Sec on 4.4
Curve Sketching
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
April 13, 2011
.

Announcements
Quiz 4 on Sec ons 3.3,
3.4, 3.5, and 3.7 this
week (April 14/15)
Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm
I am teaching Calc II MW
2:00pm and Calc III TR
2:00pm both Fall ’11 and
Spring ’12

Objectives
given a func on, graph it
completely, indica ng
zeroes (if easy)
asymptotes if applicable
cri cal points
local/global max/min
inflec on points

Why?
Graphing func ons is like
dissec on

Why?
Graphing func ons is like
dissec on … or diagramming
sentences

Why?
Graphing func ons is like
dissec on … or diagramming
sentences
You can really know a lot
about a func on when you
know all of its anatomy.

The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),
then f is decreasing on (a, b).
Example
f(x)
f(x) = x3 + x2
.

The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),
then f is decreasing on (a, b).
Example
f(x)
f′ (x)
f(x) = x3 + x2
f′ (x) = 3x2 + 2x .

Testing for Concavity
Theorem (Concavity Test)
If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward
on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave
downward on (a, b).
Example
f(x)
f(x) = x3 + x2
.

Testing for Concavity
Theorem (Concavity Test)
If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward
on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave
downward on (a, b).
Example
f′ (x) f(x)
f(x) = x3 + x2
f′ (x) = 3x2 + 2x .

Testing for Concavity
Theorem (Concavity Test)
If f′′ (x) > 0 for all x in (a, b), then the graph of f is concave upward
on (a, b) If f′′ (x) < 0 for all x in (a, b), then the graph of f is concave
downward on (a, b).
Example
f′′ (x) f′ (x) f(x)
f(x) = x3 + x2
f′ (x) = 3x2 + 2x .
f′′ (x) = 6x + 2

Graphing Checklist
To graph a func on f, follow this plan:
0. Find when f is posi ve, nega ve, zero,
not defined.

Graphing Checklist
To graph a func on f, follow this plan:
0. Find when f is posi ve, nega ve, zero,
not defined.
1. Find f′ and form its sign chart.
Conclude informa on about
increasing/decreasing and local
max/min.

Graphing Checklist
To graph a func on f, follow this plan:
0. Find when f is posi ve, nega ve, zero,
not defined.
1. Find f′ and form its sign chart.
Conclude informa on about
increasing/decreasing and local
max/min.
2. Find f′′ and form its sign chart.
Conclude concave up/concave down
and inflec on.

Graphing Checklist
To graph a func on f, follow this plan:
3. Put together a big chart to assemble
monotonicity and concavity data

Graphing Checklist
To graph a func on f, follow this plan:
3. Put together a big chart to assemble
monotonicity and concavity data
4. Graph!

Outline
Simple examples
A cubic func on
A quar c func on
More Examples
Points of nondifferen ability
Horizontal asymptotes
Ver cal asymptotes
Trigonometric and polynomial together
Logarithmic

Graphing a cubic
Example
Graph f(x) = 2x3 − 3x2 − 12x.

Graphing a cubic
Example
Graph f(x) = 2x3 − 3x2 − 12x.
(Step 0) First, let’s find the zeros. We can at least factor out one
power of x:
f(x) = x(2x2 − 3x − 12)
so f(0) = 0. The other factor is a quadra c, so we the other two
roots are
√ √
3 ± 32 − 4(2)(−12) 3 ± 105
x= =
4 4
It’s OK to skip this step for now since the roots are so complicated.

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
.

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
. x−2
2
x+1
−1 f′ (x)
−1 2 f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
x+1
−1 f′ (x)
−1 2 f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
−1 2 f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
+
−1 2 f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
+ −
−1 2 f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
+ − +
−1 2 f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
+ − +
↗−1 2 f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
+ − +
↗−1 ↘ 2 f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
+ − +
↗−1 ↘ 2 ↗ f(x)

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
+ − +
↗−1 ↘ 2 ↗ f(x)
max

Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)
We can form a sign chart from this:
− .− +
x−2
2
− + +
x+1
−1 f′ (x)
+ − +
↗−1 ↘ 2 ↗ f(x)
max min

Step 2: Concavity
f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)
Another sign chart:
.

Step 2: Concavity
f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)
Another sign chart:
.
f′′ (x)
1/2 f(x)

Step 2: Concavity
f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)
Another sign chart:
.
−− f′′ (x)
1/2 f(x)

Step 2: Concavity
f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)
Another sign chart:
.
−− ++ f′′ (x)
1/2 f(x)

Step 2: Concavity
f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)
Another sign chart:
.
−− ++ f′′ (x)
⌢ 1/2 f(x)

Step 2: Concavity
f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)
Another sign chart:
.
−− ++ f′′ (x)
⌢ 1/2 ⌣ f(x)

Step 2: Concavity
f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)
Another sign chart:
.
−− ++ f′′ (x)
⌢ 1/2 ⌣ f(x)
IP

Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
.

Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity

Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity

Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

monotonicity and concavity
II I
.
III IV

monotonicity and concavity
decreasing,
concave
down
II I
.
III IV

monotonicity and concavity
increasing, decreasing,
concave concave
down down
II I
.
III IV

monotonicity and concavity
increasing, decreasing,
concave concave
down down
II I
.
III IV
decreasing,
concave
up

monotonicity and concavity
increasing, decreasing,
concave concave
down down
II I
.
III IV
decreasing, increasing,
concave concave
up up

Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

Step 3: One sign chart to rule them all
Remember, f(x) = 2x3 − 3x2 − 12x.
+ −. − + f′ (x)
↗−1 ↘ ↘ 2 ↗ monotonicity
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

f(x)
Step 4: Graph
f(x) = 2x3 − 3x2 − 12x
( √ ) (−1, 7)
3− 105
4 ,0 (0, 0)
. ( x√ )
(1/2, −61/2)
3+ 105
4 ,0
(2, −20)
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

f(x)
Step 4: Graph
f(x) = 2x3 − 3x2 − 12x
( √ ) (−1, 7)
3− 105
4 ,0 (0, 0)
. ( x√ )
(1/2, −61/2)
3+ 105
4 ,0
(2, −20)
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

f(x)
Step 4: Graph
f(x) = 2x3 − 3x2 − 12x
( √ ) (−1, 7)
3− 105
4 ,0 (0, 0)
. ( x√ )
(1/2, −61/2)
3+ 105
4 ,0
(2, −20)
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

f(x)
Step 4: Graph
f(x) = 2x3 − 3x2 − 12x
( √ ) (−1, 7)
3− 105
4 ,0 (0, 0)
. ( x√ )
(1/2, −61/2)
3+ 105
4 ,0
(2, −20)
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

f(x)
Step 4: Graph
f(x) = 2x3 − 3x2 − 12x
( √ ) (−1, 7)
3− 105
4 ,0 (0, 0)
. ( x√ )
(1/2, −61/2)
3+ 105
4 ,0
(2, −20)
7 −61/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

Graphing a quartic
Example
Graph f(x) = x4 − 4x3 + 10

Graphing a quartic
Example
Graph f(x) = x4 − 4x3 + 10
(Step 0) We know f(0) = 10 and lim f(x) = +∞. Not too many
x→±∞
other points on the graph are evident.

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
.

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
0. 4x2
0

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0. 4x2
0

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0. +
4x2
0

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0. + +
4x2
0

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0. + +
4x2
0
0
(x − 3)
3

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− 0
(x − 3)
3

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0
(x − 3)
3

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 f′ (x)
0 0
0 3 f(x)

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 f′ (x)
−0 0
0 3 f(x)

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 f′ (x)
−0 − 0
0 3 f(x)

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 ′
−0 − 0 + f (x)
0 3 f(x)

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 ′
−0 − 0 + f (x)
↘0 3 f(x)

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 ′
−0 − 0 + f (x)
↘0 ↘ 3 f(x)

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 ′
−0 − 0 + f (x)
↘0 ↘ 3 ↗ f(x)

Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′ (x) = 4x3 − 12x2 = 4x2 (x − 3)
We make its sign chart.
+0 . + +
4x2
0
− − 0+
(x − 3)
3 ′
−0 − 0 + f (x)
↘0 ↘ 3 ↗ f(x)
min

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
.

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
.

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
0
. 12x
0

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0
. 12x
0

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0
. +
12x
0

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0
. + +
12x
0

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0
. + +
12x
0
0
x−2
2

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− 0
x−2
2

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− − 0
x−2
2

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− − 0 +
x−2
2

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− − 0 +
x−2
2 f′′ (x)
0 0
0 2 f(x)

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− − 0 +
x−2
2 f′′ (x)
++0 0
0 2 f(x)

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− −
0 +
x−2
2 f′′ (x)
++0 −− 0
0 2 f(x)

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− −
0 +
x−2
2 f′′ (x)
++0 −− 0 ++
0 2 f(x)

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− −
0 +
x−2
2 f′′ (x)
++0 −− 0 ++
⌣0 2 f(x)

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− −
0 +
x−2
2 f′′ (x)
++0 −− 0 ++
⌣0 ⌢ 2 f(x)

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− −
0 +
x−2
2 f′′ (x)
++0 −− 0 ++
⌣0 ⌢ 2 ⌣ f(x)

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− −
0 +
x−2
2 f′′ (x)
++0 −− 0 ++
⌣0 ⌢ 2 ⌣ f(x)
IP

Step 2: Concavity
f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
−0. + +
12x
0
− −0 +
x−2
2 f′′ (x)
++0 −− 0 ++
⌣0 ⌢ 2 ⌣ f(x)
IP IP

Step 3: Grand Unified Sign Chart
.
Remember, f(x) = x − 4x3 + 10.
4
−0 − −0+ f′ (x)
↘0 ↘ ↘3↗ monotonicity
f′′ (x)
++0 −− 0++ ++
⌣0 ⌢ 2⌣ ⌣ concavity
10 −6 −17 f(x)
0 2 3 shape
IP IP min

Step 3: Grand Unified Sign Chart
.
Remember, f(x) = x − 4x3 + 10.
4
−0 − −0+ f′ (x)
↘0 ↘ ↘3↗ monotonicity
f′′ (x)
++0 −− 0++ ++
⌣0 ⌢ 2⌣ ⌣ concavity
10 −6 −17 f(x)
0 2 3 shape
IP IP min

Step 3: Grand Unified Sign Chart
.
Remember, f(x) = x − 4x3 + 10.
4
−0 − −0+ f′ (x)
↘0 ↘ ↘3↗ monotonicity
f′′ (x)
++0 −− 0++ ++
⌣0 ⌢ 2⌣ ⌣ concavity
10 −6 −17 f(x)
0 2 3 shape
IP IP min

Step 3: Grand Unified Sign Chart
.
Remember, f(x) = x − 4x3 + 10.
4
−0 − −0+ f′ (x)
↘0 ↘ ↘3↗ monotonicity
f′′ (x)
++0 −− 0++ ++
⌣0 ⌢ 2⌣ ⌣ concavity
10 −6 −17 f(x)
0 2 3 shape
IP IP min

Step 3: Grand Unified Sign Chart
.
Remember, f(x) = x − 4x3 + 10.
4
−0 − −0+ f′ (x)
↘0 ↘ ↘3↗ monotonicity
f′′ (x)
++0 −− 0++ ++
⌣0 ⌢ 2⌣ ⌣ concavity
10 −6 −17 f(x)
0 2 3 shape
IP IP min

y
Step 4: Graph
f(x) = x4 − 4x3 + 10
(0, 10)
. x
(2, −6)
(3, −17)
10 −6 −17 f(x)
0 2 3 shape
IP IP min

y
Step 4: Graph
f(x) = x4 − 4x3 + 10
(0, 10)
. x
(2, −6)
(3, −17)
10 −6 −17 f(x)
0 2 3 shape
IP IP min

y
Step 4: Graph
f(x) = x4 − 4x3 + 10
(0, 10)
. x
(2, −6)
(3, −17)
10 −6 −17 f(x)
0 2 3 shape
IP IP min

y
Step 4: Graph
f(x) = x4 − 4x3 + 10
(0, 10)
. x
(2, −6)
(3, −17)
10 −6 −17 f(x)
0 2 3 shape
IP IP min

y
Step 4: Graph
f(x) = x4 − 4x3 + 10
(0, 10)
. x
(2, −6)
(3, −17)
10 −6 −17 f(x)
0 2 3 shape
IP IP min

Outline
Simple examples
A cubic func on
A quar c func on
More Examples
Points of nondifferen ability
Horizontal asymptotes
Ver cal asymptotes
Trigonometric and polynomial together
Logarithmic

Graphing a function with a cusp
Example
√
Graph f(x) = x + |x|

Graphing a function with a cusp
Example
√
Graph f(x) = x + |x|
This func on looks strange because of the absolute value. But
whenever we become nervous, we can just take cases.

Step 0: Finding Zeroes
√
f(x) = x + |x|
First, look at f by itself. We can tell that f(0) = 0 and that
f(x) > 0 if x is posi ve.

Step 0: Finding Zeroes
√
f(x) = x + |x|
First, look at f by itself. We can tell that f(0) = 0 and that
f(x) > 0 if x is posi ve.
Are there nega ve numbers which are zeroes for f?

Step 0: Finding Zeroes
√
f(x) = x + |x|
First, look at f by itself. We can tell that f(0) = 0 and that
f(x) > 0 if x is posi ve.
Are there nega ve numbers which are zeroes for f?
√ √
x + −x = 0 =⇒ −x = −x
−x = x2 =⇒ x2 + x = 0
The only solu ons are x = 0 and x = −1.

Step 0: Asymptotic behavior
√
f(x) = x + |x|
lim f(x) = ∞, because both terms tend to ∞.
x→∞

Step 0: Asymptotic behavior
√
f(x) = x + |x|
lim f(x) = ∞, because both terms tend to ∞.
x→∞
lim f(x) is indeterminate of the form −∞ + ∞. It’s the same
x→−∞ √
as lim (−y + y)
y→+∞

Step 0: Asymptotic behavior
√
f(x) = x + |x|
lim f(x) = ∞, because both terms tend to ∞.
x→∞
lim f(x) is indeterminate of the form −∞ + ∞. It’s the same
x→−∞ √
as lim (−y + y)
y→+∞
√
√ √ y+y
lim (−y + y) = lim ( y − y) · √
y→+∞ y→∞ y+y
y − y2
= lim √ = −∞
y→∞ y+y

Step 1: The derivative
√
Remember, f(x) = x + |x|.
To find f′ , first assume x > 0. Then
d ( √ ) 1
f′ (x) = x+ x =1+ √
dx 2 x

Step 1: The derivative
√
Remember, f(x) = x + |x|.
To find f′ , first assume x > 0. Then
d ( √ ) 1
f′ (x) = x+ x =1+ √
dx 2 x
No ce
f′ (x) > 0 when x > 0 (so no cri cal points here)

Step 1: The derivative
√
Remember, f(x) = x + |x|.
To find f′ , first assume x > 0. Then
d ( √ ) 1
f′ (x) = x+ x =1+ √
dx 2 x
No ce
f′ (x) > 0 when x > 0 (so no cri cal points here)
lim+ f′ (x) = ∞ (so 0 is a cri cal point)
x→0

Step 1: The derivative
√
Remember, f(x) = x + |x|.
To find f′ , first assume x > 0. Then
d ( √ ) 1
f′ (x) = x+ x =1+ √
dx 2 x
No ce
f′ (x) > 0 when x > 0 (so no cri cal points here)
lim+ f′ (x) = ∞ (so 0 is a cri cal point)
x→0
lim f′ (x) = 1 (so the graph is asympto c to a line of slope 1)
x→∞

Step 1: The derivative
√
Remember, f(x) = x + |x|.
If x is nega ve, we have
d ( √ ) 1
f′ (x) = x + −x = 1 − √
dx 2 −x
No ce
lim− f′ (x) = −∞ (other side of the cri cal point)
x→0

Step 1: The derivative
√
Remember, f(x) = x + |x|.
If x is nega ve, we have
d ( √ ) 1
f′ (x) = x + −x = 1 − √
dx 2 −x
No ce
lim− f′ (x) = −∞ (other side of the cri cal point)
x→0
lim f′ (x) = 1 (asympto c to a line of slope 1)
x→−∞

Step 1: The derivative
√
Remember, f(x) = x + |x|.
If x is nega ve, we have
d ( √ ) 1
f′ (x) = x + −x = 1 − √
dx 2 −x
No ce
lim− f′ (x) = −∞ (other side of the cri cal point)
x→0
lim f′ (x) = 1 (asympto c to a line of slope 1)
x→−∞
′
f (x) = 0 when
1 √ 1 1 1
1− √ = 0 =⇒ −x = =⇒ −x = =⇒ x = −
2 −x 2 4 4

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
f′ (x)
.
f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
0 f′ (x)
.
−1
4
f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
0 ∞ f′ (x)
.
−1
4
0 f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0 ∞ f′ (x)
.
−1
4
0 f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0− ∞ f′ (x)
.
−4 0
1 f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0− ∞ + f′ (x)
.
−4 0
1 f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0− ∞ + f′ (x)
.
↗ −4 0
1 f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0− ∞ + f′ (x)
.
↗ −4
1↘ 0 f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0− ∞ + f′ (x)
.
↗ −4
1↘ 0 ↗ f(x)

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0− ∞ + f′ (x)
.
↗ −41↘ 0 ↗ f(x)
max

Step 1: Monotonicity
 1
1 + √
 if x > 0
′
f (x) = 2 x
1 − √ 1
 if x < 0
2 −x
We can’t make a mul -factor sign chart because of the absolute
value, but we can test points in between cri cal points.
+ 0− ∞ + f′ (x)
.
↗ −41↘ 0 ↗ f(x)
max min

Step 2: Concavity
( )
d 1 1
If x > 0, then f′′ (x) = 1 + x−1/2 = − x−3/2 This is
dx 2 4
nega ve whenever x > 0.

Step 2: Concavity
( )
d 1 −1/2 1
If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is
dx 2 4
nega ve whenever x > 0. ( )
′′ d 1 −1/2 1
If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2
dx 2 4
which is also always nega ve for nega ve x.

Step 2: Concavity
( )
d 1 −1/2 1
If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is
dx 2 4
nega ve whenever x > 0. ( )
′′ d 1 −1/2 1
If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2
dx 2 4
which is also always nega ve for nega ve x.
1
In other words, f′′ (x) = − |x|−3/2 .
4

Step 2: Concavity
( )
d 1 −1/2 1
If x > 0, then f′′ (x) = 1+ x = − x−3/2 This is
dx 2 4
nega ve whenever x > 0. ( )
′′ d 1 −1/2 1
If x < 0, then f (x) = 1 − (−x) = − (−x)−3/2
dx 2 4
which is also always nega ve for nega ve x.
1
In other words, f′′ (x) = − |x|−3/2 .
4
Here is the sign chart:
−− −∞ −− f′′ (x)
.
⌢ 0 ⌢ f(x)

Step 3: Synthesis
Now we can put these things together.
√
f(x) = x + |x|
+1 + 0− ∞ + f′
+1 (x)
.
↗ ↗ 1↘ 0 ↗ ↗monotonicity
−∞ −− − −−
4 −∞ −− f′′
−∞ (x)
⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity
−∞ 0 4 0 +∞f(x)
−1 −4 0
1 shape
zero max min

Step 3: Synthesis
Now we can put these things together.
√
f(x) = x + |x|
+1 + 0− ∞ + f′
+1 (x)
.
↗ ↗ 1↘ 0 ↗ ↗monotonicity
−∞ −− − −−
4 −∞ −− f′′
−∞ (x)
⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity
−∞ 0 4 0 +∞f(x)
−1 −4 0
1 shape
zero max min

Step 3: Synthesis
Now we can put these things together.
√
f(x) = x + |x|
+1 + 0− ∞ + f′
+1 (x)
.
↗ ↗ 1↘ 0 ↗ ↗monotonicity
−∞ −− − −−
4 −∞ −− f′′
−∞ (x)
⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity
−∞ 0 4 0 +∞f(x)
−1 −4 0
1 shape
zero max min

Step 3: Synthesis
Now we can put these things together.
√
f(x) = x + |x|
+1 + 0− ∞ + f′
+1 (x)
.
↗ ↗ 1↘ 0 ↗ ↗monotonicity
−∞ −− − −−
4 −∞ −− f′′
−∞ (x)
⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity
−∞ 0 4 0 +∞f(x)
−1 −4 0
1 shape
zero max min

Step 3: Synthesis
Now we can put these things together.
√
f(x) = x + |x|
+1 + 0− ∞ + f′
+1 (x)
.
↗ ↗ 1↘ 0 ↗ ↗monotonicity
−∞ −− − −−
4 −∞ −− f′′
−∞ (x)
⌢ ⌢ 1 ⌢0 ⌢ ⌢concavity
−∞ 0 4 0 +∞f(x)
−1 −4 0
1 shape
zero max min

Graph
√
f(x) = x + |x|
. x
1
−∞0 4 0 +∞ x
−1 −1 0 shape
4
zero max min

Graph
√
f(x) = x + |x|
(−1, 0)
. x
1
−∞0 4 0 +∞ x
−1 −1 0 shape
4
zero max min

Graph
√
f(x) = x + |x|
(− 1 , 1 )
4 4
(−1, 0)
. x
1
−∞0 4 0 +∞ x
−1 −1 0 shape
4
zero max min

Graph
√
f(x) = x + |x|
(− 1 , 1 )
4 4
(−1, 0)
. x
(0, 0)
1
−∞0 4 0 +∞ x
−1 −1 0 shape
4
zero max min

Graph
√
f(x) = x + |x|
(− 1 , 1 )
4 4
(−1, 0)
. x
(0, 0)
1
−∞0 4 0 +∞ x
−1 −1 0 shape
4
zero max min

Graph
√
f(x) = x + |x|
(− 1 , 1 )
4 4
(−1, 0)
. x
(0, 0)
1
−∞0 4 0 +∞ x
−1 −1 0 shape
4
zero max min

Graph
√
f(x) = x + |x|
(− 1 , 1 )
4 4
(−1, 0)
. x
(0, 0)
1
−∞0 4 0 +∞ x
−1 −1 0 shape
4
zero max min

Example with Horizontal
Asymptotes
Example
Graph f(x) = xe−x
2

Example with Horizontal
Asymptotes
Example
Graph f(x) = xe−x
2
Before taking deriva ves, we no ce that f is odd, that f(0) = 0, and
lim f(x) = 0
x→∞

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
. √0 √
1− 2x
0 1/2
√
√ 1 + 2x
− 0 1/2 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ . √0 √
1− 2x
0 1/2
√
√ 1 + 2x
− 0 1/2 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ √
1− 2x
0 1/2
√
√ 1 + 2x
− 0 1/2 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
0 1/2
√
√ 1 + 2x
− 0 1/2 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 1/2
√
√ 1 + 2x
− 0 1/2 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
√
√ 1 + 2x
− 0 1/2 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− 0 1/2 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− − 0 1/2 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− − 0 1/2 + 0 f′ (x)
√ √
− 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− − 0 1/2 + 0 − f′ (x)
√ √
↘ − 1/2 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− − 0 1/2 + 0 − f′ (x)
√ √
↘ − 1/2 ↗ 1/2 f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− − 0 1/2 + 0 − f′ (x)
√ √
↘ − 1/2 ↗ 1/2 ↘ f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− − 0 1/2 + 0 − f′ (x)
√ √
↘ − 1/2 ↗ 1/2 ↘ f(x)

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− − 0 1/2 + 0 − f′ (x)
√ √
↘ − 1/2 ↗ 1/2 ↘ f(x)
min

Step 1: −xMonotonicity
2
If f(x) = xe , then
( ) 2
f′ (x) = 1 · e−x + xe−x (−2x) = 1 − 2x2 e−x
2 2
( √ )( √ ) −x2
= 1 − 2x 1 + 2x e
+ + 0
. √ − √
1− 2x
− 0 + 1/2
+ √
√ 1 + 2x
− − 0 1/2 + 0 − f′ (x)
√ √
↘ − 1/2 ↗ 1/2 ↘ f(x)
min max

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
0. 2x
0
0 √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− 0. 2x
0
0 √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. 2x
0
0 √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. +
2x
0
0 √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
0 √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− 0 √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − 0 √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + 3/2
√ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 √ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
− 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 f′′ (x)
√ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 ++ f′′ (x)
⌢ √ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 ++ f′′ (x)
⌢ √ ⌣ √
− 3/2 0 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 ++ f′′ (x)
⌢ √ ⌣ √
− 3/2 0 ⌢ 3/2 f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 ++ f′′ (x)
⌢ √ ⌣ √
− 3/2 0 ⌢ 3/2 ⌣ f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 ++ f′′ (x)
⌢ √ ⌣ √
− 3/2 0 ⌢ 3/2 ⌣ f(x)

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 ++ f′′ (x)
⌢ √ ⌣ √
− 3/2 0 ⌢ 3/2 ⌣ f(x)
IP

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 ++ f′′ (x)
⌢ √ ⌣ √
− 3/2 0 ⌢ 3/2 ⌣ f(x)
IP IP

Step 2: Concavity
′ 2 −x 2
If f (x) = (1 − 2x )e , we know
( ) 2
f′′ (x) = (−4x)e−x + (1 − 2x2 )e−x (−2x) = 4x3 − 6x e−x
2 2
= 2x(2x2 − 3)e−x
2
− − 0. + +
2x
0
− − − 0 + √ √
√ 2x − 3
− 0 + + 3/2 + √ √
√ 2x + 3
−− − 0 3/2
++ 0 −− 0 ++ f′′ (x)
⌢ √ ⌣ √
− 3/2 0 ⌢ 3/2 ⌣ f(x)
IP IP IP