$ . Write rejected, with its reason, next to every solution of $f'(x) = 0$ that falls outside the interval.

Which method to choose

Which algebraic gesture, by the form of f'

Read the form of f' before solving anything: the gesture decides whether the candidates survive

If a sum of powers with a negative or fractional exponent → factor out the LOWEST power, then write it as a denominator

Example: $\frac{10}{3}x^{-1/3} - \frac{10}{3}x^{2/3} = \frac{10(1 - x)}{3x^{1/3}}$ : critical points $1$ and $0$
If a term with a square root plus a term divided by it → common denominator: multiply the first term by the root over itself

Example: $2x\sqrt{5 - x} - \frac{x^2}{2\sqrt{5 - x}} = \frac{5x(4 - x)}{2\sqrt{5 - x}}$
If a trigonometric expression with $\sin^2$ , $\cos^2$ or $\sin 2x$ → apply an identity to get ONE trigonometric function, then factor, never divide

Example: $\cos x + \cos^2 x - \sin^2 x = (2\cos x - 1)(\cos x + 1)$
If an exponential factor such as $e^{-x^2/2}$ → factor it out: it is never zero, only the other factor counts

Example: $e^{-x^2/2}(1 - x^2) = 0$ gives $x = \pm 1$
If a logarithm or a quotient with a restricted domain → state the domain first, then reject every zero outside it

Example: $x(2\ln x + 1) = 0$ : $x = 0$ is rejected, $x = e^{-1/2}$ is kept
If a complex fraction from the quotient rule → multiply numerator and denominator by the inner denominator

Example: $\frac{\ln x}{\sqrt x}$ gives $\frac{2 - \ln x}{2x^{3/2}}$ after multiplying by $2\sqrt x$

The increasing and decreasing test and the second derivative test belong to sections 4.3 and 4.4. In a 4.1 question, decide with values of $f$ .

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, quotient whose denominator does not vanish on $[a, b]$ , root defined on $[a, b]$ ).
2 Compute $f'$ and rewrite it as ONE factored fraction. List where $f' = 0$ and where $f'$ is undefined while $f$ is defined.
3 Keep the critical points in the OPEN interval $(a, b)$ , and write rejected, with its reason, next to each one outside.
4 Make a table of values of $f$ (not $f'$ ) at the kept critical points and at $a$ and $b$ : exact value first, then the decimal, calculator in radians, brackets around negative inputs.
5 Conclude with each extremum as a value AND every place where it is attained.

Concluding sentence

“ $f$ is continuous on the closed interval $[0, 3]$ , so by the Extreme Value Theorem it attains an absolute maximum and minimum. The critical points are $1$ and $2$ . Comparing $f(0) = -1$ , $f(1) = 4$ , $f(2) = 3$ and $f(3) = 8$ : the absolute maximum value is $8$ , attained at $x = 3$ , and the absolute minimum value is $-1$ , attained at $x = 0$ .”

The trap: Skipping the continuity sentence: on an interval that contains a vertical asymptote the method produces a false answer, and the marker takes the first mark away before reading the rest.

Marking: Typically 1 mark for continuity, 3 for the critical points (including the undefined ones and the rejections), 3 for the 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.

Each critical point cancels the ORIGINAL f'

Substitute every critical point into f' as you first computed it, before the factoring. A slip in the factoring shows at once.

$T'(t) = -1.92 + 0.6t - 0.03t^2$ at $t = 4$ : $-1.92 + 2.4 - 0.48 = 0$ . The factored form $-0.03(t - 4)(t - 16)$ is confirmed.
One more point, between min and max

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

For $f(x) = x^{2/3}(5 - 2x)$ on $[-1, 2]$ with a claimed minimum $1.5874$ : $f(0) = 0$ exposes the forgotten cusp.
The calculator is in radians and the brackets are there

Before the table, check the mode on a known value, and look at every negative input on the screen.

$\sin\frac{\pi}{6}$ must display $0.5$ ; $(-2)^4$ must display $16$ , not $-16$ .
A bound that the answer must respect

If an algebraic rewriting bounds f, the extrema must respect the bound: a completed square, a sign, a range.

$x^4 - 2x^2 = (x^2 - 1)^2 - 1 \ge -1$ , so a minimum below $-1$ is impossible.

The typical problem, taken apart

A fractional exponent, a vertical tangent, and a negative cube root

Find the absolute maximum and minimum values of $f(x) = (x - 1)^{1/3}(x + 3)$ on $[-7, 2]$ , and where they are attained.

Give exact values; a scientific calculator is allowed.

The curve dips with a horizontal tangent at

x = 0

and becomes vertical as it crosses

x = 1

: two kinds of candidate, and the endpoints still to check.

Step 1

$(x - 1)^{1/3}$ is defined and continuous for every real $x$ , so $f$ is continuous on the closed interval $[-7, 2]$ . 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

Write $x + 3 = (x - 1) + 4$ : $f(x) = (x - 1)^{4/3} + 4(x - 1)^{1/3}$ , so $f'(x) = \frac{4}{3}(x - 1)^{1/3} + \frac{4}{3}(x - 1)^{-2/3}$ . Factor the lowest power: $f'(x) = \frac{4}{3}(x - 1)^{-2/3}\left((x - 1) + 1\right) = \frac{4x}{3(x - 1)^{2/3}}$ .

Why

Factoring out $(x - 1)^{-2/3}$ puts the hidden denominator in plain view. Setting the unfactored sum to zero and multiplying by $(x - 1)^{2/3}$ would find $0$ and lose $1$ .

Step 3

Numerator zero: $x = 0$ . Denominator zero: $x = 1$ , where $f(1) = 0$ is defined and $1$ is interior. Critical points: $0$ and $1$ .

Why

Both halves of the definition, read on the same fraction. The point $1$ is a vertical tangent: it will turn out to be neither max nor min, but it must be in the table.

Step 4

Values: $f(-7) = (-8)^{1/3}(-4) = (-2)(-4) = 8$ ; $f(0) = (-1)^{1/3} \cdot 3 = -3$ ; $f(1) = 0$ ; $f(2) = 1 \cdot 5 = 5$ .

Why

If the calculator refuses $(-8)^{1/3}$ , write $-\sqrt[3]{8} = -2$ by hand. The endpoint $-7$ carries the answer, so dropping it on an ERROR message costs the maximum.

Step 5

Maximum $8$ at $x = -7$ ; minimum $-3$ at $x = 0$ . Check at an unlisted point: $f(-3) = 0$ , between $-3$ and $8$ .

Why

The comparison is immediate once the values are exact. The critical point $1$ was a candidate dismissed by the table, exactly as Theorem 2 allows.

The conclusion, written out

“ $f$ is continuous on $[-7, 2]$ . Comparing its values at the critical points $0$ and $1$ and at the endpoints $-7$ and $2$ , the absolute maximum value is $8$ , attained at $x = -7$ , and the absolute minimum value is $-3$ , attained at $x = 0$ .”

The classic mistake on this problem: Losing $x = 1$ by multiplying through by $(x - 1)^{2/3}$ , or dropping the endpoint $-7$ because the calculator returns ERROR on $(-8)^{1/3}$ , and answering max $5$ at $x = 2$ .

Learn by heart

• Absolute: compared with the whole domain. Local: compared with the domain near $c$ ; in Thomas an endpoint can be a local extremum.
• Critical point: INTERIOR point with $f' = 0$ or $f'$ undefined. Endpoints: never critical, always candidates.
• EVT: continuous on a CLOSED interval $[a, b]$ implies an absolute max and min exist, at least once each.
• Theorem 2 has no converse: a critical point is a candidate, the values of $f$ decide.
• Algebra first: $f'$ as one factored fraction, lowest power factored out, never divide by a factor.
• Calculator: radians, brackets around negative inputs, $(-8)^{2/3} = \left((-8)^{1/3}\right)^2 = 4$ .

Frequently asked questions

What is a critical point in Thomas' Calculus?

It is an interior point of the domain of the function where the derivative is zero or does not exist. Endpoints of the domain or of the interval are never critical points in Thomas, even when the derivative is undefined there, but they are always evaluated in the closed interval method. A point outside the domain is not a critical point either.

Can an endpoint be a local maximum or minimum in MATH 203?

Yes, with the definitions of Thomas used at Concordia. A local extremum compares the value at a point with the values of the function at nearby points of its domain, and at an endpoint those points lie on one side only. So the square root function on the interval from zero to four has a local minimum at zero. Some other textbooks forbid this, so name the convention on your copy.

Why do I miss critical points when the derivative has fractional exponents?

Because a negative exponent is a denominator in disguise. If you set a sum like x to the minus one fifth plus x to the four fifths equal to zero and multiply through, the point where the denominator vanishes disappears. Factor out the lowest power first and write the derivative as one fraction: its numerator gives the zeros, its denominator gives the points where the derivative is undefined.

My calculator gives an error for a negative number to the power two thirds. What do I do?

The number exists, the calculator simply refuses fractional powers of negative numbers. Take the cube root first and square it, or square first and take the cube root: negative eight to the power two thirds equals four. Never drop a candidate from the table because of that error message, it may be the endpoint that carries the maximum.

Practise it

Corrected exercises: Extreme values of functions, MATH 203 at Concordia

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Indeterminate forms and L'Hôpital's Rule Next sheet Monotonic functions and the First Derivative Test

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-extreme-values. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: extreme values of functions (MATH 203)

The essentials

Absolute or local, with Thomas's definitions

Candidates: critical points, and what Theorem 2 does not say

The rules in table form

Critical point, candidate, or neither?

The mistakes that cost marks

1. Dividing by a factor and losing a critical point

2. Leaving a negative exponent in f' and missing the undefined point

3. Typing a negative number without brackets

4. Taking a calculator ERROR for an undefined value

5. Evaluating in degree mode

6. Making a table of f' instead of f

7. Keeping a critical point outside the interval

Which method to choose

Which algebraic gesture, by the form of f'

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 fractional exponent, a vertical tangent, and a negative cube root

Learn by heart

Frequently asked questions

What is a critical point in Thomas' Calculus?

Can an endpoint be a local maximum or minimum in MATH 203?

Why do I miss critical points when the derivative has fractional exponents?

My calculator gives an error for a negative number to the power two thirds. What do I do?

Corrected exercises: Extreme values of functions, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?