Connecting f, f' and f'' — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

A weather station logs temperature, its rate of change, and how fast that rate is changing. A forecaster reading the three together can say "still warming, but the warming is easing off, so the high is coming soon." Reading

f

,

f'

and

f''

together is the same skill, and the graph-of-

f'

question on the exam is built on it.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Where the graph of $f'$ has a high point, the graph of $f$ has

a point of inflection
a relative maximum
a relative minimum
a corner

Show the answer

$f'$ turns, so $f''$ changes sign: the concavity of $f$ changes.

Correct: a point of inflection:
a relative maximum:
a relative minimum:
a corner:

2. $f'(c) = 0$ and $f''(c) > 0$ mean $f$ has

a relative minimum at $c$
a relative maximum at $c$
a point of inflection at $c$
no conclusion

Show the answer

The Second Derivative Test.

Correct: a relative minimum at $c$ :
a relative maximum at $c$ :
a point of inflection at $c$ :
no conclusion:

3. The line tangent to $f$ at $x = a$ is

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

Show the answer

Point $(a, f(a))$ , slope $f'(a)$ .

Correct: $y = f(a) + f'(a)(x - a)$ :
$y = f'(a) + f(a)(x - a)$ :
$y = f(a)(x - a)$ :
$y = f'(a)x$ :

Part 4 · See it

See it first

Three stacked graphs on the same x-axis from 0 to 2 pi: f(x) = sin x, f'(x) = cos x and f''(x) = -sin x. Dashed lines at pi/2, pi, 3pi/2 and 2pi line up the features: f peaks at pi/2 where f' is zero and changes from positive to negative; f has an inflection point at pi where f' is lowest and f'' changes sign; f'' is negative on (0, pi), where f is concave down. — Three graphs on one $x$ -axis. Where $f$ peaks, $f'$ is zero; where $f$ inflects, $f'$ is at an extreme and $f''$ is zero; where $f''$ is negative, $f$ is concave down. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The sign of $f'$ tells whether $f$ is increasing or decreasing; a sign change of $f'$ is a relative extremum of $f$
The sign of $f''$ , or whether $f'$ is rising or fallingtells the concavity of $f$ ; a sign change of $f''$ is a point of inflection of $f$
$f(a)$ and $f'(a)$ give the tangent line, and the sign of $f''$ near $a$ says whether it overestimates or underestimates
A justification on the examnames the function whose sign or direction you read: " $f'$ changes from positive to negative", " $f'$ is decreasing"

Part 6 · Key ideas

Key ideas

Graphical analysis: given any one of $f$ , $f'$ , $f''$ (graph, table or formula), read the features of the others.
From a graph of $f'$ : sign of $f'$ (above or below the axis) gives increasing or decreasing; direction of $f'$ (rising or falling) gives concavity.
From a table: $f'(a) = 0$ and the sign of $f''(a)$ give the Second Derivative Test; a continuous $f''$ that changes from positive to negative between two inputs gives a point of inflection between them.
Words: "increasing at a decreasing rate" means $f' > 0$ and $f'' < 0$ .

Part 7 · Misconception

A common mistake

The wrong idea: On a graph of $f'$ , the points of inflection of $f$ are where the graph of $f'$ crosses the $x$ -axis.

What actually happens: Crossings of $f'$ are where $f$ has extrema. Inflection points of $f$ are where $f'$ turns (its peaks and valleys), because that is where $f''$ changes sign. Chief readers see this swap every year.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $f'(x) = x^3 - 3x^2 + 2$

The graph shows the derivative of

f

for

-1 \le x \le 3.2

. It crosses the

x

-axis at

x = 1 - \sqrt{3} \approx -0.73

,

x = 1

and

x = 1 + \sqrt{3} \approx 2.73

. Also,

f(0) = 4

.

1. At which value of $x$ does $f$ have a relative maximum?

$x = 1$
$x = 0$ , where $f'$ is highest
$x = 2$ , where $f'$ is lowest
$x \approx -0.73$ , where $f' = 0$

Show the answer

$f'$ crosses the axis going down at $x = 1$ (positive before, negative after), so $f$ has a relative maximum there.

Correct: $x = 1$ : $f'$ changes from positive to negative at $x = 1$ .
$x = 0$ , where $f'$ is highest: A high point of $f'$ is an inflection point of $f$ .
$x = 2$ , where $f'$ is lowest: A low point of $f'$ is an inflection point of $f$ .
$x \approx -0.73$ , where $f' = 0$ : $f'$ changes from negative to positive there: a relative minimum.

2. On which intervals is the graph of $f$ concave up?

$(0, 2)$
$(-0.73, 1)$ and $(2.73, 3.2)$
$(-1, -0.73)$ and $(1, 2.73)$
$(-1, 0)$ and $(2, 3.2)$

Show the answer

$f$ is concave up where $f'$ is increasing: from $x = -1$ up to its high point at 0, and after its low point at 2.

$(0, 2)$ : $f'$ is decreasing there: concave down.
$(-0.73, 1)$ and $(2.73, 3.2)$ : There $f' > 0$ : $f$ is increasing. That is direction, not concavity.
$(-1, -0.73)$ and $(1, 2.73)$ : There $f' < 0$ : $f$ is decreasing.
Correct: $(-1, 0)$ and $(2, 3.2)$ : $f'$ is increasing there.

3. Use the graph and the formula to find the slope of the graph of $f'$ at $x = 1$ , which is $f''(1)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f''(x) = 3x^2 - 6x$ , so $f''(1) = -3$ : the graph of $f'$ is falling at $x = 1$ , and $f$ is concave down there.

Answer: -3

4. Which is true of the graph of $f$ at $x = 2$ ?

It has a relative minimum
It is decreasing and has a point of inflection
It is increasing and concave up
It has a point of inflection and a relative minimum

Show the answer

$f'(2)$ $= 8 - 12 + 2$ $= -2 < 0$ , so $f$ is decreasing. $f'$ has a low point at 2 ( $f''$ changes from negative to positive), so the concavity of $f$ changes: a point of inflection.

It has a relative minimum: $f'(2) = -2 \ne 0$ : not a critical point.
Correct: It is decreasing and has a point of inflection: $f'(2) = -2 < 0$ , and $f'$ changes from decreasing to increasing at 2.
It is increasing and concave up: $f'(2) < 0$ means decreasing.
It has a point of inflection and a relative minimum: A relative minimum would need $f'$ to change sign at 2.

5. Write an equation for the line tangent to the graph of $f$ at $x = 0$ . Enter the right side.

Type an expression for $y =$ .

Show the answer

Point $(0, 4)$ , slope $f'(0) = 2$ : $y$ $= 4 + 2(x - 0)$ $= 4 + 2x$ .

Answer: $y =$ $4 + 2x$

6. On $(a, b)$ , $f'(x) > 0$ and $f''(x) < 0$ . Which describes $f$ there?

Increasing, at an increasing rate
Decreasing, at a decreasing rate
Decreasing, and concave down
Increasing, at a decreasing rate

Show the answer

$f' > 0$ means $f$ increases; $f'' < 0$ means $f'$ decreases, so the increase slows down: concave down.

Increasing, at an increasing rate: That would need $f'' > 0$ .
Decreasing, at a decreasing rate: $f' > 0$ means increasing.
Decreasing, and concave down: $f' > 0$ means increasing.
Correct: Increasing, at a decreasing rate: $f' > 0$ : increasing. $f'' < 0$ : the rate $f'$ is falling.

Part 9 · Summary

Summary

The three graphs carry one story.

f'

gives the direction of

f

, and its sign changes give extrema.

f''

(or whether

f'

rises or falls) gives the concavity of

f

, and its sign changes give inflection points. Read from whichever representation you have, and name the function in every justification.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of f′(x)=x3−3x2+2f'(x) = x^3 - 3x^2 + 2

Summary

What comes next

Connections

The graph of $f'(x) = x^3 - 3x^2 + 2$