Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: Rolle's theorem and the mean value theorem (201-SN2-RE)

Q: What is the difference between Rolle's theorem and the mean value theorem?

Rolle's theorem also requires the function to take the same value at both endpoints, and its conclusion is that the derivative is zero somewhere. The mean value theorem drops that condition and concludes that the derivative equals the slope of the secant line. Rolle's theorem is therefore the special case where the secant line is horizontal.

Q: Why is the interval closed for continuity and open for differentiability?

Because the conclusion only concerns an interior point, and the proof only needs the derivative on the inside. This asymmetry lets you apply the theorem to functions whose derivative becomes infinite at the endpoints, such as the square root of one minus x squared on the interval from minus one to one.

Q: Does Rolle's theorem give a single point c?

No. It guarantees that there is at least one point where the derivative is zero, and there can be several, or even infinitely many. Writing a single c costs a mark at every application. If the question asks for uniqueness, it is proved separately, through the strict monotonicity of the derivative.

Q: How do you find the value of c in the mean value theorem?

By solving the equation that sets the derivative equal to the average rate of change, that is, the difference of the values at the endpoints divided by the length of the interval. You then keep only the roots that lie in the open interval. The theorem itself never provides this value, and the midpoint of the interval works only in exceptional cases.

Q: Does a zero derivative prove that the function is constant?

Only on an interval. If the domain is made of several disjoint pieces, the function is constant on each one but the constants can differ: the function equal to one on the positive real numbers and minus one on the negative ones has a zero derivative wherever it is defined, and it is not constant.

These theorems are memorized in one line and lost in three words: a forgotten hypothesis, one “unique” too many, a $c$ you claim to compute. This sheet deals with exactly those three losses, with the counterexamples that make each hypothesis indisputable.

It is written for CEGEP students in the Science program (Sciences de la nature) taking Differential Calculus, 201-SN2-RE (formerly 201-NYA). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Rolle's theorem and the mean value theorem give the EXISTENCE of a $c$ , never its value or its uniqueness. Marks are lost by skipping a hypothesis, or by concluding more than the theorem allows.

This chapter is part of 201-SN2-RE, Differential Calculus

What is on this sheet

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

Topics covered

Rolle's theorem
Mean value theorem
Continuity on a closed interval
Differentiability on an open interval
Auxiliary function
Mean value inequality
Constant function on an interval
Intermediate Value Theorem
Existence and uniqueness
Strict monotonicity
Average velocity and instantaneous velocity

The essentials

Rolle's theorem: three hypotheses, an existence conclusion

• 1. $f$ CONTINUOUS on the CLOSED interval $[a, b]$ .
• 2. $f$ DIFFERENTIABLE on the OPEN interval $\left(a, b\right)$ .
• 3. $f(a)=f(b)$ .
• Conclusion: there exists AT LEAST ONE $c\in\left(a, b\right)$ such that $f'(c)=0$ .
• The $c$ is always INTERIOR: never $a$ , never $b$ .

The secant line joining

(a, f(a))

(b, f(b))

is HORIZONTAL, so there is at least one interior

c

where the tangent line is horizontal too. The theorem says such a

c

exists, not where it is.

The asymmetry, closed for continuity and open for differentiability, is deliberate: it lets you apply the theorem to functions whose derivative blows up at the endpoints, such as $\sqrt{1-x^{2}}$ on $[-1, 1]$ .

Mean value theorem: the same thing without equal values at the endpoints

• Same hypotheses 1 and 2, without hypothesis 3. Conclusion: there exists $c\in\left(a, b\right)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$ .
• Geometrically: some tangent line is PARALLEL to the secant line.
• In terms of motion: the instantaneous velocity equals the average velocity at least once.
• Rolle's theorem is the special case where the secant line is horizontal: the two theorems are really one.

[0, 4]

, the secant line of

y=\sqrt{x}

has slope

\frac{2-0}{4}=\frac{1}{2}

; at

c=1

, where

\frac{1}{2\sqrt{c}}=\frac{1}{2}

, the tangent line (green) is parallel to it.

The proof applies Rolle's theorem to $g(x)=f(x)-\left[f(a)+\frac{f(b)-f(a)}{b-a}(x-a)\right]$ , the vertical gap between the curve and its secant line, which is zero at both endpoints.

The mean value inequality, and the two consequences

• If $m\leq f'\leq M$ on $[a, b]$ , then $m(b-a)\leq f(b)-f(a)\leq M(b-a)$ .
• If $\left|f'\right|\leq M$ , then $\left|f(b)-f(a)\right|\leq M\left|b-a\right|$ , which gives for example $\left|\sin a-\sin b\right|\leq\left|a-b\right|$ .
• On an INTERVAL: $f'=0$ is equivalent to $f$ constant.
• On an INTERVAL: $f'=g'$ is equivalent to $f-g$ constant. This is what will give the “ $+C$ ” of integral calculus.

[0, 2]

\cos 2\le(\sin x)'\le 1

: the graph of

\sin x

stays inside the wedge between the two lines of slopes

m

and

M

through the origin. That is the inequality

m(b-a)\le f(b)-f(a)\le M(b-a)

The word INTERVAL is the hidden hypothesis of the last two lines. On a domain in two pieces, the two constants can differ, and that is a classic exam counterexample.

Existence and uniqueness: two arguments of different kinds

• The Intermediate Value Theorem gives the EXISTENCE of a root, through a sign change.
• Strict MONOTONICITY, justified by the sign of $f'$ , gives UNIQUENESS.
• Rolle's theorem also gives uniqueness by contradiction: two roots of $f$ would force a root of $f'$ between them.
• A complete write-up handles the two in TWO separate sentences, never in one.

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. Applying Rolle's theorem without checking equal values at the endpoints

the whole question

What not to write

“ $f(x)=x^{2}$ is continuous and differentiable on $[1, 3]$ , so by Rolle's theorem there exists $c$ with $f'(c)=0$ .”

What to write

“ $f(1)=1$ and $f(3)=9$ are not equal: Rolle's theorem does not apply. The mean value theorem is the one to use, and it gives $f'(c)=4$ .”

f(1)\neq f(3)

: the secant line rises, and

x^{2}

has no horizontal tangent line in

(1, 3)

. Rolle's theorem says nothing; the mean value theorem gives a tangent line of slope

4

Why: On $[1, 3]$ , $f'(x)=2x$ is never zero: the conclusion of Rolle's theorem is plainly FALSE here, which shows that the third hypothesis is not decorative.

2. Skipping differentiability because the function “looks smooth”

the whole question, and the credibility of the paper

What not to write

“ $f(x)=|x|$ satisfies $f(-1)=f(1)=1$ and it is continuous, so Rolle's theorem gives a $c$ with $f'(c)=0$ .”

What to write

“ $f$ is not differentiable at $0$ , which is interior to $[-1, 1]$ : hypothesis 2 fails, and indeed $f'$ equals $-1$ or $+1$ wherever it exists, never $0$ .”

Every hypothesis of Rolle's theorem but one:

|x|

is continuous on

[-1, 1]

and takes the same value at both endpoints, but it is not differentiable at

0

, and no point has a horizontal tangent line.

Why: A single missing hypothesis is enough to make the conclusion false. This counterexample is the one most often asked on exams, precisely because it fits in one drawing.

3. Concluding that the $c$ is unique

1 mark, and often the next question

What not to write

“By Rolle's theorem, there exists EXACTLY ONE $c$ such that $f'(c)=0$ .”

What to write

“By Rolle's theorem, there exists AT LEAST ONE $c$ such that $f'(c)=0$ .” Uniqueness, if the question asks for it, is proved separately, through the strict monotonicity of $f'$ .

\sin 0=\sin 3\pi=0

: Rolle's theorem promises AT LEAST one

c

, and the curve has three, at

\frac{\pi}{2}

\frac{3\pi}{2}

and

\frac{5\pi}{2}

. Uniqueness is another question.

Why: A function can have ten horizontal tangent lines between $a$ and $b$ : $f(x)=\sin(4\pi x)$ on $[0, 1]$ has four. The theorem counts at least one $c$ , never exactly one.

4. Claiming the theorem gives the value of $c$

2 marks

What not to write

“By the mean value theorem, $c=\frac{a+b}{2}$ .”

What to write

“The theorem guarantees that $c$ exists; to find it you must SOLVE $f'(c)=\frac{f(b)-f(a)}{b-a}$ , which is not always possible.”

x^{3}-x

, the midpoint

1

gives a tangent line of slope

2

(red), which is not parallel to the secant line. The right

c=\frac{2}{\sqrt{3}}

(green) comes from solving

f'(c)=3

Why: The midpoint only works for a quadratic, by a quirk of symmetry. For $f(x)=x^{3}-x$ on $[0, 2]$ , $c=\frac{2}{\sqrt{3}}\approx 1.15$ and not $1$ .

5. Declaring a function constant on a domain in two pieces

the whole question, and the same mistake will come back in integral calculus

What not to write

“ $f'(x)=0$ for every $x\neq 0$ , so $f$ is constant on $\mathbb{R}\setminus\{0\}$ .”

What to write

“ $f$ is constant ON EACH interval of its domain. The function equal to $1$ on $\left(0, +\infty\right)$ and $-1$ on $\left(-\infty, 0\right)$ has a zero derivative and is not constant.”

f'=0

everywhere on

\mathbb{R}\setminus\{0\}

, and yet

f

equals

1

on the right and

-1

on the left: it is constant on EACH interval, not on the whole domain.

Why: The consequence “ $f'=0$ , so $f$ is constant” is proved WITH the mean value theorem, applied between two points of the interval. Without an interval, the two points cannot be connected.

6. Mixing up the conclusion of Rolle's theorem with that of the IVT

1 to 2 marks

What not to write

“By Rolle's theorem, there exists $c$ such that $f(c)=0$ .”

What to write

“By Rolle's theorem, there exists $c$ such that $f'(c)=0$ .” The IVT (Intermediate Value Theorem) is the one about $f(c)=0$ , but it requires a change of SIGN, not equal values at the endpoints.

Why: The two theorems look alike in their wording and differ in their object: the IVT is about the values of $f$ , Rolle's theorem about those of $f'$ . Reread the conclusion you wrote before turning the page.

7. Bounding $f'$ at a single point to apply the inequality

2 marks, and an inequality that is false in general

What not to write

“ $f'(0)=1$ , so $\left|f(b)-f(a)\right|\leq\left|b-a\right|$ .”

What to write

Why: The mean value inequality bounds by the maximum of $\left|f'\right|$ over the WHOLE interval. A local bound says nothing about what the derivative does elsewhere.

8. Justifying continuity with “it is obvious”

1 mark for rigour, at every application

What not to write

“ $f$ is obviously continuous and differentiable, so Rolle's theorem applies.”

What to write

“ $f$ is a polynomial, so it is continuous and differentiable on $\mathbb{R}$ , in particular on $[a, b]$ and on $\left(a, b\right)$ .”

Why: The marker looks for the REASON, not the claim: a polynomial, a quotient whose denominator is never zero, a composition of continuous functions. It is the easiest mark of the chapter and the one most often lost.

Which method to choose

Which theorem, from the verb of the question

The four theorems of this block look alike. It is the exact wording of the question that points to the right one, and the choice is made before any computation.

If “show that the equation has a solution”, “that there exists a root” → Intermediate Value Theorem, with the sign change written out

Example: $x^{3}+x-1=0$ on $[0, 1]$ , since $f(0)=-1$ and $f(1)=1$
If “show that there exists $c$ such that $f'(c)=0$ ”, with $f(a)=f(b)$ → Rolle's theorem, its three hypotheses checked one by one

Example: $f(x)=x^{2}-4x+3$ on $[1, 3]$ , where $f(1)=f(3)=0$
If “there is an instant when the instantaneous velocity equals the average velocity”, or “a tangent line parallel to the secant line” → mean value theorem

Example: the average speed camera: an average speed of $115$ proves the limit was exceeded
If “bound”, “find an upper bound for $\left|f(b)-f(a)\right|$ ”, “prove an inequality” → mean value inequality, with a bound valid on the WHOLE interval

Example: $\left|\sin a-\sin b\right|\leq\left|a-b\right|$ since $\left|\cos\right|\leq 1$
If “show that $f$ is constant” or “that $f=g+C$ ” → show $f'=0$ , or $f'=g'$ , ON AN INTERVAL

Example: $\arctan x+\arctan\frac{1}{x}$ is constant on each of the two intervals, with two different constants

Checking that the domain really is an interval is part of the proof: that is where the mark is won or lost.
If “a UNIQUE solution” → IVT for existence, then strict monotonicity for uniqueness, in two separate sentences

If the question mixes the two, for example “show that $f'$ is zero at least twice”, apply Rolle's theorem TWICE, on two subintervals bounded by three points where $f$ takes the same value.

What to do when a hypothesis is missing

Before concluding that a theorem does not apply, look at WHICH hypothesis fails: some can be repaired, others cannot.

If $f(a)\neq f(b)$ → Rolle's theorem fails, but the mean value theorem applies: the conclusion becomes $f'(c)=\frac{f(b)-f(a)}{b-a}$

Example: $x^{2}$ on $[1, 3]$ gives $f'(c)=4$ , so $c=2$
If $f$ is not differentiable at an INTERIOR point → nothing to be done, the conclusion may be false; this is the $|x|$ counterexample

Example: $|x|$ on $[-1, 1]$ has no horizontal tangent line
If $f$ fails to be differentiable only AT THE ENDPOINTS → no problem, hypothesis 2 is only about the open interval

Example: $\sqrt{1-x^{2}}$ on $[-1, 1]$ , whose derivative blows up at $\pm 1$

That is exactly why the statement of the theorem distinguishes closed and open intervals.
If $f$ is not continuous at an endpoint → the theorem fails: a function that jumps at the end may have no horizontal tangent line

Example: $f(x)=x$ on $[0, 1)$ extended by $f(1)=0$
If the interval is not bounded → apply the theorem on a closed subinterval, then let it grow

Example: show that a function with zero derivative on $\mathbb{R}$ is constant

A hypothesis is checked by NAMING it and giving its reason. “ $f$ is continuous because it is a quotient of polynomials whose denominator is never zero on $[a, b]$ ” earns the mark; “ $f$ is continuous” does not.

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.

Applying the mean value theorem, written for the marker

When to use it: The question asks you to show that there is an instant, an x-value or a point where the derivative takes a given value.

1 Name the function and the interval: “Let $f(x)=x^{3}-x$ on $[0, 2]$ .”
2 Justify continuity BY ITS REASON: “ $f$ is a polynomial, so it is continuous on $[0, 2]$ .”
3 Justify differentiability on the OPEN interval: “ $f$ is differentiable on $\left(0, 2\right)$ , with derivative $f'(x)=3x^{2}-1$ .”
4 Compute the average rate of change and write it down: “ $\frac{f(2)-f(0)}{2-0}=\frac{6-0}{2}=3$ .”
5 Conclude with the standard sentence, naming the theorem and placing $c$ in the OPEN interval.

Concluding sentence

Since $f$ is continuous on $[0, 2]$ and differentiable on $\left(0, 2\right)$ , the mean value theorem guarantees that there is at least one $c\in\left(0, 2\right)$ such that $f'(c)=3$ .

The trap: Writing $c\in[0, 2]$ instead of $c\in\left(0, 2\right)$ . The $c$ is always interior, and the marker checks this detail every time because it proves the statement was read.

Marking: Usually 1 mark for justified continuity, 1 mark for differentiability on the open interval, 1 mark for the average rate of change, 1 mark for the conclusion naming the theorem.

Proving an inequality with the mean value inequality

When to use it: The question asks you to bound a gap, to trap an approximate value between two bounds, or to prove an inequality of the form $|f(a)-f(b)|\leq k|a-b|$ .

1 Identify the function $f$ and the two points, even when the question does not name them: “Let $f(t)=\sin t$ , and let $a<b$ be two real numbers.”
2 Check the hypotheses on $[a, b]$ : “ $\sin$ is continuous and differentiable on $\mathbb{R}$ .”
3 Bound $\left|f'\right|$ on the WHOLE interval, with the reason: “ $\left|f'(t)\right|=\left|\cos t\right|\leq 1$ for every $t$ .”
4 Apply the inequality and write the result: “so $\left|\sin b-\sin a\right|\leq 1\times\left|b-a\right|$ .”
5 Conclude by going back to the notation of the question, and point out the case of equality if there is one.

Concluding sentence

For all real numbers $a$ and $b$ , $\left|\sin a-\sin b\right|\leq\left|a-b\right|$ , with equality only when $a=b$ .

The trap: Bounding $\left|f'\right|$ by its value at one particular point. The bound must hold on the whole interval, otherwise the inequality obtained is false as soon as the derivative exceeds that value elsewhere.

Marking: 1 mark for the hypotheses, 1 mark for bounding the derivative on the whole interval, 1 mark for the conclusion.

Check before you hand in

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

Check that the $c$ you found lies in the OPEN interval

Compare the value obtained with the two endpoints. A root equal to $a$ or $b$ , or lying outside the interval, signals an equation solved incorrectly or one root too many.

On $[0, 2]$ , the equation $3c^{2}-1=3$ gives $c=\pm\frac{2}{\sqrt{3}}$ : only the positive root $1.155$ works, and the negative one must be rejected EXPLICITLY.
Recompute the slope of the secant line on the calculator

Evaluate $f(a)$ and $f(b)$ , then divide the difference by $b-a$ . The number obtained must be the one you set $f'(c)$ equal to.

$\frac{f(2)-f(0)}{2}=\frac{6}{2}=3$ . A slope found to be $6$ signals a forgotten division by $b-a$ , the most common mistake on this question.
Go over the hypotheses before handing in

Three boxes for Rolle's theorem, two for the mean value theorem. For each one, a reason written on the paper, not only in your head.

“A polynomial, so continuous” and “a polynomial, so differentiable” are enough; “obviously continuous” earns nothing and costs the mark for the hypothesis.
Test an inequality you obtained on a particular case

Pick two simple values and check numerically. An inequality that fails in one case is simply false, and that shows in ten seconds.

$\left|\sin 1-\sin 0\right|=0.841\leq 1$ : consistent. Had you written $\leq\frac{1}{2}\left|a-b\right|$ , the test would have disproved it at once.

The typical problem, taken apart

The mean value theorem on $f(x)=x^{3}-x$ , down to the exact value of $c$

Let $f(x)=x^{3}-x$ on the interval $[0, 2]$ .

1. Check the hypotheses of the mean value theorem.

2. Find every value of $c$ whose existence the theorem guarantees.

The secant line from

(0, 0)

(2, 6)

has slope

3

: the theorem guarantees at least one point of

(0, 2)

where the tangent line is PARALLEL to it. It does not say which one.

Step 1

$f$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $\left(0, 2\right)$ , with $f'(x)=3x^{2}-1$ .

Why

One sentence, but it carries two marks of the grading scheme and gives the REASON. “Polynomial” is the word the marker looks for: it justifies both hypotheses at once.

Step 2

$f(0)=0$ and $f(2)=8-2=6$ , so the average rate of change is $\dfrac{f(2)-f(0)}{2-0}=\dfrac{6}{2}=3$ .

Why

Write the quotient BEFORE simplifying it: it is the quotient that will be set equal to $f'(c)$ , and forgetting to divide by $b-a$ is the number one mistake on this question.

Step 3

The mean value theorem therefore guarantees that there is at least one $c\in\left(0, 2\right)$ such that $f'(c)=3$ .

Why

State the conclusion BEFORE looking for $c$ . The mark for the theorem is earned even if the solving that follows fails, and many papers lose that guarantee by reversing the order.

Step 4

$3c^{2}-1=3$ gives $c^{2}=\dfrac{4}{3}$ , so $c=\pm\dfrac{2}{\sqrt{3}}$ . Only the positive root $c=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\approx 1.155$ belongs to $\left(0, 2\right)$ .

Why

The quadratic equation gives two roots and the interval rules one out. Rejecting the negative root EXPLICITLY is worth a mark: it is the only trace that the interval was reread.

Step 5

Check: $f'\!\left(\dfrac{2}{\sqrt{3}}\right)=3\times\dfrac{4}{3}-1=4-1=3$ , and $0<1.155<2$ .

Why

Two independent and immediate checks: the value of the derivative, then membership in the open interval. They catch a computing error as well as a badly chosen root.

Step 6

Remark: the midpoint $\dfrac{0+2}{2}=1$ does not work, since $f'(1)=2\neq 3$ .

Why

The midpoint only works for a quadratic. Noting it here defuses the most widespread mistake of the chapter, which is to believe the theorem more generous than it is.

The conclusion, written out

Since $f$ is continuous on $[0, 2]$ and differentiable on $\left(0, 2\right)$ , the mean value theorem guarantees that there is a $c\in\left(0, 2\right)$ such that $f'(c)=3$ ; here this $c$ is unique and equals $\frac{2\sqrt{3}}{3}\approx 1.155$ .

The classic mistake on this problem: Solving $f'(c)=6$ instead of $f'(c)=3$ , by taking the difference $f(2)-f(0)$ as the average rate of change. You then get $c=\sqrt{\frac{7}{3}}\approx 1.53$ , which does lie in the interval: nothing flags the mistake, except checking the slope of the secant line.

Learn by heart

• Rolle's theorem: continuous on the CLOSED interval, differentiable on the OPEN interval, $f(a)=f(b)$ . Conclusion: $f'(c)=0$ for at least one INTERIOR $c$ .
• Mean value theorem: same hypotheses without the third, conclusion $f'(c)=\frac{f(b)-f(a)}{b-a}$ .
• Both theorems give EXISTENCE, never uniqueness and never the value of $c$ .
• A hypothesis is justified by its REASON: “a polynomial, so continuous”, not “obviously continuous”.
• $m\leq f'\leq M$ on $[a, b]$ gives $m(b-a)\leq f(b)-f(a)\leq M(b-a)$ , with a bound valid EVERYWHERE.
• $f'=0$ on an INTERVAL is equivalent to $f$ constant. On a domain in two pieces, it is false.
• The IVT is about $f(c)$ , Rolle's theorem is about $f'(c)$ : reread the conclusion you wrote before handing in.
• Existence by the IVT, uniqueness by strict monotonicity: two sentences, never one.

Frequently asked questions

What is the difference between Rolle's theorem and the mean value theorem?

Rolle's theorem also requires the function to take the same value at both endpoints, and its conclusion is that the derivative is zero somewhere. The mean value theorem drops that condition and concludes that the derivative equals the slope of the secant line. Rolle's theorem is therefore the special case where the secant line is horizontal.

Why is the interval closed for continuity and open for differentiability?

Because the conclusion only concerns an interior point, and the proof only needs the derivative on the inside. This asymmetry lets you apply the theorem to functions whose derivative becomes infinite at the endpoints, such as the square root of one minus x squared on the interval from minus one to one.

Does Rolle's theorem give a single point c?

No. It guarantees that there is at least one point where the derivative is zero, and there can be several, or even infinitely many. Writing a single c costs a mark at every application. If the question asks for uniqueness, it is proved separately, through the strict monotonicity of the derivative.

How do you find the value of c in the mean value theorem?

By solving the equation that sets the derivative equal to the average rate of change, that is, the difference of the values at the endpoints divided by the length of the interval. You then keep only the roots that lie in the open interval. The theorem itself never provides this value, and the midpoint of the interval works only in exceptional cases.

Does a zero derivative prove that the function is constant?

Only on an interval. If the domain is made of several disjoint pieces, the function is constant on each one but the constants can differ: the function equal to one on the positive real numbers and minus one on the negative ones has a zero derivative wherever it is defined, and it is not constant.

Practise it

Corrected exercises: Rolle's theorem and the mean value theorem, 201-SN2-RE

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.

16 corrected exercises
160 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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Revision sheet: Rolle's theorem and the mean value theorem (201-SN2-RE)

The essentials

Rolle's theorem: three hypotheses, an existence conclusion

Mean value theorem: the same thing without equal values at the endpoints

The mean value inequality, and the two consequences

Existence and uniqueness: two arguments of different kinds

The mistakes that cost marks

1. Applying Rolle's theorem without checking equal values at the endpoints

2. Skipping differentiability because the function “looks smooth”

3. Concluding that the ccc is unique

4. Claiming the theorem gives the value of ccc

5. Declaring a function constant on a domain in two pieces

6. Mixing up the conclusion of Rolle's theorem with that of the IVT

7. Bounding f′f'f′ at a single point to apply the inequality

8. Justifying continuity with “it is obvious”

Which method to choose

Which theorem, from the verb of the question

What to do when a hypothesis is missing

How the answer is expected to be written

Applying the mean value theorem, written for the marker

Proving an inequality with the mean value inequality

Check before you hand in

The typical problem, taken apart

The mean value theorem on f(x)=x3−xf(x)=x^{3}-xf(x)=x3−x, down to the exact value of ccc

Learn by heart

Frequently asked questions

What is the difference between Rolle's theorem and the mean value theorem?

Why is the interval closed for continuity and open for differentiability?

Does Rolle's theorem give a single point c?

How do you find the value of c in the mean value theorem?

Does a zero derivative prove that the function is constant?

Corrected exercises: Rolle's theorem and the mean value theorem, 201-SN2-RE

See also

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3. Concluding that the $c$ is unique

4. Claiming the theorem gives the value of $c$

7. Bounding $f'$ at a single point to apply the inequality

The mean value theorem on $f(x)=x^{3}-x$ , down to the exact value of $c$