MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: Rolle's theorem and the Mean Value Theorem (MATH 140)

This sheet is not a summary of section 4.2 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on Rolle's theorem and the Mean Value Theorem in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter is short and its statements fit in two lines, which is exactly why it is marked severely: the examiner is not testing whether you can solve $f'(c) = m$ , but whether you check the hypotheses, keep the right $c$ , and use an existence theorem as an existence theorem. Every value below is exact and done by hand, as on the exam.

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The thread of the chapter

The Mean Value Theorem is an existence statement bought with hypotheses: continuity on the CLOSED $[a, b]$ and differentiability on the OPEN $(a, b)$ , each checked with a reason. It then promises a $c$ it never locates, and it is used for what it FORBIDS: a second root, a slope never reached, a non-constant function with zero derivative, a change larger than the bound on $f'$ .

This chapter is part of MATH 140, Calculus 1 (McGill)

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
Functions with zero derivative
Mean Value Inequality
Intermediate Value Theorem
Counting real roots
Average rate of change
Proof by contradiction

The essentials

Two theorems, the same hypotheses, the same kind of conclusion

• Hypothesis (1): $f$ continuous on the CLOSED interval $[a, b]$ , endpoints included. Hypothesis (2): $f$ differentiable on the OPEN interval $(a, b)$ ; the endpoints are not required.
• Rolle adds hypothesis (3), $f(a) = f(b)$ , and concludes: $f'(c) = 0$ for at least one $c$ in $(a, b)$ .
• Mean Value Theorem, with (1) and (2) only: $f(b) - f(a) = f'(c)(b - a)$ for at least one $c$ in $(a, b)$ . The tangent at $c$ is parallel to the chord.
• Both are EXISTENCE statements: at least one $c$ , strictly inside, never located. There may be several, and $c$ is the midpoint only for a quadratic.
• Hypotheses are sufficient, not necessary: if one fails the theorem guarantees nothing, and the conclusion may hold or fail.

[-2, 2]

the chord of

y = x^3

has slope

4

, and the curve has TWO tangents of slope

4

, at

\pm\frac{2\sqrt 3}{3}

: the theorem promises at least one, never exactly one.

A marker gives the method mark for each hypothesis stated WITH its reason: polynomial, hence continuous and differentiable on $\mathbb{R}$ . A bare list of the hypotheses earns less; skipping them earns nothing, even with the right $c$ .

The four uses: what the theorem forbids

• Zero derivative: $f' = 0$ on an INTERVAL gives $f$ constant there; $f' = g'$ on an interval gives $f = g + C$ . On two pieces, one constant per piece.
• Bounds: $m \le f'(x) \le M$ on $(a, b)$ gives $m(b - a) \le f(b) - f(a) \le M(b - a)$ . Bound $f'$ on the whole interval, never compute $c$ .
• At most one root: if $f(r_1) = f(r_2) = 0$ , Rolle gives a zero of $f'$ between them. If $f'$ never vanishes, two roots are impossible.
• Counting: if $f'$ has $n$ distinct zeros, $f$ has at most $n + 1$ distinct roots. The converse direction is false.
• Exactly one root: existence by the Intermediate Value Theorem, then at most one by Rolle. Two theorems, two paragraphs.

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. Reading f(a) = f(b) and skipping the other two hypotheses

the whole question

What not to write

“ $\tan 0 = \tan \pi = 0$ , so by Rolle there is $c$ in $(0, \pi)$ with $\sec^2 c = 0$ .”

What to write

“ $\tan x$ is not defined at $\frac{\pi}{2} \in [0, \pi]$ , so it is not continuous on $[0, \pi]$ : Rolle's theorem does not apply, and indeed $\sec^2 x \ge 1$ .”

\tan 0 = \tan \pi = 0

, but the graph breaks at

x = \frac{\pi}{2}

into two rising branches: no horizontal tangent anywhere, because continuity on

[0, \pi]

fails.

Why: Equal values at the ends are the least demanding of the three hypotheses. A function that is undefined or jumps somewhere in $[a, b]$ owes you nothing.

2. Applying Rolle to a function with a cusp

2 to 3 marks

What not to write

“ $f(x) = 1 - x^{2/3}$ has $f(-1) = f(1) = 0$ and is continuous, so $f'(c) = 0$ for some $c$ in $(-1, 1)$ .”

What to write

“ $f'(x) = -\frac{2}{3x^{1/3}}$ for $x \ne 0$ and $f'(0)$ does not exist, so hypothesis (2) fails at $0$ : Rolle does not apply, and $f'$ is never $0$ .”

f(-1) = f(1) = 0

and the top is at

x = 0

, but it is a cusp, not a horizontal tangent: one missing derivative and Rolle's conclusion is gone.

Why: Differentiability must hold at EVERY point of $(a, b)$ ; one corner or cusp is enough to lose the conclusion. Check the points where the formula for $f'$ divides by zero.

3. Keeping a value of c that is not strictly inside

1 mark

What not to write

“ $f(x) = x^3 + x$ on $[-1, 2]$ : slope $\frac{10 - (-2)}{3} = 4$ , $3c^2 + 1 = 4$ , so $c = -1$ or $c = 1$ .”

What to write

“ $c^2 = 1$ gives $c = \pm 1$ ; $c = -1$ is the endpoint $a$ , not in $(-1, 2)$ , so it is rejected, and $c = 1$ .”

Why: The theorem speaks about $c$ in the OPEN interval. Solving $f'(c) = m$ is algebra; selecting the solutions in $(a, b)$ is the theorem, and an endpoint is never an answer.

4. Refusing the theorem because f' does not exist at an endpoint

the whole question

What not to write

“ $f(x) = \sqrt x$ is not differentiable at $0$ , so the Mean Value Theorem does not apply on $[0, 4]$ .”

What to write

“ $\sqrt x$ is continuous on $[0, 4]$ and differentiable on $(0, 4)$ , which is all the theorem asks. Slope $\frac{2 - 0}{4} = \frac{1}{2}$ , and $\frac{1}{2\sqrt c} = \frac{1}{2}$ gives $c = 1$ .”

Why: Differentiability is required on the OPEN interval only. A vertical tangent at an endpoint is allowed; continuity at that endpoint is what is needed.

5. Declaring constant a function whose domain is not an interval

2 marks

What not to write

“ $f(x) = \frac{x}{|x|}$ has $f'(x) = 0$ for every $x \ne 0$ , so $f$ is constant.”

What to write

“ $f' = 0$ on $(-\infty, 0)$ and on $(0, \infty)$ , two intervals, so $f$ is constant on EACH: $f = -1$ on the left, $f = 1$ on the right.”

Why: The proof applies the Mean Value Theorem between two points, which needs the whole segment between them in the domain. Across a gap there is one constant per piece.

6. Computing c in a proof of an inequality

the method mark, and usually a false line

What not to write

“ $\sin b - \sin a = \cos c\,(b - a)$ with $c = \frac{a + b}{2}$ , so ...”

What to write

“ $\sin b - \sin a = \cos c\,(b - a)$ for some $c$ between $a$ and $b$ ; whatever $c$ is, $|\cos c| \le 1$ , so $|\sin b - \sin a| \le |b - a|$ .”

Why: $c$ is unknown and depends on $a$ and $b$ . The proof bounds $f'$ on the whole interval; it never needs, and cannot have, the value of $c$ .

7. Claiming exactly one root from the Intermediate Value Theorem alone

the whole question

What not to write

“ $f(x) = x^3 - x$ : $f(-2) = -6 < 0 < 6 = f(2)$ , so $f$ has exactly one root in $(-2, 2)$ .”

What to write

“By the IVT, $f$ has AT LEAST one root in $(-2, 2)$ .” In fact $x^3 - x$ has three, $-1$ , $0$ and $1$ .

Why: The IVT gives existence only. Uniqueness needs a second argument, Rolle by contradiction, which requires $f' \ne 0$ ; here $f'(x) = 3x^2 - 1$ vanishes twice.

8. Reading Rolle backwards

the whole question

What not to write

“ $f'(x) = 2x$ vanishes at $0$ , so by Rolle $f(x) = x^2 + 1$ has two roots around $0$ .”

What to write

“Rolle goes from two roots of $f$ to one root of $f'$ , not back. $x^2 + 1 \ge 1$ has no real root at all.”

Why: A zero of $f'$ says nothing about the roots of $f$ . The correct direction bounds the count: $n$ zeros of $f'$ allow at most $n + 1$ roots of $f$ .

Which method to choose

Which theorem, by the WORDS of the question

Read the verb and the conclusion asked for: they pick the theorem

If show there is $c$ with $f'(c) = 0$ , and $f(a) = f(b)$ is given or computable → Rolle, three hypotheses checked

Example: $\sin x + \cos x$ on $\left[0, \frac{\pi}{2}\right]$ : $c = \frac{\pi}{4}$
If find $c$ with $f'(c)$ equal to a slope, or $f(b) - f(a) = f'(c)(b - a)$ → Mean Value Theorem, then keep only the c in (a, b)

Example: $x^3 + x$ on $[-1, 2]$ : $c = 1$ , and $c = -1$ rejected
If $f(0) \ne f(5)$ but a horizontal tangent is asked → Intermediate Value Theorem to create two equal values, then Rolle

Example: $f(0) = 1$ , $f(2) = 7$ , $f(5) = 4$ : $f(d) = 4$ for some $d$ in $(0, 2)$ , Rolle on $[d, 5]$
If prove an inequality between $f(b) - f(a)$ and $b - a$ → Mean Value Theorem, then bound f' on the whole interval

Example: $\frac{x}{1 + x^2} < \arctan x < x$ for $x > 0$
If at least one root → Intermediate Value Theorem, with two values of opposite signs

Example: $x^5 + 3x + 1$ : $f(-1) = -3$ , $f(0) = 1$
If at most one root, or at most $n$ roots → Rolle by contradiction, from the zeros of f'

Example: $x^4 + 4x + k$ : $f'$ has one zero, so at most two roots
If $f$ is constant, or $f = g + C$ , or an identity to prove → derivative zero on an interval, constant read at one point

Example: $\arcsin x + \arccos x = \frac{\pi}{2}$

Exactly one root takes two branches, existence then uniqueness, and the write-up has two paragraphs. The increasing and decreasing test belongs to the next chapter: in this one, uniqueness is proved by Rolle.

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.

Proving that an equation has exactly one real root

When to use it: Any question with the words exactly one, a unique solution, or one and only one

1 Write the equation as $f(x) = 0$ and state that $f$ is continuous and differentiable on $\mathbb{R}$ , with the reason (polynomial, sum of differentiable functions).
2 Existence: compute two values of opposite signs and name the Intermediate Value Theorem on that closed interval.
3 Uniqueness: suppose two roots $r_1 < r_2$ ; name Rolle's theorem with its three hypotheses on $[r_1, r_2]$ to get $f'(c) = 0$ .
4 Show that $f'$ never vanishes, by bounding it: $5x^4 + 3 \ge 3$ , $3 - \sin x \ge 2$ .
5 Write the contradiction and the conclusion, with the location of the root.

Concluding sentence

“By the Intermediate Value Theorem, $f$ has at least one root in $(-1, 0)$ . If it had two, $r_1 < r_2$ , then $f$ would be continuous on $[r_1, r_2]$ , differentiable on $(r_1, r_2)$ with $f(r_1) = f(r_2)$ , and Rolle's theorem would give $f'(c) = 0$ ; but $f'(x) \ge 3 > 0$ . So $f$ has exactly one real root, in $(-1, 0)$ .”

The trap: Stopping after the IVT, or proving uniqueness first and forgetting that a function with $f' > 0$ may still have no root at all.

Marking: Typically 3 marks for existence, 4 for the Rolle argument with its hypotheses, 2 for the bound on f', 1 for the conclusion.

Proving an inequality with the Mean Value Theorem

When to use it: Prove that $f(x) < g(x)$ for $x > 0$ , or bound $|f(a) - f(b)|$ , where the difference is a change in one function

1 Choose the function and the interval: for $\sqrt{1 + x} < 1 + \frac{x}{2}$ , take $f(t) = \sqrt{1 + t}$ on $[0, x]$ .
2 Check continuity on $[0, x]$ and differentiability on $(0, x)$ , with the reason.
3 Write the conclusion with the unknown $c$ and its interval: $\sqrt{1 + x} - 1 = \frac{x}{2\sqrt{1 + c}}$ , $0 < c < x$ .
4 Bound $f'(c)$ using ONLY $0 < c < x$ : $\sqrt{1 + c} > 1$ , so $\frac{1}{2\sqrt{1 + c}} < \frac{1}{2}$ .
5 Multiply by the positive length and conclude, keeping the strictness that comes from $c$ being strictly inside.

Concluding sentence

“By the Mean Value Theorem there is $c$ in $(0, x)$ with $\sqrt{1 + x} - 1 = \frac{x}{2\sqrt{1 + c}}$ . Since $c > 0$ , $\sqrt{1 + c} > 1$ , so $\sqrt{1 + x} - 1 < \frac{x}{2}$ .”

The trap: Multiplying an inequality by $b - a$ without saying it is positive, or bounding $f'(c)$ by a value at an endpoint that $c$ never reaches.

Marking: Typically 2 marks for the hypotheses, 3 for the conclusion with c, 3 for the bound on f'(c), 2 for the final inequality.

Check before you hand in

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

Every c strictly inside

Compare each value of c with a and b before writing it as an answer. An endpoint or a value outside the interval is a solution of the algebra, not of the theorem.

$x^3 + x$ on $[-1, 2]$ gives $c = \pm 1$ : $-1$ equals $a$ , so only $c = 1$ survives.
Plug c back into f'

Recompute f'(c) with the value found and compare with the chord slope computed separately. A mismatch means an error in the slope or in f'.

$\sqrt x$ on $[0, 4]$ : $f'(1) = \frac{1}{2}$ and the slope is $\frac{2}{4} = \frac{1}{2}$ . A slope of $\frac{2}{3}$ would reveal a wrong endpoint value.
Count the hypotheses you wrote

Rolle needs three lines, the Mean Value Theorem two, each with a reason. Count them on your paper before moving to the conclusion.

“ $f$ is a polynomial, hence continuous on $[0, 2]$ and differentiable on $(0, 2)$ ”: two hypotheses in one sentence, both justified.
Test an inequality at one easy point

Before proving an inequality in general, check it at a value where everything is exact. If it fails there, the direction is wrong.

$\sqrt{1 + x} < 1 + \frac{x}{2}$ at $x = 3$ : $2 < 2.5$ . The reverse inequality would already fail here.

The typical problem, taken apart

Exactly one solution, then a guaranteed location

Prove that the equation $e^x = 2 - x$ has exactly one real solution $r$ , that $0 < r < \frac{1}{2}$ , and then, with the Mean Value Theorem, that $\frac{1}{1 + \sqrt e} < r$ .

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

The rising curve

y = e^x

and the falling line

y = 2 - x

cross once, between

0

and

\frac{1}{2}

: the picture suggests it, the proof must show it.

Step 1

Let $f(x) = e^x + x - 2$ , so the equation is $f(x) = 0$ . $f$ is a sum of differentiable functions, hence continuous and differentiable on $\mathbb{R}$ .

Why

Putting everything on one side turns an equation into the root of one function, the form the theorems speak about. The regularity sentence is the permit for the three theorems that follow.

Step 2

$f(0) = 1 + 0 - 2 = -1 < 0$ and $f\left(\frac{1}{2}\right) = \sqrt e - \frac{3}{2} > 0$ , because $e > \frac{9}{4}$ gives $\sqrt e > \frac{3}{2}$ . By the Intermediate Value Theorem, $f$ has a root $r$ in $\left(0, \frac{1}{2}\right)$ .

Why

Existence and location together. The sign of $f\left(\frac{1}{2}\right)$ comes from $e > 2.25$ , an inequality anyone can justify, not from a decimal value of $\sqrt e$ .

Step 3

Suppose $f$ had two roots $r_1 < r_2$ . $f$ is continuous on $[r_1, r_2]$ , differentiable on $(r_1, r_2)$ , and $f(r_1) = f(r_2) = 0$ , so by Rolle's theorem $f'(c) = 0$ for some $c$ . But $f'(x) = e^x + 1 > 1$ . Contradiction: $r$ is the only root.

Why

The uniqueness paragraph is where the marks for the theorem are. Naming the three hypotheses of Rolle inside the contradiction is expected, even though they are obvious here.

Step 4

Mean Value Theorem on $[0, r]$ : $f(r) - f(0) = f'(c)\,r$ for some $c$ in $(0, r)$ , that is $1 = (e^c + 1)\,r$ , so $r = \frac{1}{e^c + 1}$ .

Why

The theorem converts the known change of $f$ , from $-1$ to $0$ , into an exact formula for $r$ with one unknown, $c$ , whose interval is known.

Step 5

From $0 < c < r < \frac{1}{2}$ and $e^x$ increasing: $1 < e^c < \sqrt e$ , so $2 < e^c + 1 < 1 + \sqrt e$ , and taking reciprocals: $\frac{1}{1 + \sqrt e} < r < \frac{1}{2}$ .

Why

Only the interval of $c$ is used, never its value. Reversing the inequalities when taking reciprocals of positive numbers is the step to write explicitly.

Step 6

Check: $\frac{1}{1 + \sqrt e} < \frac{1}{2}$ because $\sqrt e > 1$ , so the bracket is not empty; and with $\sqrt e < \frac{7}{4}$ (since $e < \frac{49}{16}$ ), $\frac{1}{1 + \sqrt e} > \frac{4}{11}$ .

Why

A bracket whose lower end exceeds its upper end reveals a reversed inequality. The last line turns the exact bound into a fraction readable without a calculator.

The conclusion, written out

“The equation $e^x = 2 - x$ has exactly one real solution $r$ , and $\frac{1}{1 + \sqrt e} < r < \frac{1}{2}$ ; in particular $\frac{4}{11} < r < \frac{1}{2}$ .”

The classic mistake on this problem: Proving existence by the figure, or writing $f' > 0$ so there is one root, without the Intermediate Value Theorem: a function with positive derivative, such as $e^x$ , may have no root at all.

Learn by heart

• Rolle: continuous on $[a, b]$ , differentiable on $(a, b)$ , $f(a) = f(b)$ $\Rightarrow$ $f'(c) = 0$ , $c \in (a, b)$ .
• MVT: continuous on $[a, b]$ , differentiable on $(a, b)$ $\Rightarrow$ $f(b) - f(a) = f'(c)(b - a)$ , $c \in (a, b)$ .
• At least one $c$ , strictly inside, never located. Hypotheses sufficient, not necessary.
• $f' = 0$ on an INTERVAL: constant. $f' = g'$ on an interval: $f = g + C$ . One constant per piece.
• Inequality: bound $f'$ on the whole interval; never compute $c$ .
• Roots: IVT for at least one, Rolle by contradiction for at most one; $n$ zeros of $f'$ allow at most $n + 1$ roots.

Frequently asked questions

What is the difference between Rolle's theorem and the Mean Value Theorem?

Rolle's theorem is the special case of the Mean Value Theorem where the function takes the same value at both ends of the interval. Both need continuity on the closed interval and differentiability on the open interval. Rolle then gives a point where the derivative is zero; the Mean Value Theorem gives a point where the derivative equals the slope of the chord joining the endpoints.

Do I have to check the hypotheses of the Mean Value Theorem on an exam?

Yes, every time, with a reason. Write that the function is continuous on the closed interval and differentiable on the open interval, and say why, for instance because it is a polynomial or because its only problem point lies outside the interval. Without that sentence the conclusion is not justified, and most of the method marks are lost even if the number c is right.

How do I prove an equation has exactly one real root?

Use two theorems. First the Intermediate Value Theorem, with two values of opposite signs, to show that at least one root exists. Then suppose there are two roots and apply Rolle's theorem between them: the derivative would vanish somewhere, so if you can show the derivative is never zero, you get a contradiction and at most one root. Together they give exactly one.

Why can't I find the value of c in the Mean Value Theorem proof of an inequality?

Because the theorem only says that such a c exists somewhere strictly between a and b; it gives no formula for it, and the point changes when a and b change. You do not need it: bound the derivative on the whole interval, for instance the cosine is between minus one and one, and that bound holds at c wherever it is.

Practise it

Corrected exercises: Rolle's theorem and the Mean Value Theorem, MATH 140 at McGill

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

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-mean-value-theorem. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: Rolle's theorem and the Mean Value Theorem (MATH 140)

The essentials

Two theorems, the same hypotheses, the same kind of conclusion

The four uses: what the theorem forbids

The mistakes that cost marks

1. Reading f(a) = f(b) and skipping the other two hypotheses

2. Applying Rolle to a function with a cusp

3. Keeping a value of c that is not strictly inside

4. Refusing the theorem because f' does not exist at an endpoint

5. Declaring constant a function whose domain is not an interval

6. Computing c in a proof of an inequality

7. Claiming exactly one root from the Intermediate Value Theorem alone

8. Reading Rolle backwards

Which method to choose

Which theorem, by the WORDS of the question

How the answer is expected to be written

Proving that an equation has exactly one real root

Proving an inequality with the Mean Value Theorem

Check before you hand in

The typical problem, taken apart

Exactly one solution, then a guaranteed location

Learn by heart

Frequently asked questions

What is the difference between Rolle's theorem and the Mean Value Theorem?

Do I have to check the hypotheses of the Mean Value Theorem on an exam?

How do I prove an equation has exactly one real root?

Why can't I find the value of c in the Mean Value Theorem proof of an inequality?

Corrected exercises: Rolle's theorem and the Mean Value Theorem, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?