Differential Calculus 201-SN2-RE • CEGEP, Montreal

Revision sheet: applications of the derivative (201-SN2-RE)

This sheet does not state the theorems again: they fit in three lines in your notes. It deals with what decides the mark, that is, checking the hypotheses, telling what a theorem gives from what it does not give, and the exact write-up the 201-SN2-RE (formerly 201-NYA) marker looks for before even reading the computation.

It is written for CEGEP students in the Science program (Sciences de la nature). Once the sheet is read, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

The theorems of this chapter do not compute anything, they GUARANTEE: Rolle's Theorem and the Mean Value Theorem assert that a point exists without ever giving it. Their hypotheses are not decorative, and a solution that applies the theorem without checking them is worth nothing, even when its conclusion is true.

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

The essentials

Rolle and the Mean Value Theorem: same hypotheses, one conclusion

• Common hypotheses: $f$ CONTINUOUS on the closed interval $\left[a, b\right]$ , and DIFFERENTIABLE on the open interval $\left(a, b\right)$ . The two intervals are different, and this is not a detail.
• Rolle's Theorem adds $f(a)=f(b)$ and concludes: there is a $c\in\left(a, b\right)$ such that $f'(c)=0$ .
• Mean Value Theorem (Lagrange): there is a $c\in\left(a, b\right)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$ .
• In words: there is an instant when the instantaneous speed equals the average speed.
• Both theorems give the EXISTENCE of $c$ , never its value. Finding $c$ is a separate question.

The chord joins

(0, 0)

(2, 2)

and has slope

1

; the tangent line at

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

is PARALLEL to it. This is exactly what the Mean Value Theorem asserts, and it asserts nothing more.

Rolle's Theorem is the special case of the Mean Value Theorem where the chord is horizontal. Remembering one theorem is enough.

The consequences, the ones you really use

• $f'>0$ on an interval $\Rightarrow$ $f$ strictly increasing. This is a CONSEQUENCE of the Mean Value Theorem, not a definition.
• $f'=0$ on an interval $\Rightarrow$ $f$ constant on that interval.
• $f'=g'$ on an interval $\Rightarrow$ $f-g$ is constant. This is what justifies the $+C$ of antiderivatives.
• Strictly increasing $\Rightarrow$ one-to-one $\Rightarrow$ UNIQUENESS of a root, once the Intermediate Value Theorem has given its existence.

f(x)=\frac{1}{x}

has a negative derivative everywhere, and goes down on EACH branch; yet

f(-1)=-1<f(1)=1

. Without an interval, the consequence of the Mean Value Theorem says nothing.

“On an INTERVAL” is compulsory: $f(x)=\frac{1}{x}$ has a strictly negative derivative and is not decreasing on $\mathbb{R}\setminus\{0\}$ .

Rectilinear motion: six words not to confuse

• Position $s(t)$ , velocity $v=s'$ , acceleration $a=v'=s''$ .
• The object CHANGES DIRECTION where $v$ changes SIGN, not where $v=0$ without a change of sign.
• It SPEEDS UP when $v$ and $a$ have the SAME sign, and slows down when their signs are opposite.
• Displacement on $\left[t_{1}, t_{2}\right]$ : $s(t_{2})-s(t_{1})$ , a signed difference.
• Distance travelled: the sum of the $\left|s(t_{i+1})-s(t_{i})\right|$ between the instants when the direction changes.

s(t)=t^{3}-6t^{2}+9t

(blue) rises, falls, rises again: the object changes direction at

t=1

and

t=3

, where

v=s'

(dashed) changes sign.

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 differentiability

the whole question, and the conclusion is false anyway

What not to write

“ $f(x)=|x|$ on $\left[-1, 1\right]$ is continuous and $f(-1)=f(1)$ , so there is a $c$ such that $f'(c)=0$ .”

What to write

“ $f$ is not differentiable at $0$ , which belongs to $\left(-1, 1\right)$ : the differentiability hypothesis fails and the theorem does not apply.”

|x|

is continuous on

[-1, 1]

and

f(-1)=f(1)

, but no tangent line there is horizontal: the corner at

0

breaks the differentiability hypothesis, and Rolle's Theorem says nothing any more.

Why: $|x|$ has a derivative equal to $-1$ on the left and $+1$ on the right, never $0$ . It is the counterexample that proves the differentiability hypothesis is not decorative, and it is the one the marker expects when the question asks you to discuss the hypotheses.

2. Taking a critical number for an extremum

2 marks, and a wrong sign chart

What not to write

“ $f(x)=x^{3}$ satisfies $f'(0)=0$ , so $f$ has an extremum at $0$ .”

What to write

“ $f'(x)=3x^{2}\ge 0$ does NOT change sign at $0$ : there is no extremum, but an inflection point with a horizontal tangent.”

Two horizontal tangents, two opposite situations: at

0

x^{3}

keeps rising and there is no extremum; at

3

, the parabola turns around and the minimum is real.

Why: A critical number is a CANDIDATE, never a conclusion. It is the CHANGE OF SIGN of $f'$ that decides, and the Second Derivative Test is silent when $f''(c)=0$ , as here. The only test that always settles it is the First Derivative Test.

3. Confusing displacement and distance travelled

the whole question

What not to write

“ $s(t)=t^{2}-4t$ on $\left[0, 5\right]$ : the distance travelled is $s(5)-s(0)=5$ .”

What to write

“The object changes direction at $t=2$ , where $v=2t-4=0$ . Distance $=\left|s(2)-s(0)\right|+\left|s(5)-s(2)\right|=4+9=13$ .”

s(t)=t^{2}-4t

: the object moves back

4

until

t=2

, then forward

9

. Distance travelled

4+9=13

; the displacement

s(5)-s(0)=5

wipes out the round trip.

Why: The displacement is indeed $5$ , but the object first went $4$ units in the other direction. The words “distance travelled” force you to split at the instants where $v$ changes sign, exactly as an area forces you to split at the zeros of the function.

4. Confusing speeding up with having a positive acceleration

1 mark, and the final interpretation of the problem

What not to write

“ $a(t)>0$ , so the object is speeding up.”

What to write

“The object speeds up when $v$ and $a$ have the SAME sign. With $v<0$ and $a>0$ , it is slowing down.”

v=t-4

is negative and rising (

a=1>0

), but its absolute value (dashed) goes down: the object is SLOWING DOWN. Speeding up means

v

and

a

of the same sign.

Why: The acceleration increases the SIGNED velocity, not necessarily its absolute value. An object moving backward with a positive acceleration brakes before setting off forward again. It is the most frequent interpretation question at the end of a motion problem.

5. Applying L'Hôpital's rule to a form that is not indeterminate

the whole question, since the answer is wrong

What not to write

$\lim_{x\to 0}\frac{\cos x}{x+1}=\lim_{x\to 0}\frac{-\sin x}{1}=0$

What to write

The form is $\frac{1}{1}=1$ : it is NOT indeterminate, so substitute, and the limit is $1$ .

\frac{\cos x}{x+1}

(blue) arrives at

1

x\to 0

: the form

\frac{1}{1}

is not indeterminate at all. Differentiating top and bottom produces

-\sin x

(red), which equals

0

Why: L'Hôpital's rule only holds for $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . Applied anywhere else, it gives a number that has nothing to do with the limit. Writing the indeterminate form BEFORE differentiating, in full, is what protects you from this error, and it earns half a mark.

6. Believing that Rolle's Theorem or the Mean Value Theorem gives the value of $c$

2 marks

What not to write

“The Mean Value Theorem gives $c=\frac{a+b}{2}$ .”

What to write

“The theorem guarantees the EXISTENCE of such a $c$ ; finding it requires solving $f'(c)=\frac{f(b)-f(a)}{b-a}$ , which is a separate question.”

At the midpoint

x=1

, the tangent line is horizontal (red), not parallel to the chord. The

c

of the theorem is

\frac{2}{\sqrt{3}}\approx 1.15

(green): you find it by solving

f'(c)=1

Why: The midpoint only works for a parabola, by chance. For $f(x)=x^{3}-3x$ on $\left[0, 2\right]$ , the average slope is $1$ and $c=\frac{2}{\sqrt{3}}\approx 1.155$ , which is not $1$ . An existence theorem never provides a formula.

7. Starting Newton's method from any point at all

the whole question, or an endless loop on the calculator

What not to write

“ $x_{0}=0$ for $f(x)=x^{3}-2x+2$ , the method will converge.”

What to write

“From $x_{0}=0$ , the sequence oscillates between $0$ and $1$ without ever converging: you must choose $x_{0}$ in an interval where $f'$ does not vanish and where the sign of $f$ changes.”

x^{3}-2x+2

, the tangent at

0

sends you to

1

(solid line), the one at

1

sends you back to

0

(dashed): the sequence goes round in circles, far from the root near

-1.77

Why: $x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$ diverges as soon as $f'(x_{n})$ is close to zero, and it can fall into a cycle. The choice of starting point is justified by the Intermediate Value Theorem on a small interval, not by intuition.

Which method to choose

Which theorem, from the verb of the question

The questions of this chapter all look alike. The VERB names the tool, and there is no ambiguity.

If “show that there is a solution” on an interval → Intermediate Value Theorem, with continuity and the change of sign written out

Example: $x^{3}+x-1=0$ has a root in $\left[0, 1\right]$
If the question adds “a UNIQUE solution” → Intermediate Value Theorem for existence, THEN strict monotonicity for uniqueness, in two separate sentences

Example: $f'(x)=3x^{2}+1>0$ gives uniqueness
If “show that there is a $c$ such that $f'(c)=0$ ” with $f(a)=f(b)$ → Rolle's Theorem, hypotheses checked one by one

Example: show that a polynomial equation has at most three roots
If “show that at some instant the instantaneous speed equals the average speed” → Mean Value Theorem

Example: a $200$ km trip in $2$ h forces an instant at exactly $100$ km/h
If “approximate” or “give an approximate value of a root” → Newton's method, with the starting point justified

Example: $x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$ from $x_{0}=1$
If a limit of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ → L'Hôpital's rule, after WRITING the indeterminate form

Example: $\lim_{x\to 0}\frac{\sin x}{x}$ gives $1$ again

“Show that there is” always calls for an existence theorem; “compute” calls for a computation. Confusing the two makes you write a computation where checked hypotheses were expected.

Distance or displacement, in a motion problem

Two very similar questions, two different answers as soon as the object changes direction.

If the question says “displacement”, “change in position” or “where is it” → compute $s(t_{2})-s(t_{1})$ , a plain signed difference

Example: $s(5)-s(0)=5-0=5$
If the question says “distance travelled”, “total path” or “how many kilometres” → find the instants where $v$ CHANGES sign, split there, and add the absolute values

Example: $\left|s(2)-s(0)\right|+\left|s(5)-s(2)\right|=4+9=13$
If the question asks whether the object is speeding up → compare the SIGNS of $v$ and $a$ : same sign, it speeds up

Example: $v<0$ and $a>0$ : it slows down

Split at the changes of SIGN of $v$ , not at its zeros: $v(t)=t^{2}$ vanishes at $0$ without the object turning back.

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.

Checking the hypotheses of a theorem, written for the marker

When to use it: Every application of Rolle's Theorem, the Mean Value Theorem or the Intermediate Value Theorem.

1 Name the function and the interval: “Let $f(x)=x^{3}-3x$ on $\left[0, 2\right]$ .”
2 Justify continuity by its REASON: “ $f$ is a polynomial, hence continuous on $\mathbb{R}$ , in particular on $\left[0, 2\right]$ .”
3 Justify differentiability on the OPEN interval, naming it: “ $f$ is differentiable on $\left(0, 2\right)$ .”
4 Check the specific hypothesis: equal values at the endpoints for Rolle's Theorem, nothing more for the Mean Value Theorem.
5 Conclude by citing the theorem by name and writing what it gives, nothing more.

Concluding sentence

Since $f$ is continuous on $\left[0, 2\right]$ and differentiable on $\left(0, 2\right)$ , the Mean Value Theorem guarantees that there is a real number $c\in\left(0, 2\right)$ such that $f'(c)=\frac{f(2)-f(0)}{2-0}=1$ .

The trap: Writing “differentiable on $\left[0, 2\right]$ ”. The closed interval for continuity, the OPEN interval for differentiability: this is what lets the theorem apply to $\sqrt{x}$ on $\left[0, 1\right]$ , whose derivative is infinite at $0$ .

Marking: Usually 1 mark for continuity with its justification, 1 mark for differentiability on the open interval, 1 mark for a conclusion that names the theorem.

Writing up a rectilinear motion problem

When to use it: The question gives a position $s(t)$ and asks several questions about the motion.

1 Differentiate once for $v$ , twice for $a$ , and write the three expressions together.
2 Solve $v(t)=0$ and make the SIGN CHART of $v$ on the interval asked.
3 Read the chart: direction of motion, turning instants, corresponding positions.
4 Answer each question by citing the line of the chart that justifies it.

Concluding sentence

The object moves backward on $\left[0, 2\right]$ , turns back at $t=2$ at the position $s(2)=-4$ , then moves forward on $\left[2, 5\right]$ : it therefore travels $4+9=13$ units for a net displacement of $5$ .

The trap: Answering the questions in order without building the chart. The sign chart of $v$ alone answers the four usual questions, and building it once costs less than four separate arguments.

Marking: The sign chart is often worth as much as all the answers it gives.

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 is in the OPEN interval

Rolle's Theorem and the Mean Value Theorem guarantee $c\in\left(a, b\right)$ , endpoints excluded. A $c$ equal to $a$ or to $b$ signals a solving error.

On $\left[0, 2\right]$ , $c=\frac{2}{\sqrt{3}}\approx 1.155$ is indeed strictly between $0$ and $2$ ✓. A solution $c=2$ would have had to be rejected.
Check an extremum with the sign chart, not with $f'(c)=0$

Evaluate $f'$ just before and just after the critical number. A change of sign gives an extremum, no change gives none.

For $x^{3}$ at $0$ : $f'(-0.1)=0.03>0$ and $f'(0.1)=0.03>0$ . No change, so no extremum.
Compare distance and displacement

The distance travelled is always GREATER THAN OR EQUAL TO the absolute value of the displacement. A smaller distance is impossible.

Distance $13$ , displacement $5$ : consistent ✓. A distance of $3$ for a displacement of $5$ would be an immediate contradiction.
Check an approximate root by substituting

After Newton's method, substitute the value found into $f$ : the result must be close to zero, and the number of correct decimals should roughly double at each iteration.

$f(0.6823)=0.6823^{3}+0.6823-1\approx-0.00007$ ✓, which confirms four correct decimals.
Write the indeterminate form before applying L'Hôpital's rule

Substitute first. If the result is neither $\frac{0}{0}$ nor $\frac{\infty}{\infty}$ , the rule is forbidden and the limit is read directly.

$\frac{\cos x}{x+1}$ at $0$ gives $\frac{1}{1}$ : a determinate form, limit $1$ , no differentiation.

The typical problem, taken apart

Existence, uniqueness, then an approximate value

Let $f(x)=x^{3}+x-1$ .

Show that the equation $f(x)=0$ has a unique solution in $\left[0, 1\right]$ , then give an approximate value of it to within $10^{-3}$ by Newton's method.

The curve goes from negative to positive between

0

and

1

and never comes back down: it is the picture of the two arguments, a change of sign for existence, monotonicity for uniqueness.

Step 1

$f$ is a polynomial, hence continuous on $\mathbb{R}$ , in particular on $\left[0, 1\right]$ . Moreover $f(0)=-1$ and $f(1)=1$ .

Why

Continuity is justified by its REASON, never by “it is obvious”. It is the most important hypothesis of the Intermediate Value Theorem, and the one that earns the first mark.

Step 2

$0$ lies between $f(0)=-1$ and $f(1)=1$ , so the Intermediate Value Theorem guarantees that there is a real number $\alpha\in\left(0, 1\right)$ such that $f(\alpha)=0$ .

Why

The theorem gives EXISTENCE and nothing else. Stopping here would be incomplete if the question says “unique”, and the word “unique” is almost always there.

Step 3

$f'(x)=3x^{2}+1$ , strictly positive for every real number. So $f$ is strictly increasing on $\left[0, 1\right]$ , hence one-to-one, hence the solution is UNIQUE.

Why

Uniqueness is a SEPARATE sentence, resting on monotonicity and not on the previous theorem. It is the consequence of the Mean Value Theorem recalled in the essentials, and the marker counts it separately.

Step 4

Newton's method: $x_{n+1}=x_{n}-\frac{x_{n}^{3}+x_{n}-1}{3x_{n}^{2}+1}$ . Start from $x_{0}=1$ , where $f'(1)=4$ is far from zero.

Why

The starting point is justified: $f'$ does not vanish on $\left[0, 1\right]$ , so there is no risk of dividing by a number close to zero. A solution that starts at random loses the justification mark even if the computation converges.

Step 5

$x_{1}=1-\frac{1}{4}=0.75$ ; $x_{2}=0.75-\frac{0.171875}{2.6875}\approx 0.68605$ ; $x_{3}\approx 0.68234$ ; $x_{4}\approx 0.68233$ .

Why

Stop when two iterations agree on the decimals required. The number of correct decimals roughly doubles at every step, which explains why so few are needed.

Step 6

Check: $f(0.682)\approx-0.0008$ and $f(0.683)\approx 0.0016$ , so $\alpha\in\left(0.682, 0.683\right)$ ✓, which confirms the rounding to $10^{-3}$ .

Why

Bracketing by a change of sign is the only check that proves the precision announced. Substituting a single value shows that the result is small, not that it is precise enough.

The conclusion, written out

The equation $x^{3}+x-1=0$ has a unique solution $\alpha$ in $\left(0, 1\right)$ , and $\alpha\approx 0.682$ to within $10^{-3}$ .

The classic mistake on this problem: Concluding uniqueness from the Intermediate Value Theorem. That theorem gives ONLY existence; uniqueness comes from strict monotonicity, and it must be stated in a separate sentence with its own justification.

Learn by heart

• Rolle and the Mean Value Theorem: CONTINUOUS on the closed interval, DIFFERENTIABLE on the open one. Two different intervals.
• Both theorems give the EXISTENCE of $c$ , never its value.
• $f'>0$ on an INTERVAL gives strict increase, hence the uniqueness of a root.
• A critical number is a candidate: it is the change of SIGN of $f'$ that decides.
• Motion: $v$ and $a$ of the same sign, the object speeds up; of opposite signs, it slows down.
• Distance travelled: split at the changes of sign of $v$ and add the absolute values.
• L'Hôpital's rule only applies to $\frac{0}{0}$ and $\frac{\infty}{\infty}$ : write the form before differentiating.
• Newton: $x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$ , with an $x_{0}$ where $f'$ does not vanish.

Frequently asked questions

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

They have the same hypotheses, continuity on the closed interval and differentiability on the open one. Rolle's Theorem adds that the function takes the same value at both endpoints, and concludes that the derivative is zero somewhere. The Mean Value Theorem drops this equality and concludes that the derivative equals the slope of the chord somewhere. Rolle's Theorem is therefore the special case with a horizontal chord.

Why is differentiability required only on the open interval?

Because the theorem must still apply to functions whose derivative is infinite at the endpoints. The square root on the interval from zero to one is continuous everywhere there and differentiable only on the open interval, and the theorem applies to it perfectly. Writing differentiable on the closed interval needlessly excludes this case and costs a mark.

How do you show that an equation has a unique solution?

In two separate sentences. The first uses the Intermediate Value Theorem, with continuity justified and the change of sign between the endpoints, to get existence. The second uses strict monotonicity, proved by the sign of the derivative, to get uniqueness. The Intermediate Value Theorem never gives uniqueness on its own.

What is the difference between distance travelled and displacement?

The displacement is the difference between the final position and the initial position, and it can be negative. The distance travelled counts the whole path, round trips included. As soon as the velocity changes sign, you must split at that instant and add up the absolute values of the changes in position.

When is a moving object really speeding up?

When the velocity and the acceleration have the same sign. A positive acceleration on an object moving backward makes it brake, not speed up, because it brings its signed velocity back towards zero. It is the absolute value of the velocity, the speed, that must increase for the object to be speeding up in the everyday sense.

When can you apply L'Hôpital's rule?

Only to an indeterminate form zero over zero or infinity over infinity. On any other form, the rule gives a number unrelated to the limit sought. The protective reflex is to substitute first and to write the form obtained in full before differentiating anything at all.

Practise it

Corrected exercises: Applications of the derivative, 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.

15 corrected exercises
150 points
225 minutes

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Revision sheet: applications of the derivative (201-SN2-RE)

The essentials

Rolle and the Mean Value Theorem: same hypotheses, one conclusion

The consequences, the ones you really use

Rectilinear motion: six words not to confuse

The mistakes that cost marks

1. Applying Rolle's Theorem without checking differentiability

2. Taking a critical number for an extremum

3. Confusing displacement and distance travelled

4. Confusing speeding up with having a positive acceleration

5. Applying L'Hôpital's rule to a form that is not indeterminate

6. Believing that Rolle's Theorem or the Mean Value Theorem gives the value of ccc

7. Starting Newton's method from any point at all

Which method to choose

Which theorem, from the verb of the question

Distance or displacement, in a motion problem

How the answer is expected to be written

Checking the hypotheses of a theorem, written for the marker

Writing up a rectilinear motion problem

Check before you hand in

The typical problem, taken apart

Existence, uniqueness, then an approximate value

Learn by heart

Frequently asked questions

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

Why is differentiability required only on the open interval?

How do you show that an equation has a unique solution?

What is the difference between distance travelled and displacement?

When is a moving object really speeding up?

When can you apply L'Hôpital's rule?

Corrected exercises: Applications of the derivative, 201-SN2-RE

See also

Looking for a 201-SN2-RE tutor in Montreal?

6. Believing that Rolle's Theorem or the Mean Value Theorem gives the value of $c$