Lesson 21: Curve Sketching (Section 041 slides)

.

Section 4.4
Curve Sketching

V63.0121.041, Calculus I

New York University

November 17, 2010

Announcements
Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7
There is class on November 24

. . . . . .

Announcements

Quiz 4 this week in
recitation on 3.3, 3.4, 3.5,
3.7
There is class on
November 24

. . . . . .

V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 2 / 55

Objectives

given a function, graph it
completely, indicating
zeroes (if easy)
asymptotes if applicable
critical points
local/global max/min
inflection points

. . . . . .


Why?

Graphing functions is like
dissection

. . . . . .


Why?

dissection … or diagramming
sentences

. . . . . .


Why?

dissection … or diagramming
sentences
You can really know a lot about
a function 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
Here f(x) = x3 + x2 , and f′ (x) = 3x2 + 2x.

f(x)
f′ (x)

.

. . . . . .


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
Here f(x) = x3 + x2 , f′ (x) = 3x2 + 2x, and f′′ (x) = 6x + 2.
f′′ (x) f(x)
f′ (x)

.

. . . . . .


Graphing Checklist

To graph a function f, follow this plan:
0. Find when f is positive, negative, zero,
not defined.
1. Find f′ and form its sign chart. Conclude
information about increasing/decreasing
and local max/min.
2. Find f′′ and form its sign chart. Conclude
concave up/concave down and inflection.
3. Put together a big chart to assemble
monotonicity and concavity data
4. Graph!

. . . . . .


Outline

Simple examples
A cubic function
A quartic function

More Examples
Points of nondifferentiability
Horizontal asymptotes
Vertical 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 quadratic, 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:

.

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


. x−2
2
x+1
−1
f′ (x)
−1 2 f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
x+1
−1
f′ (x)
−1 2 f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
− + +
x+1
−1
f′ (x)
−1 2 f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
− + +
x+1
−1
+ f′ (x)
−1 2 f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
− + +
x+1
−1
+ − f′ (x)
−1 2 f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
− + +
x+1
−1
+ − + f′ (x)
−1 2 f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
− + +
x+1
−1
+ − + f′ (x)
↗ −1 2 f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
− + +
x+1
−1
+ − + f′ (x)
↗ −1 ↘ 2 f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
− + +
x+1
−1
+ − + f′ (x)
↗ −1 ↘ 2 ↗ f(x)

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
x−2
2
− + +
x+1
−1
+ − + f′ (x)
↗ −1 ↘ 2 ↗ f(x)
max

. . . . . .



f(x) = 2x3 − 3x2 − 12x
=⇒ f′ (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2)


− . − +
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)

f′′ (x)
1/2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)


−− f′′ (x)
1/2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)


−− ++ f′′ (x)
1/2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)


−− ++ f′′ (x)
⌢ 1/2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)


−− ++ f′′ (x)
⌢ 1/2 ⌣ f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 6x2 − 6x − 12
=⇒ f′′ (x) = 12x − 6 = 6(2x − 1)


−− ++ f′′ (x)
⌢ 1/2 ⌣ f(x)
IP

. . . . . .


Step 3: One sign chart to rule them all

Remember, f(x) = 2x3 − 3x2 − 12x.

.

. . . . . .



Remember, f(x) = 2x3 − 3x2 − 12x.

+ −. − + f′ (x)
↗ −1 ↘ ↘ 2 ↗ monotonicity

. . . . . .



Remember, f(x) = 2x3 − 3x2 − 12x.

+ −. − + f′ (x)
−− −− ++ ++ f′′ (x)
⌢ ⌢ 1/2 ⌣ ⌣ concavity

. . . . . .



Remember, f(x) = 2x3 − 3x2 − 12x.

+ −. − + f′ (x)
−− −− ++ ++ f′′ (x)
7 −6 1/2 −20 f(x)
−1 1/2 2 shape of f
max IP min

. . . . . .


Combinations of monotonicity and concavity

II I

.

III IV

. . . . . .



decreasing,
concave
down

II I

.

III IV

. . . . . .



increasing, decreasing,
concave concave
down down

II I

.

III IV

. . . . . .



concave concave
down down

II I

.

III IV

decreasing,
concave up
. . . . . .



concave concave
down down

II I

.

III IV

decreasing, increasing,
concave up concave up
. . . . . .



Remember, f(x) = 2x3 − 3x2 − 12x.

+ −. − + f′ (x)
−− −− ++ ++ f′′ (x)
7 −6 1/2 −20 f(x)
max IP min

. . . . . .


Step 4: Graph
f(x)

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)
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 other
x→±∞
points on the graph are evident.

. . . . . .



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

. . . . . .



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

We make its sign chart.

.

. . . . . .



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


0
. 4x2
0

. . . . . .



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


+ 0
. 4x2
0

. . . . . .



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


+ 0
. +
4x2
0

. . . . . .



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


+ 0
. + +
4x2
0

. . . . . .



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


+ 0
. + +
4x2
0
0
(x − 3)
3

. . . . . .



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


+ 0
. + +
4x2
0
− 0
(x − 3)
3

. . . . . .



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


+ 0
. + +
4x2
0
− − 0
(x − 3)
3

. . . . . .



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


+ 0
. + +
4x2
0
− − 0 +
(x − 3)
3

. . . . . .



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


+ 0
. + +
4x2
0
− − 0 +
(x − 3)
3
0 0 f′ (x)
0 3 f(x)

. . . . . .



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


+ 0
. + +
4x2
0
− − 0 +
(x − 3)
3
− 0 0 f′ (x)
0 3 f(x)

. . . . . .



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


+ 0
. + +
4x2
0
− − 0 +
(x − 3)
3
− 0 − 0 f′ (x)
0 3 f(x)

. . . . . .



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


+ 0
. + +
4x2
0
− − 0 +
(x − 3)
3
− 0 − 0 + f′ (x)
0 3 f(x)

. . . . . .



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


+ 0
. + +
4x2
0
− − 0 +
(x − 3)
3
− 0 − 0 + f′ (x)
↘ 0 3 f(x)

. . . . . .



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


+ 0
. + +
4x2
0
− − 0 +
(x − 3)
3
− 0 − 0 + f′ (x)
↘ 0 ↘ 3 f(x)

. . . . . .



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


+ 0
. + +
4x2
0
− − 0 +
(x − 3)
3
− 0 − 0 + f′ (x)
↘ 0 ↘ 3 ↗ f(x)

. . . . . .



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


+ 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)


0
. 12x
0

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. 12x
0

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. +
12x
0

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
0
x−2
2

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− 0
x−2
2

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0
x−2
2

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
0 0 f′′ (x)
0 2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
++ 0 0 f′′ (x)
0 2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
++ 0 −− 0 f′′ (x)
0 2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
++ 0 −− 0 ++ f′′ (x)
0 2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
++ 0 −− 0 ++ f′′ (x)
⌣ 0 2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
++ 0 −− 0 ++ f′′ (x)
⌣ 0 ⌢ 2 f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
++ 0 −− 0 ++ f′′ (x)
⌣ 0 ⌢ 2 ⌣ f(x)

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
++ 0 −− 0 ++ f′′ (x)
⌣ 0 ⌢ 2 ⌣ f(x)
IP

. . . . . .


Step 2: Concavity

f′ (x) = 4x3 − 12x2
=⇒ f′′ (x) = 12x2 − 24x = 12x(x − 2)


− 0
. + +
12x
0
− − 0 +
x−2
2
++ 0 −− 0 ++ f′′ (x)
⌣ 0 ⌢ 2 ⌣ f(x)
IP IP

. . . . . .


Step 3: Grand Unified Sign Chart

.

Remember, f(x) = x4 − 4x3 + 10.

− 0 − − 0 + f′ (x)
↘ 0 ↘ ↘ 3 ↗ monotonicity
++ 0 −− 0 ++ ++ f′′ (x)
⌣ 0 ⌢ 2 ⌣ ⌣ concavity
10 −6 −17 f(x)
0 2 3 shape
IP IP min

. . . . . .


Step 4: Graph
y

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 function
A quartic function

More Examples
Points of nondifferentiability
Horizontal asymptotes
Vertical 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 function 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 positive.

. . . . . .



√
f(x) = x + |x|
x is positive.
Are there negative numbers which are zeroes for f?

. . . . . .



√
f(x) = x + |x|
x is positive.
Are there negative numbers which are zeroes for f?
√
x + −x = 0
√
−x = −x
−x = x2
x2 + x = 0

The only solutions are x = 0 and x = −1.

. . . . . .


Step 0: Asymptotic behavior

√
f(x) = x + |x|
lim f(x) = ∞, because both terms tend to ∞.
x→∞

. . . . . .



√
f(x) = x + |x|
x→∞
lim f(x) is indeterminate of the form −∞ + ∞. It’s the same as
x→−∞ √
lim (−y + y)
y→+∞

. . . . . .



√
f(x) = x + |x|
x→∞
lim f(x) is indeterminate of the form −∞ + ∞. It’s the same as
x→−∞ √
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

. . . . . .



√
d ( √ ) 1
f′ (x) = x+ x =1+ √
dx 2 x

Notice
f′ (x) > 0 when x > 0 (so no critical points here)

. . . . . .



√
d ( √ ) 1
f′ (x) = x+ x =1+ √
dx 2 x

Notice
lim f′ (x) = ∞ (so 0 is a critical point)
x→0+

. . . . . .



√
d ( √ ) 1
f′ (x) = x+ x =1+ √
dx 2 x

Notice
lim f′ (x) = ∞ (so 0 is a critical point)
x→0+
lim f′ (x) = 1 (so the graph is asymptotic to a line of slope 1)
x→∞

. . . . . .


√
If x is negative, we have

d ( √ ) 1
f′ (x) = x + −x = 1 − √
dx 2 −x

Notice
lim f′ (x) = −∞ (other side of the critical point)
x→0−

. . . . . .


√

d ( √ ) 1
f′ (x) = x + −x = 1 − √
dx 2 −x

Notice
x→0−
lim f′ (x) = 1 (asymptotic to a line of slope 1)
x→−∞

. . . . . .


√

d ( √ ) 1
f′ (x) = x + −x = 1 − √
dx 2 −x

Notice
x→0−
lim f′ (x) = 1 (asymptotic 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

. . . . . .



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

f′ (x)
.
f(x)

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

0 f′ (x)
.
−1
4
f(x)

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

0 ∞ f′ (x)
.
−1
4
0 f(x)

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

+ 0 ∞ f′ (x)
.
−1
4
0 f(x)

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

+ 0 − ∞ f′ (x)
.
−1 4
0 f(x)

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

+ 0 − ∞ + f′ (x)
.
−1 4
0 f(x)

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

+ 0 − ∞ + f′ (x)
.
↗ −1 0 f(x)
4

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

+ 0 − ∞ + f′ (x)
.
↗ −1 ↘ 0 f(x)
4

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

+ 0 − ∞ + f′ (x)
.
↗ −1 ↘ 0 ↗ f(x)
4

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

+ 0 − ∞ + f′ (x)
.
↗ −1 ↘ 0 ↗ f(x)
4
max

. . . . . .



 1
1 + √
 if x > 0
f′ (x) = 2 x
1 − √
 1
if x < 0
2 −x

+ 0 − ∞ + f′ (x)
.
↗ −1 ↘ 0 ↗ f(x)
4
max min

. . . . . .


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

. . . . . .


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

. . . . . .


Lesson 21: Curve Sketching (Section 041 slides)

Lesson 21: Curve Sketching (Section 041 slides)

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