Connecting f, f′, and f″

What the graph of f′ shows

Reading f from the graph of f′

Let

f

be differentiable on an open interval

I

, so that

f

is also continuous on

I

. Then for points and subintervals of

I

When the graph of $f'$	the function $f$
is above the $x$ -axis on an interval	is increasing there
is below the $x$ -axis on an interval	is decreasing there
crosses the axis from above to below at $c$	has a relative maximum at $c$
crosses the axis from below to above at $c$	has a relative minimum at $c$
touches the axis at $c$ without crossing	has no relative extremum at $c$
rises on an interval	is concave up there
falls on an interval	is concave down there
changes from rising to falling, or from falling to rising, at $c$	has a point of inflection at $c$

The first five rows come from the sign of

f'

: the test for increasing and the First Derivative Test. The last three come from the direction of

f'

, which is how concavity is defined. When

f''

exists, it's the slope of the graph of

f'

, so where

f'' > 0

that graph rises, and where

f'' < 0

it falls.

Whether

f

increases depends on where the graph of

f'

sits, above or below the axis. The concavity of

f

depends on which way that graph heads, up or down.

Example 1

The function

f

is twice differentiable on

[-3, 6]

, and the graph of its derivative

f'

is shown. The graph of

f'

crosses the

x

-axis only at

x = -2

and

x = 1

, and it has horizontal tangents only at

x = -1

x = 3

, and

x = 5

The graph of f′ on [−3, 6]. It passes through (−3, 3), (−1, −2), (3, 2), (5, 0.5), and (6, 1.5).

(a) Find the open intervals on which

f

is increasing and those on which it's decreasing. Give a reason for your answer.

(b) Find the

x

-coordinate of each relative extremum of

f

(-3, 6)

, classify each one, and justify your answers.

(a) The graph of

f'

is above the axis on

(-3, -2)

and

(1, 6)

, so

f'(x) > 0

there, and

f

is increasing on

(-3, -2)

and

(1, 6)

. The graph is below the axis on

(-2, 1)

, so

f'(x) < 0

there, and

f

is decreasing on

(-2, 1)

(b) The critical points of

f

(-3, 6)

are the zeros of

f'

x = -2

and

x = 1

, since

f'

exists everywhere there. The function

f

has a relative maximum at

x = -2

, because

f'

changes from positive to negative there. It has a relative minimum at

x = 1

, because

f'

changes from negative to positive there. The low point of the graph of

f'

x = 5

isn't a critical point, since

f'(5) = 0.5

Example 2

For the function

f

of Example 1:

(a) Find the open intervals on which the graph of

f

is concave up and those on which it's concave down.

(b) Find the

x

-coordinate of each point of inflection of the graph of

f

. Give a reason for your answer.

x

[-3, 6]

at which the line tangent to the graph of

f

has the greatest slope.

(a) The graph of

f

is concave up where

f'

is increasing, and concave down where

f'

is decreasing. The graph of

f'

rises on

(-1, 3)

and

(5, 6)

, and falls on

(-3, -1)

and

(3, 5)

. So

f

is concave up on

(-1, 3)

and

(5, 6)

, and concave down on

(-3, -1)

and

(3, 5)

(b) The graph of

f

has points of inflection at

x = -1

x = 3

, and

x = 5

. At

-1

and

5

f'

changes from decreasing to increasing, and at

3

, from increasing to decreasing. At

x = 5

f

is increasing on both sides: a point of inflection needn't be a critical point.

(c) The slope of the tangent line at

x

f'(x)

, so we want the absolute maximum of

f'

[-3, 6]

. Since

f'

is differentiable, it's continuous on

[-3, 6]

. By the Extreme Value Theorem its maximum exists, and it occurs at an endpoint or where

f''(x) = 0

, at a horizontal tangent of the graph of

f'

. The candidates are

$x$	$-3$	$-1$	$3$	$5$	$6$
$f'(x)$	$3$	$-2$	$2$	$0.5$	$1.5$

The largest value is

f'(-3) = 3

. So the tangent line has its greatest slope at the left endpoint,

x = -3

. The relative maximum of

f'

at the point of inflection

x = 3

is smaller.

Example 3

For the same function

f

f(1) = -1

. Make a table of the signs of

f'

and

f''

, and use it to sketch the graph of

f

The points found so far,

-2

-1

1

3

, and

5

, split

(-3, 6)

into six intervals. On each, the sign of

f'

comes from where its graph sits, and the sign of

f''

from which way it heads.

Interval	$f'$	$f''$	$f$
$(-3, -2)$	$+$	$-$	increasing, concave down
$(-2, -1)$	$-$	$-$	decreasing, concave down
$(-1, 1)$	$-$	$+$	decreasing, concave up
$(1, 3)$	$+$	$+$	increasing, concave up
$(3, 5)$	$+$	$-$	increasing, concave down
$(5, 6)$	$+$	$+$	increasing, concave up

Read the table from left to right. The graph rises while bending down to the relative maximum at

x = -2

. It falls, bending down and then up past the point of inflection at

x = -1

, to the relative minimum at

(1, -1)

, the one point we know. Then it rises all the way to

x = 6

, bending up, then down, then up again, with points of inflection at

x = 3

and

x = 5

The graph of

f'

fixes the shape of

f

but not its heights. Apart from

f(1) = -1

, those need integration, which comes in a later unit, so a sketch shows them only roughly.

The graph of f through (1, −1). Amber: the relative maximum and minimum; purple: the points of inflection.

Sketching f′ from the graph of f

The table also runs the other way. To sketch

f'

from a graph of

f

, read the slope of

f

at each

x

and plot it as a height. Where the graph of

f

has a horizontal tangent,

f' = 0

. Where

f

rises, the graph of

f'

is above the axis, and where

f

falls, below it. At a point of inflection of

f

, the slope stops growing and starts shrinking, or the reverse, so

f'

has a relative extremum there.

Example 4

The graph of

f

is shown for

-1.6 \le x \le 4.6

. It has horizontal tangents only at

x = -1

x = 1

, and

x = 4

, and points of inflection only at

x \approx -0.4

x = 1

, and

x \approx 3.1

. Sketch the graph of

f'

The graph of f. Its tangent is level at x = −1, x = 1, and x = 4.

First the zeros and the signs:

f'

0

-1

1

, and

4

. The graph of

f

falls on

(-1.6, -1)

and

(4, 4.6)

, so

f'

is negative there. It rises on

(-1, 1)

and

(1, 4)

, so

f'

is positive there.

x = 1

the tangent is level, but

f

rises on both sides. So

f'(1) = 0

with

f'

positive on both sides, and the graph of

f'

touches the axis there without crossing.

Then the turning points. The slope of

f

grows until

x \approx -0.4

and shrinks until

x = 1

, so

f'

has a relative maximum at

-0.4

and a relative minimum,

0

, at

1

. The slope grows again until

x \approx 3.1

and then shrinks, so

f'

has a second relative maximum there. The second one is higher, since

f

is much steeper near

x = 3

than near

x = 0

. Near both ends,

f

falls steeply, so

f'

is well below the axis.

The graph of f′. It's 0 at x = −1, 1, and 4, and touches the axis at x = 1. Its peaks are at x ≈ −0.4 and x ≈ 3.1.

From a table of signs

Example 5

The function

f

is twice differentiable on

[0, 7]

. The table gives values of

f

, and the signs of

f'

and

f''

(

+

for positive,

-

for negative).

$x$	$f(x)$	$f'(x)$	$f''(x)$
$0$	$2$	$-$	$+$
$0 < x < 2$		$-$	$+$
$2$	$-1$	$0$	$+$
$2 < x < 4$		$+$	$+$
$4$	$2$	$+$	$0$
$4 < x < 5$		$+$	$-$
$5$	$4$	$0$	$-$
$5 < x < 7$		$-$	$-$
$7$	$1$	$-$	$-$

(a) Find the

x

-coordinates of the relative extrema and the points of inflection of

f

(0, 7)

, and justify your answers.

(b) Sketch a graph of

f

consistent with the table.

f

[0, 7]

. Justify your answer.

(a)

f

has a relative minimum at

x = 2

, because

f'

changes from negative to positive there. It has a relative maximum at

x = 5

, because

f'

changes from positive to negative there. Since

f''

changes from positive to negative at

x = 4

, the graph of

f

has a point of inflection there. There's no other sign change of

f''

, so there are no other points of inflection.

(b) Plot the five given points, and join them with the shape each row gives. The graph falls while bending up to the minimum at

(2, -1)

, then rises while bending up to

(4, 2)

. It keeps rising, now bending down, to the maximum at

(5, 4)

, and then falls while bending down to

(7, 1)

One graph consistent with the table. Amber: the relative minimum and maximum; purple: the point of inflection.

(c) Since

f

is differentiable, it's continuous. On

[0, 2]

f' < 0

, so

f

is decreasing, from

f(0) = 2

f(2) = -1

. By the Intermediate Value Theorem it takes the value

0

there, and since it's decreasing, it does so exactly once.

[2, 5]

f

is increasing from

-1

4

, so it has exactly one zero there as well. On

[5, 7]

it's decreasing from

4

1

, so

f(x) \ge 1

and there's no zero. So

f

has exactly

2

zeros on

[0, 7]

Matching f, f′, and f″

Three graphs can be sorted into

f

f'

, and

f''

with the same table. The horizontal tangents of a graph line up with the zeros of its derivative. Where a graph rises, its derivative is positive, and where it falls, negative.

Example 6

The graphs I, II, and III are drawn for

-2.6 \le x \le 5.6

, each with its own vertical scale. One is the graph of

f

, one is the graph of

f'

, and one is the graph of

f''

. Identify each one, and give a reason.

Graph I

Graph II

Graph III

Graph II has horizontal tangents at

x = -2

x = 1

, and

x = 5

, which is where graph I crosses the axis. Graph I is negative where II falls, on

x < -2

and

1 < x < 5

, and positive where II rises. So I is the derivative of II.

Graph I has horizontal tangents at about

x = -0.7

and

x = 3.4

, which is where graph III crosses the axis. Graph III is negative between them, where I falls, and positive where I rises. So III is the derivative of I.

No other order works. Graph III has a horizontal tangent at about

x = 1.3

, where neither of the others is

0

, so neither one is the derivative of III. So

f

is graph II,

f'

is graph I, and

f''

is graph III.