Print the chapter

Every lesson of the chapter, ready for paper. Exercise sets are lettered A, B, C… and answered at the end.

Include answers

To save a PDF, choose “Save as PDF” in the print dialog.

1. L'Hôpital's rule

Warm-up

Three quick questions from earlier lessons.

Discover

Near $x = 0$ , $\sin(3x) \approx 3x$ and $\tan x \approx x$ : each is close to its tangent line. What does $\dfrac{\sin(3x)}{\tan x}$ approach as $x \to 0$ ?

Key idea

For $0/0$ (or $\infty/\infty$ ), compare the rates instead of the values.

L'Hôpital's rule: if

f(a) = g(a) = 0

(or both tend to

\pm\infty

), then

\displaystyle\lim_{x \to a} \dfrac{f(x)}{g(x)} = \lim_{x \to a} \dfrac{f'(x)}{g'(x)}

, when the second limit exists. Differentiate top and bottom separately, not with the quotient rule.

Worked example

Find $\displaystyle\lim_{x \to 0} \dfrac{e^{2x} - 1}{\sin x}$ .

Form

\dfrac{0}{0}

Check the form: e⁰ − 1 = 0 and sin 0 = 0.

\displaystyle\lim_{x \to 0} \dfrac{2e^{2x}}{\cos x}

Differentiate top and bottom.

= \dfrac{2}{1} = 2

Now substitute.

Your turn, guided

Your turn: $\displaystyle\lim_{x \to 1} \dfrac{\ln x}{x - 1}$ .

\displaystyle\lim_{x \to 1} \dfrac{1/x}{1}

0/0: differentiate.

= 1

Substitute x = 1.

Practice ★

Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 5x} - 1}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 4x} - 1}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 2x} - 1}{x}$ .

Watch out!

Practice ★★

Find $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos(6x)}{x^2}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{1 - \cos(5x)}{x^2}$ .

Find $\displaystyle\lim_{x \to 1} \dfrac{x^{ 7 } - 1}{\ln x}$ .
Find $\displaystyle\lim_{x \to 1} \dfrac{x^{ 8 } - 1}{\ln x}$ .

Exit ticket

Three questions to check you have got it.

Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 5x} - 1}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 4x} - 1}{x}$ .
Find $\displaystyle\lim_{x \to 0} \dfrac{e^{ 2x} - 1}{x}$ .

Summary card

L'Hôpital's rule

Check the form is $0/0$ or $\infty/\infty$ .
Differentiate top and bottom separately.
Take the new limit (repeat if still $0/0$ ).

Watch out: no indeterminate form, no L'Hôpital.

2. Critical numbers and the first derivative test

Warm-up

Three quick questions from earlier lessons.

Discover

On a graph, a hilltop has a horizontal tangent. Just before it the graph rises; just after it falls. What does $f'$ do at a hilltop?

It goes from $+$ to $-$
It goes from $-$ to $+$

Key idea

Extremes happen where $f'$ changes sign.

A critical number of

f

is a

c

where

f'(c) = 0

f'(c)

does not exist. First derivative test: if

f'

changes from

+

-

c

f

has a local maximum; from

-

+

, a local minimum; no sign change, no extremum.

Worked example

$f(x) = x^3 - 12x$ . Find the local extremes.

f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)

Factor f'.

Signs:

+\ |\ -\ |\ +

Sign chart: + before −2, − between, + after 2.

Local max at

x = -2

, local min at

x = 2

+ to − at −2; − to + at 2.

Your turn, guided

Your turn: $f(x) = x^2 - 6x$ .

critical number

x = 3

f'(x) = 2x − 6 = 0.

local minimum at

x = 3

− before 3, + after: a minimum.

Practice ★

$f(x) = x^3 - 12x$ . At which $x$ does $f$ have a local maximum?
$f(x) = x^3 - 27x$ . At which $x$ does $f$ have a local maximum?

Here is the graph of the derivative $y = f'(x)$ . Tap the point of this graph at the $x$ -value where $f$ has a local maximum.
Here is the graph of the derivative $y = f'(x)$ . Tap the point of this graph at the $x$ -value where $f$ has a local maximum.

Watch out!

Practice ★★

$f'(x) = (x - -3)(x - 5)^2$ . Tick all the true statements.
1. $f$ has a local minimum at $x = -3$
2. $f$ is increasing for $x > -3$
3. $f$ has a local maximum at $x = -3$
4. $f$ has a local extremum at $x = 5$
$f'(x) = (x - 0)(x - 4)^2$ . Tick all the true statements.
1. $f$ has a local minimum at $x = 0$
2. $f$ is increasing for $x > 0$
3. $f$ has a local maximum at $x = 0$
4. $f$ has a local extremum at $x = 4$

Exit ticket

Three questions to check you have got it.

$f(x) = x^3 - 12x$ . At which $x$ does $f$ have a local maximum?
$f(x) = x^3 - 27x$ . At which $x$ does $f$ have a local maximum?
$f(x) = x^3 - 3x$ . At which $x$ does $f$ have a local maximum?

Summary card

First derivative test

Find the critical numbers.
Sign chart of $f'$ .
$+ \to -$ : max; $- \to +$ : min; no change: neither.

Watch out: $f'(c) = 0$ alone does not make an extremum.

3. The second derivative test and inflection points

Warm-up

Three quick questions from earlier lessons.

Discover

At a critical point the graph is flat. If the graph is concave up there, like a cup, is the point a maximum or a minimum?

A minimum
A maximum

Key idea

The second derivative can classify a critical point, and locates changes in concavity.

Second derivative test: if

f'(c) = 0

and

f''(c) > 0

, local minimum;

f''(c) < 0

, local maximum;

f''(c) = 0

, no conclusion. An inflection point is where

f''

changes sign.

Worked example

$f(x) = x^3 - 6x^2$ . Classify the critical points and find the inflection point.

Critical numbers

0

and

4

f' = 3x² − 12x = 3x(x − 4).

Local max at

0

, local min at

4

f'' = 6x − 12: f''(0) = −12, f''(4) = 12.

Inflection point at

x = 2

f'' changes sign at x = 2.

Your turn, guided

Your turn: $f(x) = x^3 - 9x^2$ . Where is the inflection point?

x = 3

f''(x) = 6x − 18.

Practice ★

$f(x) = x^3 - 3x^2$ . At which $x$ does $f$ have an inflection point?
$f(x) = x^3 - 12x^2$ . At which $x$ does $f$ have an inflection point?

$f'(8) = 0$ and $f''(8) = 3$ . What happens at $x = 8$ ?
1. A local minimum
2. A local maximum
3. An inflection point

Watch out!

Practice ★★

$f'(1) = 0$ and $f''(1) = 0$ . What can you conclude about $x = 1$ ?
1. Nothing yet: it could be a maximum, a minimum or neither
2. It is an inflection point
3. It is a local maximum

Exit ticket

Three questions to check you have got it.

$f(x) = x^3 - 3x^2$ . At which $x$ does $f$ have an inflection point?
$f(x) = x^3 - 12x^2$ . At which $x$ does $f$ have an inflection point?
$f(x) = x^3 - 15x^2$ . At which $x$ does $f$ have an inflection point?

Summary card

Second derivative test

$f' = 0$ , $f'' > 0$ : min. $f' = 0$ , $f'' < 0$ : max. $f'' = 0$ : test fails.
Inflection: $f''$ changes sign.

Watch out: $f''(c) = 0$ is not enough for an inflection point.

4. Global extrema on a closed interval

Warm-up

Three quick questions from earlier lessons.

Discover

$f(x) = x^2$ on $[-1, 3]$ . Its only critical point is $x = 0$ . Is $f(0) = 0$ the largest value on the interval?

No, $f(3) = 9$ is larger
Yes

Key idea

On a closed interval, compare a short list of candidates.

Extreme Value Theorem: a function continuous on

[a, b]

has a maximum and a minimum value there. Closed interval method: evaluate

f

at the critical numbers in

(a, b)

and at

a

and

b

; the largest value is the maximum, the smallest the minimum.

Worked example

Extremes of $f(x) = x^3 - 3x$ on $[0, 3]$ .

Critical number

1

f'(x) = 3x² − 3 = 0 at x = 1 (−1 is outside).

f(0) = 0

f(1) = -2

f(3) = 18

Evaluate the candidates.

Minimum

-2

(at

x = 1

), maximum

18

(at

x = 3

)

Compare.

Your turn, guided

Your turn: $f(x) = x^2 - 4x$ on $[0, 5]$ . Candidates: $f(0) = 0$ , $f(2) = -4$ , $f(5) = 5$ .

Maximum value

5

The largest candidate.

Practice ★

Find the maximum value of $f(x) = x^2 - 6x$ on $[0, 8]$ .
Find the maximum value of $f(x) = x^2 - 2x$ on $[0, 7]$ .

Find the minimum value of $f(x) = x^3 - 3x$ on $[0, 2]$ .
Find the minimum value of $f(x) = x^3 - 3x$ on $[0, 5]$ .

Watch out!

Practice ★★

Which functions are guaranteed by the Extreme Value Theorem to have a maximum value on the interval? Tick all that are.
1. $x^2$ on $[-7, 7]$
2. $\sin x$ on $[0, 7]$
3. $x$ on $(0, 7)$
4. $\dfrac{1}{x}$ on $[-1, 1]$

Exit ticket

Three questions to check you have got it.

Find the minimum value of $f(x) = x^3 - 3x$ on $[0, 2]$ .
Find the minimum value of $f(x) = x^3 - 3x$ on $[0, 5]$ .
Find the minimum value of $f(x) = x^3 - 27x$ on $[0, 6]$ .

Summary card

Closed interval method

Critical numbers inside $(a, b)$ .
Evaluate $f$ there and at $a$ and $b$ .
Largest = maximum, smallest = minimum.

Watch out: the EVT needs continuity on a closed interval. Give the value, not just where it occurs, when asked.

5. The Mean Value Theorem

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = x^3$ on $[0, 2]$ with its secant line, of slope 4. The blue line is the tangent at $x = c$ . Move $c$ until the tangent is parallel to the secant.

The curve y = x³, its secant line y = 4x on [0, 2], and its tangent at x = c. Near c ≈ 1.15 the tangent is parallel to the secant.

Is there a point in $(0, 2)$ where the tangent's slope equals 4?

Yes, near $c \approx 1.15$
Yes, at the midpoint $c = 1$

Key idea

Somewhere, the instantaneous rate equals the average rate.

Mean Value Theorem: if

f

is continuous on

[a, b]

and differentiable on

(a, b)

, there is a

c

(a, b)

with

f'(c) = \dfrac{f(b) - f(a)}{b - a}

Worked example

$f(x) = x^2$ on $[1, 5]$ . Find $c$ .

\dfrac{f(5) - f(1)}{5 - 1} = 6

Average rate: (25 − 1)/4.

2c = 6

f'(c) = 2c.

c = 3

Check: 3 is in (1, 5).

Your turn, guided

Your turn: $f(x) = x^2$ on $[2, 8]$ .

c = 5

(64 − 4)/6 = 10, so 2c = 10.

Practice ★

$f(x) = x^2$ on $[2, 5]$ . Find the $c$ given by the Mean Value Theorem.
$f(x) = x^2$ on $[3, 9]$ . Find the $c$ given by the Mean Value Theorem.
$f(x) = x^2$ on $[1, 7]$ . Find the $c$ given by the Mean Value Theorem.

Watch out!

Practice ★★

$f(x) = x^3$ on $[0, 4]$ . Find the $c$ in $(0, 4)$ given by the Mean Value Theorem.
$f(x) = x^3$ on $[0, 6]$ . Find the $c$ in $(0, 6)$ given by the Mean Value Theorem.

A car drove 276 miles in 4 hours. Must its speedometer have read exactly 69 mph at some moment?
1. Yes, by the Mean Value Theorem
2. Only if it drove at a constant speed
3. No: it may have gone faster and slower without hitting 69 exactly

Exit ticket

Three questions to check you have got it.

$f(x) = x^2$ on $[2, 5]$ . Find the $c$ given by the Mean Value Theorem.
$f(x) = x^2$ on $[3, 9]$ . Find the $c$ given by the Mean Value Theorem.
$f(x) = x^2$ on $[1, 7]$ . Find the $c$ given by the Mean Value Theorem.

Summary card

Mean Value Theorem

Check: continuous on $[a, b]$ , differentiable on $(a, b)$ .
Average rate $\dfrac{f(b) - f(a)}{b - a}$ .
Solve $f'(c) =$ that rate, with $c$ in $(a, b)$ .

Watch out: $c$ is not always the midpoint.

6. Motion on a line: velocity, acceleration, distance

Warm-up

Three quick questions from earlier lessons.

Discover

A ball moving left has velocity $-4$ m/s; its acceleration is $-2$ m/s². Is it getting faster or slower?

Faster
Slower, since the acceleration is negative

Key idea

For a particle on a line with position $s(t)$ :

v(t) = s'(t)

a(t) = v'(t) = s''(t)

. At rest:

v = 0

. It changes direction where

v

changes sign. Speeding up when

v

and

a

have the same sign, slowing down when opposite. Total distance adds the distances of each leg between turns.

Worked example

$s(t) = t^2 - 4t$ on $[0, 5]$ . Find the total distance.

It turns at

t = 2

v = 2t − 4 = 0.

Legs:

|-4 - 0| = 4

and

|5 - (-4)| = 9

s(0) = 0, s(2) = −4, s(5) = 5.

Total distance

13

Add the legs. (The displacement is only 5.)

Your turn, guided

Your turn: $s(t) = t^2 - 6t$ . When is it at rest?

t = 3

v(t) = 2t − 6 = 0.

Practice ★

A particle's position is $s(t) = t^2 - 2t$ . At what time $t$ is it at rest?
A particle's position is $s(t) = t^2 - 6t$ . At what time $t$ is it at rest?

At $t = 3$ , a particle has velocity $-8$ m/s and acceleration $-2$ m/s². Is it speeding up or slowing down?
1. Speeding up: velocity and acceleration have the same sign
2. Slowing down, because the acceleration is negative
3. Neither: its speed stays the same

Watch out!

Practice ★★

A particle's position is $s(t) = t^2 - 4t$ meters. Find the total distance it travels from $t = 0$ to $t = 9$ .
A particle's position is $s(t) = t^2 - 4t$ meters. Find the total distance it travels from $t = 0$ to $t = 6$ .

Exit ticket

Three questions to check you have got it.

A particle's position is $s(t) = t^2 - 2t$ . At what time $t$ is it at rest?
A particle's position is $s(t) = t^2 - 6t$ . At what time $t$ is it at rest?
A particle's position is $s(t) = t^2 - 10t$ . At what time $t$ is it at rest?

Summary card

Motion on a line

$v = s'$ , $a = v'$ ; at rest when $v = 0$ .
Same signs of $v$ and $a$ : speeding up.
Distance: add each leg between turns.

Watch out: displacement and distance differ when the particle turns back.

7. Optimization

Warm-up

Three quick questions from earlier lessons.

Discover

A farmer has $p$ m of fence for three sides of a rectangle against a wall. The graph shows the area $A(x) = x(p - 2x)$ for a width $x$ . Move $p$ .

The area A(x) = x(p − 2x) of a rectangle fenced on three sides with p meters. Each parabola peaks at x = p/4.

For $p = 40$ , where is the area largest?

At $x = 10$
At $x \approx 13.3$ , a square

Key idea

Turn the story into one function of one variable, then use the closed interval method.

Optimization: (1) name the quantity to optimize and the variables; (2) use the constraint to write it as a function of one variable; (3) find its domain; (4) find the critical numbers and compare with the endpoints; (5) answer the question asked, with units.

Worked example

Two positive numbers add to 20. Maximize their product.

P(x) = x(20 - x)

0 < x < 20

Constraint y = 20 − x.

x = 10

, a maximum (

P'' = -2

)

P'(x) = 20 − 2x.

Largest product

P(10) = 100

The question asks for the product.

Your turn, guided

Your turn: same problem with sum 30.

Largest product

225

x = 15 = y.

Practice ★

Two positive numbers add up to 35. What is the largest possible value of their product?
Two positive numbers add up to 20. What is the largest possible value of their product?

A farmer has 104 m of fence for three sides of a rectangle; a long wall forms the fourth side. Find the largest possible area, in m².
A farmer has 40 m of fence for three sides of a rectangle; a long wall forms the fourth side. Find the largest possible area, in m².

Watch out!

Practice ★★

Squares of side $x$ are cut from the corners of a 9-cm square sheet, and the sides are folded up into an open box. Which $x$ gives the largest volume?
Squares of side $x$ are cut from the corners of a 15-cm square sheet, and the sides are folded up into an open box. Which $x$ gives the largest volume?

Boss problem. Squares of side $x$ are cut from the corners of a 18-cm square sheet to fold an open box.

Exit ticket

Three questions to check you have got it.

A farmer has 104 m of fence for three sides of a rectangle; a long wall forms the fourth side. Find the largest possible area, in m².
A farmer has 40 m of fence for three sides of a rectangle; a long wall forms the fourth side. Find the largest possible area, in m².
A farmer has 60 m of fence for three sides of a rectangle; a long wall forms the fourth side. Find the largest possible area, in m².

Summary card

Optimization

Quantity and constraint.
One-variable function and its domain.
Critical numbers vs endpoints.
Answer the question: the value or the place, with units.

Watch out: not every critical number is the optimum.

Answers: Applications of derivatives

1. L'Hôpital's rule

$18$
$\frac{25}{2}$

2. Critical numbers and the first derivative test

$-2$
$-3$

$(-3, 0)$
$(-1, 0)$

$f$ has a local minimum at $x = -3$ · $f$ is increasing for $x > -3$
$f$ has a local minimum at $x = 0$ · $f$ is increasing for $x > 0$

$-2$
$-3$
$-1$

3. The second derivative test and inflection points

A local minimum

Nothing yet: it could be a maximum, a minimum or neither

4. Global extrema on a closed interval

$16$
$35$

$-2$
$-2$

$x^2$ on $[-7, 7]$ · $\sin x$ on $[0, 7]$

$-2$
$-2$
$-54$

5. The Mean Value Theorem

$\frac{7}{2}$
$6$
$4$

$\frac{4 \sqrt{3}}{3}$
$2 \sqrt{3}$

Yes, by the Mean Value Theorem

$\frac{7}{2}$
$6$
$4$

6. Motion on a line: velocity, acceleration, distance

Speeding up: velocity and acceleration have the same sign

$53$
$20$

7. Optimization

$\frac{1225}{4}$
$100$

$1352$
$200$

$\frac{3}{2}$
$\frac{5}{2}$

$432$

$1352$
$200$
$450$