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. Average and instantaneous velocity

Warm-up

Three quick questions from earlier lessons.

Discover

A ball is thrown up. Its height after $t$ seconds is $s(t)$ meters. Here is the line through the points of the graph at $t = 1$ and $t = 1 + h$ (the curve is in violet). Make $h$ smaller.

The height s(t) = 20t − 5t² and the line through its points at t = 1 and t = 1 + h. As h shrinks, the line settles on the tangent at t = 1, of slope 10.

The slope of that line is the average velocity on $[1, 1 + h]$ . As $h$ shrinks, what does it approach?

About 10 m/s
15

Key idea

Velocity at one instant is the limit of average velocities over shorter and shorter intervals.

Average velocity on

[a, a + h]

\dfrac{s(a + h) - s(a)}{h}

. Instantaneous velocity at

t = a

\displaystyle\lim_{h \to 0} \dfrac{s(a + h) - s(a)}{h}

Worked example

$s(t) = 20t - 5t^2$ . Find the velocity at $t = 1$ .

s(1 + h) = 15 + 10h - 5h^2

Substitute 1 + h and expand: 20 + 20h − 5(1 + 2h + h²).

\dfrac{s(1 + h) - s(1)}{h} = 10 - 5h

Subtract s(1) = 15 and divide by h ≠ 0.

velocity at

t = 1

10

m/s

Let h → 0.

Your turn, guided

Your turn: same ball, $t = 3$ . The average velocity on $[3, 3 + h]$ simplifies to $-10 - 5h$ .

velocity at

t = 3

-10

m/s

Let h → 0: the ball is on its way down.

Practice ★

A ball's position is $s(t) = 4t^2 + 9t$ meters after $t$ seconds. Find its average velocity on $[0, 6]$ , in m/s.
A ball's position is $s(t) = 1t^2 + 4t$ meters after $t$ seconds. Find its average velocity on $[0, 5]$ , in m/s.
A ball's position is $s(t) = 3t^2 + 9t$ meters after $t$ seconds. Find its average velocity on $[0, 4]$ , in m/s.

Watch out!

Practice ★★

$s(t) = 5t^2$ . Simplify the average velocity on $[3, 3 + h]$ , $\dfrac{s(3 + h) - s(3)}{h}$ for $h \ne 0$ . Enter an expression in $h$ .
$s(t) = 2t^2$ . Simplify the average velocity on $[2, 2 + h]$ , $\dfrac{s(2 + h) - s(2)}{h}$ for $h \ne 0$ . Enter an expression in $h$ .

$s(t) = 3t^2 + 7t$ meters. The average velocity on $[6, 6 + h]$ simplifies to $43 + 3h$ m/s. Find the instantaneous velocity at $t = 6$ , in m/s.
$s(t) = 1t^2 + 6t$ meters. The average velocity on $[6, 6 + h]$ simplifies to $18 + 1h$ m/s. Find the instantaneous velocity at $t = 6$ , in m/s.

Exit ticket

Three questions to check you have got it.

$s(t) = 3t^2 + 7t$ meters. The average velocity on $[6, 6 + h]$ simplifies to $43 + 3h$ m/s. Find the instantaneous velocity at $t = 6$ , in m/s.
$s(t) = 1t^2 + 6t$ meters. The average velocity on $[6, 6 + h]$ simplifies to $18 + 1h$ m/s. Find the instantaneous velocity at $t = 6$ , in m/s.
$s(t) = 5t^2 + 5t$ meters. The average velocity on $[5, 5 + h]$ simplifies to $55 + 5h$ m/s. Find the instantaneous velocity at $t = 5$ , in m/s.

Summary card

Velocity

Average: $\dfrac{s(a + h) - s(a)}{h}$ , change in position ÷ change in time.
Instantaneous: the limit of that as $h \to 0$ .

Watch out: divide by the time; a change in position alone is not a velocity.

2. The derivative at a point: the limit definition

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = x^2$ . The blue line goes through its points at $x = a$ and $x = a + h$ . Move $a$ , then shrink $h$ .

The parabola y = x² and a secant line through its points at x = a and x = a + h. As h shrinks, the secant turns into the tangent line at x = a, of slope 2a.

With $a = 1$ , the slope of the blue line is $2 + h$ . What does it approach as $h \to 0$ ?

Key idea

The derivative at a point is the limit of the secant slopes.

Step 1h = 2: the secant through P and Q has slope 2.
Step 2h = 1: slope 1.5.
Step 3h = 0.5: slope 1.25.
Step 4h = 0.1: slope 1.05 — the slopes approach 1, the slope of the tangent at P.

f'(a) = \displaystyle\lim_{h \to 0} \dfrac{f(a + h) - f(a)}{h}

, when this limit exists. It is the instantaneous rate of change of

f

a

and the slope of the tangent line at

(a, f(a))

Worked example

$f(x) = x^2 + 3x$ . Find $f'(2)$ .

f(2 + h) - f(2) = 7h + h^2

f(2 + h) = 4 + 4h + h² + 6 + 3h; f(2) = 10.

\dfrac{f(2 + h) - f(2)}{h} = 7 + h

Divide by h ≠ 0.

f'(2) = 7

Let h → 0.

Your turn, guided

Your turn: $f(x) = x^2 + 3x$ at $a = 1$ .

f(1 + h) - f(1) = 5h + h^2

f(1 + h) = 1 + 2h + h² + 3 + 3h; f(1) = 4.

f'(1) = 5

Divide by h, then let h → 0.

Practice ★

$f(x) = x^2 + 4x$ . Use $f'(4) = \displaystyle\lim_{h \to 0} \dfrac{f(4 + h) - f(4)}{h}$ to find $f'(4)$ .
$f(x) = x^2 + 4x$ . Use $f'(2) = \displaystyle\lim_{h \to 0} \dfrac{f(2 + h) - f(2)}{h}$ to find $f'(2)$ .
$f(x) = x^2 + 5x$ . Use $f'(5) = \displaystyle\lim_{h \to 0} \dfrac{f(5 + h) - f(5)}{h}$ to find $f'(5)$ .

Watch out!

Practice ★★

Which expression is $f'(2)$ ?
1. $\displaystyle\lim_{h \to 0} \dfrac{f(2 + h) - f(2)}{h}$
2. $\displaystyle\lim_{h \to 0} \big(f(2 + h) - f(2)\big)$
3. $\dfrac{f(2 + 1) - f(2)}{1}$
4. $f(2)$

$f(x) = \dfrac{ 4 }{x}$ . Use the limit definition to find $f'(3)$ .
$f(x) = \dfrac{ 6 }{x}$ . Use the limit definition to find $f'(4)$ .

Exit ticket

Three questions to check you have got it.

$f(x) = x^2 + 4x$ . Use $f'(4) = \displaystyle\lim_{h \to 0} \dfrac{f(4 + h) - f(4)}{h}$ to find $f'(4)$ .
$f(x) = x^2 + 4x$ . Use $f'(2) = \displaystyle\lim_{h \to 0} \dfrac{f(2 + h) - f(2)}{h}$ to find $f'(2)$ .
$f(x) = x^2 + 5x$ . Use $f'(5) = \displaystyle\lim_{h \to 0} \dfrac{f(5 + h) - f(5)}{h}$ to find $f'(5)$ .

Summary card

The derivative at a point

Write $f(a + h) - f(a)$ and simplify.
Divide by $h$ .
Let $h \to 0$ : the result is $f'(a)$ .

Watch out: $f'(a)$ is a slope, not the value $f(a)$ .

3. The tangent line

Warm-up

Three quick questions from earlier lessons.

Discover

The blue line is the tangent to $y = x^2$ at $x = a$ . Move $a$ .

The parabola y = x² and its tangent line at x = a, which touches the curve at (a, a²) with slope 2a.

At $a = 1$ , the tangent has slope 2 and touches the curve at $(1, 1)$ . Which equation is it?

$y = 1 + 2(x - 1)$
$y = 2x$

Key idea

The tangent line touches the graph at $(a, f(a))$ with slope $f'(a)$ .

Tangent line to

y = f(x)

x = a

y = f(a) + f'(a)(x - a)

Worked example

$f(2) = 5$ and $f'(2) = -3$ . Write the tangent line at $x = 2$ .

y = 5 + (-3)(x - 2)

The point (2, 5) and the slope −3.

y = -3x + 11

Expand if you like.

Your turn, guided

Your turn: $f(4) = 1$ and $f'(4) = 2$ .

y = 1 + 2(x - 4)

Point (4, 1), slope 2.

y = 2x - 7

Expanded: 2x − 8 + 1.

Practice ★

For $f(x) = x^2$ , $f'(x) = 2x$ . Draw the tangent line to the graph at $x = 1$ : place two points on it.
For $f(x) = x^2$ , $f'(x) = 2x$ . Draw the tangent line to the graph at $x = -2$ : place two points on it.

$f(1) = 2$ and $f'(1) = 3$ . Write the tangent line to the graph of $f$ at $x = 1$ : enter the right-hand side of $y = \ldots$
$f(2) = -5$ and $f'(2) = 3$ . Write the tangent line to the graph of $f$ at $x = 2$ : enter the right-hand side of $y = \ldots$

Watch out!

Practice ★★

For $f(x) = x^3$ , the difference quotient at $x = -3$ simplifies to $27 + -9h + h^2$ . What is the slope of the tangent line at $x = -3$ ?
For $f(x) = x^3$ , the difference quotient at $x = -2$ simplifies to $12 + -6h + h^2$ . What is the slope of the tangent line at $x = -2$ ?

Boss problem. $f(x) = x^2 - 4x$ . Find the tangent line to its graph at $x = 5$ , from the definition.

Exit ticket

Three questions to check you have got it.

$f(1) = 2$ and $f'(1) = 3$ . Write the tangent line to the graph of $f$ at $x = 1$ : enter the right-hand side of $y = \ldots$
$f(2) = -5$ and $f'(2) = 3$ . Write the tangent line to the graph of $f$ at $x = 2$ : enter the right-hand side of $y = \ldots$
$f(1) = -4$ and $f'(1) = -3$ . Write the tangent line to the graph of $f$ at $x = 1$ : enter the right-hand side of $y = \ldots$

Summary card

The tangent line

Point: $(a, f(a))$ . Slope: $f'(a)$ .
$y = f(a) + f'(a)(x - a)$ .

Watch out: $y = f'(a)\,x$ has the right slope but usually misses the point.

4. The derivative function

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = x^3 - 3x$ . The blue line is its tangent at $x = a$ . Slide $a$ from $-2$ to $2$ and watch the slope.

The cubic y = x³ − 3x and its tangent at x = a. The tangent is horizontal at x = −1 and x = 1, slopes down between them, and slopes up outside.

Between $x = -1$ and $x = 1$ , the graph falls. What is the sign of the slope there?

Negative
Positive

Key idea

Computing $f'(a)$ at every $a$ gives a new function.

The derivative function is

f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}

. Where

f

rises,

f'(x) > 0

; where

f

falls,

f'(x) < 0

; where the tangent is horizontal,

f'(x) = 0

Worked example

$f(x) = x^2 - 4x$ . Find $f'(x)$ .

f(x + h) - f(x) = 2xh + h^2 - 4h

Expand (x + h)² − 4(x + h) and subtract x² − 4x.

\dfrac{f(x + h) - f(x)}{h} = 2x + h - 4

Divide by h ≠ 0.

f'(x) = 2x - 4

Let h → 0. Check: f' = 0 at x = 2, the vertex.

Your turn, guided

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

\dfrac{f(x + h) - f(x)}{h} = 2x + h + 6

The same steps.

f'(x) = 2x + 6

Let h → 0.

Practice ★

Here is the graph of $y = f(x)$ . Tap the point of the graph where $f'(x) = 0$ .
Here is the graph of $y = f(x)$ . Tap the point of the graph where $f'(x) = 0$ .

$f(x) = (x - 3)^2 + 3$ , a parabola with its vertex at $(3, 3)$ . Tick all the true statements.
1. $f'(3) = 0$
2. $f'(x) < 0$ for $x < 3$
3. $f'(x) > 0$ for $x > 3$
4. $f'(3) = 3$
5. $f'(x) > 0$ for every $x$ , because $f(x) > 0$

Watch out!

Practice ★★

$f(x) = 3x^2 + 8x$ . Use $f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$ to find $f'(x)$ .
$f(x) = 3x^2 + 5x$ . Use $f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$ to find $f'(x)$ .
$f(x) = 5x^2 + 2x$ . Use $f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$ to find $f'(x)$ .

Exit ticket

Three questions to check you have got it.

$f(x) = 3x^2 + 8x$ . Use $f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$ to find $f'(x)$ .
$f(x) = 3x^2 + 5x$ . Use $f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$ to find $f'(x)$ .
$f(x) = 5x^2 + 2x$ . Use $f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}$ to find $f'(x)$ .

Summary card

The derivative function

$f'(x)$ is the slope of $f$ at each $x$ .
$f$ rises: $f' > 0$ . $f$ falls: $f' < 0$ . Horizontal tangent: $f' = 0$ .

Watch out: $f' = 0$ (flat) is not the same as $f = 0$ (on the axis).

5. Interpreting the derivative: rates and units

Warm-up

Three quick questions from earlier lessons.

Discover

$T(m)$ is the temperature of a cup of coffee, in °F, $m$ minutes after it is poured. $T'(10) = -2$ . What are the units of $-2$ ?

°F per minute
°F

Key idea

A derivative is a rate, with units.

f(x)

is measured in output units and

x

in input units, then

f'(a)

is in output units per input unit.

f'(a) = r

means: at

x = a

f

changes by about

r

for each extra unit of

x

Worked example

Interpret $T'(10) = -2$ .

Ten minutes after it is poured…

Input 10 minutes, output a rate in °F per minute.

…the coffee is cooling by about 2 °F per minute

Negative: the temperature goes down.

T(11) \approx T(10) - 2

A rate lets you estimate a small change.

Your turn, guided

Your turn: $s(t)$ meters, $t$ seconds. $s(4) = 30$ and $s(4.5) = 34$ .

s'(4) \approx \dfrac{34 - 30}{0.5}

4 meters in 0.5 seconds.

s'(4) \approx 8

m/s

4 ÷ 0.5.

Practice ★

$V(t)$ is the volume of water in a tank, in liters, $t$ minutes after a tap is opened. $V'(2) = 8$ . What does this mean?
1. At $t = 2$ min, the volume is increasing at about 8 liters per minute.
2. After 2 minutes, the tank holds 8 liters.
3. In the first 2 minutes, 8 liters flowed in.
4. It takes 8 minutes to add 2 liters.
$V(t)$ is the volume of water in a tank, in liters, $t$ minutes after a tap is opened. $V'(7) = 11$ . What does this mean?
1. At $t = 7$ min, the volume is increasing at about 11 liters per minute.
2. After 7 minutes, the tank holds 11 liters.
3. In the first 7 minutes, 11 liters flowed in.
4. It takes 11 minutes to add 7 liters.

$C(q)$ is the cost, in dollars, of making $q$ bikes. What are the units of $C'(125)$ ?
1. dollars per bike
2. dollars
3. bikes per dollar

Watch out!

Practice ★★

A car's position is $s(t)$ meters after $t$ seconds. $s(9) = 83$ and $s(9.5) = 88$ . Estimate $s'(9)$ , in m/s.
A car's position is $s(t)$ meters after $t$ seconds. $s(3) = 31$ and $s(3.5) = 37$ . Estimate $s'(3)$ , in m/s.
A car's position is $s(t)$ meters after $t$ seconds. $s(7) = 49$ and $s(7.5) = 53$ . Estimate $s'(7)$ , in m/s.

Exit ticket

Three questions to check you have got it.

$V(t)$ is the volume of water in a tank, in liters, $t$ minutes after a tap is opened. $V'(2) = 8$ . What does this mean?
1. At $t = 2$ min, the volume is increasing at about 8 liters per minute.
2. After 2 minutes, the tank holds 8 liters.
3. In the first 2 minutes, 8 liters flowed in.
4. It takes 8 minutes to add 2 liters.
$V(t)$ is the volume of water in a tank, in liters, $t$ minutes after a tap is opened. $V'(7) = 11$ . What does this mean?
1. At $t = 7$ min, the volume is increasing at about 11 liters per minute.
2. After 7 minutes, the tank holds 11 liters.
3. In the first 7 minutes, 11 liters flowed in.
4. It takes 11 minutes to add 7 liters.
$V(t)$ is the volume of water in a tank, in liters, $t$ minutes after a tap is opened. $V'(3) = 19$ . What does this mean?
1. At $t = 3$ min, the volume is increasing at about 19 liters per minute.
2. After 3 minutes, the tank holds 19 liters.
3. In the first 3 minutes, 19 liters flowed in.
4. It takes 19 minutes to add 3 liters.

Summary card

Reading a derivative

Units: output units per input unit.
$f'(a) = r$ : at $x = a$ , $f$ changes by about $r$ per extra unit of $x$ .

Watch out: $f'(a)$ is a rate at one moment, not a value and not a total change.

6. The second derivative, concavity and acceleration

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = x^3$ . The blue line is its tangent at $x = a$ . Slide $a$ from $-2$ to $2$ .

The cubic y = x³ and its tangent at x = a. For a < 0 the curve bends down and lies below the tangent; for a > 0 it bends up and lies above it. The bending changes at x = 0.

For $a > 0$ , the curve lies above its tangent lines and the slopes $3a^2$ grow. How does the graph bend there?

It bends up (concave up)
It bends down (concave down)

Key idea

The derivative of the derivative tells how the slope changes.

f''(x)

is the derivative of

f'(x)

. If

f'' > 0

f'

increases and

f

is concave up; if

f'' < 0

f'

decreases and

f

is concave down. A point where the concavity changes is an inflection point. For motion: acceleration

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

Worked example

$v(t) = t^2 + 4t$ m/s. Find the acceleration at $t = 3$ .

a(t) = v'(t) = 2t + 4

From the derivative-function lesson: the derivative of t² + 4t.

a(3) = 10

m/s²

Substitute t = 3.

Your turn, guided

Your turn: same $v(t)$ , at $t = 1$ .

a(1) = 6

m/s²

2 · 1 + 4.

Practice ★

On $(2, 8)$ , the graph of $f$ is increasing and concave down. Tick all the true statements about that interval.
1. $f'(x) > 0$
2. $f''(x) < 0$
3. $f'$ is decreasing: the graph rises more and more slowly
4. $f''(x) > 0$ , because $f$ is increasing
5. $f'(x) < 0$ , because the graph bends down

A particle's velocity is $v(t) = t^2 + 4t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².
A particle's velocity is $v(t) = t^2 + 1t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².

Watch out!

Practice ★★

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

Exit ticket

Three questions to check you have got it.

A particle's velocity is $v(t) = t^2 + 4t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².
A particle's velocity is $v(t) = t^2 + 1t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².
A particle's velocity is $v(t) = t^2 + 6t$ m/s. Its acceleration is $a(t) = v'(t)$ . Find $a(6)$ , in m/s².

Summary card

The second derivative

$f'' > 0$ : slopes grow, concave up. $f'' < 0$ : slopes shrink, concave down.
Inflection point: the concavity changes; there $f'$ has a maximum or minimum.
Acceleration is the derivative of velocity.

Watch out: $f'$ says rising or falling; $f''$ says how it bends.

7. Differentiability and continuity

Warm-up

Three quick questions from earlier lessons.

Discover

$f(x) = |x|$ . Just left of 0 the slope is $-1$ ; just right of 0 it is $1$ . Does the graph have one tangent line at the origin?

No
Yes, since the graph has no break

Key idea

Differentiable means the limit defining $f'(a)$ exists.

f

is differentiable at

a

, then

f

is continuous at

a

. The converse is false: a continuous function fails to be differentiable at a corner, a cusp or a vertical tangent. A function is also not differentiable where it is not continuous.

Worked example

Is $f(x) = |x - 2| + 1$ differentiable at $x = 2$ ?

f

is continuous at

2

No jump or hole at 2.

Slope

-1

on the left,

1

on the right

The two pieces are 1 − (x − 2) and 1 + (x − 2).

Not differentiable at

x = 2

: a corner

The one-sided slopes differ.

Your turn, guided

Your turn: $f(x) = x^2$ for $x \le 1$ and $f(x) = 2x - 1$ for $x > 1$ .

The pieces meet at

(1, 1)

Both pieces give 1 at x = 1.

Slope from the left:

2

, from the right:

2

, so

f

is differentiable at

1

The slope of x² at 1 is 2 · 1.

Practice ★

Is $f(x) = |x - -1|$ differentiable at $x = -1$ ?
1. No: the graph has a corner there (slopes −1 and 1).
2. Yes, because $f$ is continuous at $x = -1$ .
3. No, because $f$ is not continuous at $x = -1$ .
Is $f(x) = |x - 1|$ differentiable at $x = 1$ ?
1. No: the graph has a corner there (slopes −1 and 1).
2. Yes, because $f$ is continuous at $x = 1$ .
3. No, because $f$ is not continuous at $x = 1$ .

Watch out!

Practice ★★

Which functions are differentiable at $x = 0$ ? Tick all that are.
1. $x^2 + 8x$
2. $\sin(8x)$
3. $8|x|$
4. $\sqrt[3]{x}$
5. $f(x) = 0$ for $x < 0$ and $f(x) = 8$ for $x \ge 0$

$f(x) = x^2$ for $x \le 4$ and $f(x) = mx - 16$ for $x > 4$ . Which value of $m$ makes $f$ differentiable at $x = 4$ ?
$f(x) = x^2$ for $x \le 5$ and $f(x) = mx - 25$ for $x > 5$ . Which value of $m$ makes $f$ differentiable at $x = 5$ ?

Exit ticket

Three questions to check you have got it.

$f(x) = x^2$ for $x \le 4$ and $f(x) = mx - 16$ for $x > 4$ . Which value of $m$ makes $f$ differentiable at $x = 4$ ?
$f(x) = x^2$ for $x \le 5$ and $f(x) = mx - 25$ for $x > 5$ . Which value of $m$ makes $f$ differentiable at $x = 5$ ?
$f(x) = x^2$ for $x \le 6$ and $f(x) = mx - 36$ for $x > 6$ . Which value of $m$ makes $f$ differentiable at $x = 6$ ?

Summary card

Differentiability

Differentiable ⇒ continuous; not the other way round.
No derivative at a corner, a cusp, a vertical tangent, or a break.
To join two pieces smoothly, match the heights and the slopes.

Watch out: an unbroken graph can still have no tangent line.

8. Local linearization

Warm-up

Three quick questions from earlier lessons.

Discover

The violet curve is $y = \sqrt{x}$ ; the blue line is its tangent at $x = a$ . Move $a$ : near the point of contact, curve and line almost agree.

The curve y = √x and its tangent line at x = a. Near x = a the two are almost the same; farther away the line rises above the curve.

At $a = 4$ the tangent is $L(x) = 2 + \tfrac{1}{4}(x - 4)$ . Use it to estimate $\sqrt{4.4}$ .

Key idea

Near $x = a$ , a differentiable function is almost its tangent line.

The local linearization of

f

x = a

L(x) = f(a) + f'(a)(x - a)

, and

f(x) \approx L(x)

for

x

near

a

. If

f

is concave up near

a

L

underestimates; if concave down, it overestimates.

Worked example

$f(x) = \sqrt{x}$ , $f'(x) = \dfrac{1}{2\sqrt{x}}$ . Estimate $\sqrt{9.6}$ .

f(9) = 3

f'(9) = \tfrac{1}{6}

A nearby point with an exact root: a = 9.

L(9.6) = 3 + \tfrac{1}{6}(0.6) = 3.1

L(x) = f(9) + f'(9)(x − 9).

\sqrt{9.6} \approx 3.1

(true value

3.098…

)

√x is concave down: the estimate is a little too big.

Your turn, guided

Your turn: $f(2) = 7$ and $f'(2) = 3$ . Estimate $f(2.1)$ .

f(2.1) \approx 7 + 3(0.1) = 7.3

L(x) = 7 + 3(x − 2).

Practice ★

$f(2) = 3$ and $f'(2) = 2$ . Write the local linearization $L(x)$ of $f$ at $x = 2$ .
$f(4) = -5$ and $f'(4) = -4$ . Write the local linearization $L(x)$ of $f$ at $x = 4$ .
$f(4) = 4$ and $f'(4) = -1$ . Write the local linearization $L(x)$ of $f$ at $x = 4$ .

Watch out!

Practice ★★

$f(x) = \sqrt{x}$ has $f'(x) = \dfrac{1}{2\sqrt{x}}$ . Use the local linearization at $x = 4$ to estimate $\sqrt{ 6 }$ . Give an exact fraction.
$f(x) = \sqrt{x}$ has $f'(x) = \dfrac{1}{2\sqrt{x}}$ . Use the local linearization at $x = 4$ to estimate $\sqrt{ 8 }$ . Give an exact fraction.

Near $x = 4$ , the graph of $f$ is concave up. Does the local linearization at $x = 4$ over- or underestimate $f(x)$ for $x$ near $4$ ?
1. Underestimate: the tangent line lies below the curve.
2. Overestimate: the tangent line lies above the curve.
3. Neither: near $x = 4$ the linearization is exact.

Exit ticket

Three questions to check you have got it.

$f(2) = 3$ and $f'(2) = 2$ . Write the local linearization $L(x)$ of $f$ at $x = 2$ .
$f(4) = -5$ and $f'(4) = -4$ . Write the local linearization $L(x)$ of $f$ at $x = 4$ .
$f(4) = 4$ and $f'(4) = -1$ . Write the local linearization $L(x)$ of $f$ at $x = 4$ .

Summary card

Local linearization

$L(x) = f(a) + f'(a)(x - a)$ , the tangent line used as an estimate.
Good near $a$ ; concave up underestimates, concave down overestimates.

Watch out: multiply the step $x - a$ by the slope $f'(a)$ .

Answers: The derivative

1. Average and instantaneous velocity

$33$
$9$
$21$

$5 h + 30$
$2 h + 8$

$43$
$18$

$43$
$18$
$55$

2. The derivative at a point: the limit definition

$12$
$8$
$15$

$\displaystyle\lim_{h \to 0} \dfrac{f(2 + h) - f(2)}{h}$

$- \frac{4}{9}$
$- \frac{3}{8}$

$12$
$8$
$15$

3. The tangent line

$y = 2 x - 1$
$y = - 4 x - 4$

$3 x - 1$
$3 x - 11$

$27$
$12$

$6 x - 25$

$3 x - 1$
$3 x - 11$
$- 3 x - 1$

4. The derivative function

$(-1, 1)$
$(-2, 1)$

$f'(3) = 0$ · $f'(x) < 0$ for $x < 3$ · $f'(x) > 0$ for $x > 3$

$6 x + 8$
$6 x + 5$
$10 x + 2$

$6 x + 8$
$6 x + 5$
$10 x + 2$

5. Interpreting the derivative: rates and units

At $t = 2$ min, the volume is increasing at about 8 liters per minute.
At $t = 7$ min, the volume is increasing at about 11 liters per minute.

dollars per bike

$10$
$12$
$8$

At $t = 2$ min, the volume is increasing at about 8 liters per minute.
At $t = 7$ min, the volume is increasing at about 11 liters per minute.
At $t = 3$ min, the volume is increasing at about 19 liters per minute.

6. The second derivative, concavity and acceleration

$f'(x) > 0$ · $f''(x) < 0$ · $f'$ is decreasing: the graph rises more and more slowly

$16$
$13$

$(-1, 4)$
$(-1, 1)$

$16$
$13$
$18$

7. Differentiability and continuity

No: the graph has a corner there (slopes −1 and 1).
No: the graph has a corner there (slopes −1 and 1).

$x^2 + 8x$ · $\sin(8x)$

$8$
$10$

$8$
$10$
$12$

8. Local linearization

$2 x - 1$
$11 - 4 x$
$8 - x$

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

Underestimate: the tangent line lies below the curve.

$2 x - 1$
$11 - 4 x$
$8 - x$