5.9 - Connecting a Function, Its First Derivative, and Its Second Derivative

Comprehensive Sign Charts

To fully understand how a function behaves, combine info from both derivatives into organized sign charts.

Building Individual Charts

Example: analyze

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

Step 1: Critical points.

f'(x) = 3x^2 + 2x - 1 = (3x - 1)(x + 1)

Critical points:

x = -1

and

x = \frac{1}{3}

Step 2: Possible inflection points.

f''(x) = 6x + 2 = 0 \quad \Rightarrow \quad x = -\frac{1}{3}

Step 3: Individual sign charts.

Chart for

f'(x)

Interval	$(-\infty, -1)$	$x = -1$	$\left(-1, \frac{1}{3}\right)$	$x = \frac{1}{3}$	$\left(\frac{1}{3}, \infty\right)$
Sign of $f'(x)$	$(+)$	$0$	$(-)$	$0$	$(+)$

Chart for

f''(x)

Interval	$\left(-\infty, -\frac{1}{3}\right)$	$x = -\frac{1}{3}$	$\left(-\frac{1}{3}, \infty\right)$
Sign of $f''(x)$	$(-)$	$0$	$(+)$

Step 4: Conclusions.

From the

f'(x)

chart:

Increasing on $(-\infty, -1) \cup \left(\frac{1}{3}, \infty\right)$

Decreasing on $\left(-1, \frac{1}{3}\right)$

Relative maximum at $x = -1$

Relative minimum at $x = \frac{1}{3}$

From the

f''(x)

chart:

Concave down on $\left(-\infty, -\frac{1}{3}\right)$

Concave up on $\left(-\frac{1}{3}, \infty\right)$

Point of inflection at $x = -\frac{1}{3}$

Lining Up the Graphs of $f$ , $f'$ , and $f''$

Here are the graphs of

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

f'(x) = 3x^2 + 2x - 1

, and

f''(x) = 6x + 2

from the example above, stacked so the

x

-values line up.

5.9 - Connecting a Function, Its First Derivative, and Its Second Derivative

The relative max and min of $f$ (at $x = -1$ and $x = \frac{1}{3}$ ) are exactly where $f'$ crosses the $x$ -axis.

The point of inflection of $f$ (at $x = -\frac{1}{3}$ ) is where $f'$ has its minimum and where $f''$ crosses the $x$ -axis.

Where $f'$ is decreasing, $f''$ is below the $x$ -axis and $f$ is concave down. Where $f'$ is increasing, $f''$ is above the $x$ -axis and $f$ is concave up.

Quick reference: the same feature shows up differently on each graph.

On the graph of $f$	On the graph of $f'$	On the graph of $f''$
Increasing	Above the $x$ -axis	Doesn’t tell you
Decreasing	Below the $x$ -axis	Doesn’t tell you
Relative maximum	Crosses the $x$ -axis from above to below	A negative value there confirms it (5.7)
Relative minimum	Crosses the $x$ -axis from below to above	A positive value there confirms it (5.7)
Concave up	Increasing	Above the $x$ -axis
Concave down	Decreasing	Below the $x$ -axis
Point of inflection	Relative max or min	Crosses the $x$ -axis (changes sign)

Combined Sign Chart Method

When a question asks about overlapping conditions (like “concave up AND decreasing”), build one comprehensive chart.

List ALL critical values, from both $f'(x) = 0$ and $f''(x) = 0$

Make intervals using all of them

Test the signs of both $f'(x)$ and $f''(x)$ in each interval

Find where the conditions you want overlap

Example: Combined Conditions

Same function,

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

. Critical values:

x = -1, -\frac{1}{3}, \frac{1}{3}

Interval	$(-\infty, -1)$	$x = -1$	$\left(-1, -\frac{1}{3}\right)$	$x = -\frac{1}{3}$	$\left(-\frac{1}{3}, \frac{1}{3}\right)$	$x = \frac{1}{3}$	$\left(\frac{1}{3}, \infty\right)$
$f'(x)$	$(+)$	$0$	$(-)$	$(-)$	$(-)$	$0$	$(+)$
$f''(x)$	$(-)$	$(-)$	$(-)$	$0$	$(+)$	$(+)$	$(+)$

Note that

f'(x)

only changes sign at

x = -1

and

x = \frac{1}{3}

, so it keeps the same sign across the intervals in between. Same idea for

f''(x)

, which only changes sign at

x = -\frac{1}{3}

Question: when is

f

both concave up AND decreasing?

Concave up: $f''(x) > 0$ on $\left(-\frac{1}{3}, \infty\right)$

Decreasing: $f'(x) < 0$ on $\left(-1, \frac{1}{3}\right)$

Overlap: $\left(-\frac{1}{3}, \frac{1}{3}\right)$

Answer:

f

is concave up and decreasing on

\left(-\frac{1}{3}, \frac{1}{3}\right)

Question: when is

f

both concave down AND increasing?

Increasing: $f'(x) > 0$ on $(-\infty, -1) \cup \left(\frac{1}{3}, \infty\right)$

Concave down: $f''(x) < 0$ on $\left(-\infty, -\frac{1}{3}\right)$

Overlap: $(-\infty, -1)$

Answer:

f

is concave down and increasing on

(-\infty, -1)

Particle Motion: Speeding Up vs. Slowing Down

For a position function

s(t)

Velocity: $v(t) = s'(t)$

Acceleration: $a(t) = s''(t) = v'(t)$

\text{Speeding up: } v(t) \text{ and } a(t) \text{ have the SAME sign}

\text{Slowing down: } v(t) \text{ and } a(t) \text{ have DIFFERENT signs}

Critical point: an object with negative velocity is NOT automatically slowing down. It’s only slowing down if acceleration and velocity point opposite ways.

Example: Particle Motion Analysis

Given position

s(t) = \dfrac{t^3}{3} - 4t^2 + 12t + 5

Step 1: Velocity and acceleration.

v(t) = t^2 - 8t + 12 = (t-2)(t-6)

a(t) = 2t - 8

Step 2: Zeros.

v(t) = 0

t = 2

and

t = 6

a(t) = 0

t = 4

Step 3: Combined chart.

Interval	$(0, 2)$	$t = 2$	$(2, 4)$	$t = 4$	$(4, 6)$	$t = 6$	$(6, \infty)$
$v(t)$	$(+)$	$0$	$(-)$	$(-)$	$(-)$	$0$	$(+)$
$a(t)$	$(-)$	$(-)$	$(-)$	$0$	$(+)$	$(+)$	$(+)$
Speed	Slowing		Speeding		Slowing		Speeding

Test points:

$t = 1$ : $v(1) = 5 > 0$ , $a(1) = -6 < 0$

$t = 3$ : $v(3) = -3 < 0$ , $a(3) = -2 < 0$

$t = 5$ : $v(5) = -3 < 0$ , $a(5) = 2 > 0$

$t = 7$ : $v(7) = 5 > 0$ , $a(7) = 6 > 0$

Step 4: Conclusions.

Speeding up on

(2, 4) \cup (6, \infty)

because

v(t)

and

a(t)

have the same sign.

Slowing down on

(0, 2) \cup (4, 6)

because

v(t)

and

a(t)

have different signs.

Sign Combinations

$v(t)$	$a(t)$	Direction	Speed Change	Intuition
$(+)$	$(+)$	Forward	Speeding up	Moving forward, accelerating forward
$(-)$	$(-)$	Backward	Speeding up	Moving backward, accelerating backward
$(+)$	$(-)$	Forward	Slowing down	Moving forward, accelerating backward
$(-)$	$(+)$	Backward	Slowing down	Moving backward, accelerating forward

Analyzing Functions from Tables of Values

When you only get a table of values for

f(x)

, you can still determine behavior IF you know that

f'

and

f''

don’t change signs.

Requirements

$f'(x)$ doesn’t change sign (always increasing or always decreasing)

$f''(x)$ doesn’t change sign (always concave up or always concave down)

Method: Average Rate of Change

$x$	1	2	3	4
$f(x)$	3	5	8	13

Step 1: Increasing or decreasing? Since

3 < 5 < 8 < 13

, it’s increasing.

Step 2: Average rates of change between consecutive points (here

\Delta x = 1

x = 1 \text{ to } 2: \frac{5-3}{1} = 2

x = 2 \text{ to } 3: \frac{8-5}{1} = 3

x = 3 \text{ to } 4: \frac{13-8}{1} = 5

Step 3: Analyze the rates. They’re

2, 3, 5

, which are increasing.

Step 4: Conclusion about concavity. The average rates of change approximate values of

f'(x)

. Since those rates are increasing,

f'(x)

is increasing. And if

f'(x)

is increasing, then

f''(x) > 0

. So

f

is concave up.

Example 2: A Decreasing Function

$x$	0	1	2	3
$g(x)$	20	15	11	8

Behavior: decreasing.

Average rates:

0 to 1: $\frac{15-20}{1} = -5$

1 to 2: $\frac{11-15}{1} = -4$

2 to 3: $\frac{8-11}{1} = -3$

The rates

-5, -4, -3

are increasing (getting less negative).

Conclusion: since

g'(x)

is increasing,

g''(x) > 0

, so

g

is concave up.

A decreasing function can absolutely still be concave up, it’s just decreasing more and more slowly.

Important Caveat

This method only works when:

You know $f'$ doesn’t change sign on the interval

You know $f''$ doesn’t change sign on the interval

The x-values are evenly spaced (makes the math easier)

Without those guarantees, the function could hide extrema or inflection points between the given points.

Key Takeaways

Make separate sign charts for $f'$ and $f''$ to get the basic behavior

Use a combined chart with all critical values from both derivatives for overlapping conditions

For overlapping conditions, find where two properties hold on the same interval

Speeding up: $v(t)$ and $a(t)$ have the same sign

Slowing down: $v(t)$ and $a(t)$ have opposite signs

Negative velocity does NOT mean slowing down, you have to check acceleration

Justify with “because $v(t)$ and $a(t)$ have the same/different signs”

From tables, check whether the average rates of change are increasing or decreasing

Increasing rates mean concave up; decreasing rates mean concave down

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