The document provides information about an upcoming calculus class, including announcements of an upcoming quiz and that class will be held on November 24. It then outlines the objectives and importance of curve sketching functions by analyzing properties like critical points, maxima/minima, and inflection points. The remainder of the document provides examples and steps for sketching curves of specific cubic and quartic functions through analyzing monotonicity using the first derivative and concavity using the second derivative.
College algebra in context 5th edition harshbarger solutions manualAnnuzzi19
College Algebra in Context 5th Edition Harshbarger Solutions Manual
Full download:https://goo.gl/eJxUd2
People also search:
college algebra in context with applications for the managerial life and social sciences
pearson
chegg
College algebra in context 5th edition harshbarger solutions manualAnnuzzi19
College Algebra in Context 5th Edition Harshbarger Solutions Manual
Full download:https://goo.gl/eJxUd2
People also search:
college algebra in context with applications for the managerial life and social sciences
pearson
chegg
In this paper we proved some new theorems related with Cubic Harmonious Labeling. A (n,m) graph G =(V,E) is said to be Cubic Harmonious Graph(CHG) if there exists an injective function f:V(G)?{1,2,3,………m3+1} such that the induced mapping f *chg: E(G)? {13,23,33,……….m3} defined by f *chg (uv) = (f(u)+f(v)) mod (m3+1) is a bijection. In this paper, focus will be given on the result “cubic harmonious labeling of star, the subdivision of the edges of the star K1,n , the subdivision of the central edge of the bistar Bm,n, Pm ? nK1”.
In this paper we proved some new theorems related with Cubic Harmonious Labeling. A (n,m) graph G =(V,E) is said to be Cubic Harmonious Graph(CHG) if there exists an injective function f:V(G)?{1,2,3,………m3+1} such that the induced mapping f *chg: E(G)? {13,23,33,……….m3} defined by f *chg (uv) = (f(u)+f(v)) mod (m3+1) is a bijection. In this paper, focus will be given on the result “cubic harmonious labeling of star, the subdivision of the edges of the star K1,n , the subdivision of the central edge of the bistar Bm,n, Pm ? nK1”.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few more good examples.
This learner's module discusses or talks about the topic of Quadratic Functions. It also discusses what is Quadratic Functions. It also shows how to transform or rewrite the equation f(x)=ax2 + bx + c to f(x)= a(x-h)2 + k. It will also show the different characteristics of Quadratic Functions.
Lesson 19: The Mean Value Theorem (Section 021 slides)Matthew Leingang
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 041 ...Matthew Leingang
The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
Auto-multiple-choice (AMC) is an open-source optical mark recognition software package built with Perl, LaTeX, XML, and sqlite. I use it for all my in-class quizzes and exams. Unique papers are created for each student, fixed-response items are scored automatically, and free-response problems, after manual scoring, have marks recorded in the same process. In the first part of the talk I will discuss AMC’s many features and why I feel it’s ideal for a mathematics course. My contributions to the AMC workflow include some scripts designed to automate the process of returning scored papers
back to students electronically. AMC provides an email gateway, but I have written programs to return graded papers via the DAV protocol to student’s dropboxes on our (Sakai) learning management systems. I will also show how graded papers can be archived, with appropriate metadata tags, into an Evernote notebook.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Lesson 24: Areas and Distances, The Definite Integral (handout)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 24: Areas and Distances, The Definite Integral (slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
1. ..
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
. . . . . .
2. . . . . . .
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
3. . . . . . .
Objectives
given a function, graph it
completely, indicating
zeroes (if easy)
asymptotes if applicable
critical points
local/global max/min
inflection points
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 3 / 55
4. . . . . . .
Why?
Graphing functions is like
dissection
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
5. . . . . . .
Why?
Graphing functions is like
dissection … or diagramming
sentences
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
6. . . . . . .
Why?
Graphing functions is like
dissection … or diagramming
sentences
You can really know a lot about
a function when you know all of
its anatomy.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
7. . . . . . .
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)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 5 / 55
8. . . . . . .
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)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 6 / 55
9. . . . . . .
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!
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 7 / 55
10. . . . . . .
Outline
Simple examples
A cubic function
A quartic function
More Examples
Points of nondifferentiability
Horizontal asymptotes
Vertical asymptotes
Trigonometric and polynomial together
Logarithmic
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 8 / 55
11. . . . . . .
Graphing a cubic
Example
Graph f(x) = 2x3
− 3x2
− 12x.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 9 / 55
12. . . . . . .
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
x =
3 ±
√
32
− 4(2)(−12)
4
=
3 ±
√
105
4
It’s OK to skip this step for now since the roots are so complicated.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 9 / 55
13. . . . . . .
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:
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
14. . . . . . .
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)
.
f(x)
..
2
..
−1
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
15. . . . . . .
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)
.
f(x)
..
2
..
−1
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
16. . . . . . .
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)
.
f(x)
..
2
..
−1
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
17. . . . . . .
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)
.
f(x)
..
2
..
−1
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
18. . . . . . .
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)
.
f(x)
..
2
..
−1
.
+
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
19. . . . . . .
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)
.
f(x)
..
2
..
−1
.
+
.
−
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
35. . . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 12 / 55
36. . . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
37. . . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,
concave
down
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
38. . . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,
concave
down
.
increasing,
concave
down
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
39. . . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,
concave
down
.
increasing,
concave
down
.
decreasing,
concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
40. . . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,
concave
down
.
increasing,
concave
down
.
decreasing,
concave up
.
increasing,
concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
41. . . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
42. . . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
..
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
43. . . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
...
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
44. . . . . . .
Step 3: One sign chart to rule them all
Remember, f(x) = 2x3
− 3x2
− 12x.
..
f′
(x)
.
monotonicity
..
−1
..
2
. +.
↗
.− .
↘
. −.
↘
.+ .
↗
.
f′′
(x)
.
concavity
..
1/2
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
....
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 14 / 55
50. . . . . . .
Graphing a quartic
Example
Graph f(x) = x4
− 4x3
+ 10
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 16 / 55
51. . . . . . .
Graphing a quartic
Example
Graph f(x) = x4
− 4x3
+ 10
(Step 0) We know f(0) = 10 and lim
x→±∞
f(x) = +∞. Not too many other
points on the graph are evident.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 16 / 55
99. . . . . . .
Outline
Simple examples
A cubic function
A quartic function
More Examples
Points of nondifferentiability
Horizontal asymptotes
Vertical asymptotes
Trigonometric and polynomial together
Logarithmic
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 21 / 55
100. . . . . . .
Graphing a function with a cusp
Example
Graph f(x) = x +
√
|x|
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 22 / 55
101. . . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 22 / 55
102. . . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
103. . . . . . .
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.
Are there negative numbers which are zeroes for f?
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
104. . . . . . .
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.
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.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 23 / 55
105. . . . . . .
Step 0: Asymptotic behavior
f(x) = x +
√
|x|
lim
x→∞
f(x) = ∞, because both terms tend to ∞.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
106. . . . . . .
Step 0: Asymptotic behavior
f(x) = x +
√
|x|
lim
x→∞
f(x) = ∞, because both terms tend to ∞.
lim
x→−∞
f(x) is indeterminate of the form −∞ + ∞. It’s the same as
lim
y→+∞
(−y +
√
y)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
107. . . . . . .
Step 0: Asymptotic behavior
f(x) = x +
√
|x|
lim
x→∞
f(x) = ∞, because both terms tend to ∞.
lim
x→−∞
f(x) is indeterminate of the form −∞ + ∞. It’s the same as
lim
y→+∞
(−y +
√
y)
lim
y→+∞
(−y +
√
y) = lim
y→∞
(
√
y − y) ·
√
y + y
√
y + y
= lim
y→∞
y − y2
√
y + y
= −∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
108. . . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
To find f′
, first assume x > 0. Then
f′
(x) =
d
dx
(
x +
√
x
)
= 1 +
1
2
√
x
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
109. . . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
To find f′
, first assume x > 0. Then
f′
(x) =
d
dx
(
x +
√
x
)
= 1 +
1
2
√
x
Notice
f′
(x) > 0 when x > 0 (so no critical points here)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
110. . . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
To find f′
, first assume x > 0. Then
f′
(x) =
d
dx
(
x +
√
x
)
= 1 +
1
2
√
x
Notice
f′
(x) > 0 when x > 0 (so no critical points here)
lim
x→0+
f′
(x) = ∞ (so 0 is a critical point)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
111. . . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
To find f′
, first assume x > 0. Then
f′
(x) =
d
dx
(
x +
√
x
)
= 1 +
1
2
√
x
Notice
f′
(x) > 0 when x > 0 (so no critical points here)
lim
x→0+
f′
(x) = ∞ (so 0 is a critical point)
lim
x→∞
f′
(x) = 1 (so the graph is asymptotic to a line of slope 1)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 25 / 55
112. . . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
If x is negative, we have
f′
(x) =
d
dx
(
x +
√
−x
)
= 1 −
1
2
√
−x
Notice
lim
x→0−
f′
(x) = −∞ (other side of the critical point)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
113. . . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
If x is negative, we have
f′
(x) =
d
dx
(
x +
√
−x
)
= 1 −
1
2
√
−x
Notice
lim
x→0−
f′
(x) = −∞ (other side of the critical point)
lim
x→−∞
f′
(x) = 1 (asymptotic to a line of slope 1)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
114. . . . . . .
Step 1: The derivative
Remember, f(x) = x +
√
|x|.
If x is negative, we have
f′
(x) =
d
dx
(
x +
√
−x
)
= 1 −
1
2
√
−x
Notice
lim
x→0−
f′
(x) = −∞ (other side of the critical point)
lim
x→−∞
f′
(x) = 1 (asymptotic to a line of slope 1)
f′
(x) = 0 when
1 −
1
2
√
−x
= 0 =⇒
√
−x =
1
2
=⇒ −x =
1
4
=⇒ x = −
1
4
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
115. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
116. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
117. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
118. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞.+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
119. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞.+ .−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
120. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞.+ .− . +
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
121. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞.+ .− . +.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
122. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞.+ .− . +.
↗
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
123. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞.+ .− . +.
↗
.
↘
.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
124. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞.+ .− . +.
↗
.
↘
.
↗
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
125. . . . . . .
Step 1: Monotonicity
f′
(x) =
1 +
1
2
√
x
if x > 0
1 −
1
2
√
−x
if x < 0
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
4
.0 ..
0
.∞.+ .− . +.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 27 / 55
126. . . . . . .
Step 2: Concavity
If x > 0, then
f′′
(x) =
d
dx
(
1 +
1
2
x−1/2
)
= −
1
4
x−3/2
This is negative whenever x > 0.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
127. . . . . . .
Step 2: Concavity
If x > 0, then
f′′
(x) =
d
dx
(
1 +
1
2
x−1/2
)
= −
1
4
x−3/2
This is negative whenever x > 0.
If x < 0, then
f′′
(x) =
d
dx
(
1 −
1
2
(−x)−1/2
)
= −
1
4
(−x)−3/2
which is also always negative for negative x.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
128. . . . . . .
Step 2: Concavity
If x > 0, then
f′′
(x) =
d
dx
(
1 +
1
2
x−1/2
)
= −
1
4
x−3/2
This is negative whenever x > 0.
If x < 0, then
f′′
(x) =
d
dx
(
1 −
1
2
(−x)−1/2
)
= −
1
4
(−x)−3/2
which is also always negative for negative x.
In other words, f′′
(x) = −
1
4
|x|−3/2
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
129. . . . . . .
Step 2: Concavity
If x > 0, then
f′′
(x) =
d
dx
(
1 +
1
2
x−1/2
)
= −
1
4
x−3/2
This is negative whenever x > 0.
If x < 0, then
f′′
(x) =
d
dx
(
1 −
1
2
(−x)−1/2
)
= −
1
4
(−x)−3/2
which is also always negative for negative x.
In other words, f′′
(x) = −
1
4
|x|−3/2
.
Here is the sign chart:
..
f′′
(x)
.
f(x)
..
0
.−∞.−− .
⌢
... −−.
⌢
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 28 / 55
130. . . . . . .
Step 3: Synthesis
Now we can put these things together.
f(x) = x +
√
|x|
..
f′
(x)
.
monotonicity
..
−1
4
.0 ..
0
.∞.+1 .
↗
.+ .
↗
.− .
↘
. +.
↗
. +1.
↗
.
f′′
(x)
.
concavity
..
0
.
−∞
.
−−
.
⌢
.
−−
.
⌢
.
−−
.
⌢
.
−∞
.
⌢
.
−∞
.
⌢
.
f(x)
.
shape
..
−1
.
0
.
zero
..
−1
4
.
1
4
.
max
..
0
.
0
.
min
.
−∞
.
+∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
142. . . . . . .
Example with Horizontal Asymptotes
Example
Graph f(x) = xe−x2
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 31 / 55
143. . . . . . .
Example with Horizontal Asymptotes
Example
Graph f(x) = xe−x2
Before taking derivatives, we notice that f is odd, that f(0) = 0, and
lim
x→∞
f(x) = 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 31 / 55
144. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.
1 +
√
2x
..
−
√
1/2
.
0
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
145. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .
1 +
√
2x
..
−
√
1/2
.
0
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
146. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+.
1 +
√
2x
..
−
√
1/2
.
0
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
147. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
148. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
149. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
150. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
151. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
152. . . . . . .
Step 1: Monotonicity
If f(x) = xe−x2
, then
f′
(x) = 1 · e−x2
+ xe−x2
(−2x) =
(
1 − 2x2
)
e−x2
=
(
1 −
√
2x
) (
1 +
√
2x
)
e−x2
The factor e−x2
is always positive so it doesn’t figure into the sign of
f′
(x). So our sign chart looks like this:
.. 1 −
√
2x.. √
1/2
. 0.+ .+. −.
1 +
√
2x
..
−
√
1/2
.
0
.
−
.
+
.
+
.
f′
(x)
.
f(x)
.
−
.
+
..
−
√
1/2
.
0
..
√
1/2
.
0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
190. . . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
191. . . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
192. . . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
193. . . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
194. . . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
195. . . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
196. . . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(
−
√
1/2, − 1√
2e
)
..
(√
1/2, 1√
2e
)
..
(
−
√
3/2, −
√
3
2e3
)
..
(0, 0)
..
(
√
3/2,
√
3
2e3
)
.
f(x)
.
shape
..
−
√
1/2
.
− 1√
2e
.
min
.. √
1/2
.
1√
2e
.
max
..
−
√
3/2
.
−
√
3
2e3
.
IP
..
0
.0.
IP
.. √
3/2
.
√
3
2e3
.
IP
......
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
197. . . . . . .
Example with Vertical Asymptotes
Example
Graph f(x) =
1
x
+
1
x2
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 36 / 55
198. . . . . . .
Step 0
Find when f is positive, negative, zero, not defined.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
199. . . . . . .
Step 0
Find when f is positive, negative, zero, not defined. We need to factor f:
f(x) =
1
x
+
1
x2
=
x + 1
x2
.
This means f is 0 at −1 and has trouble at 0. In fact,
lim
x→0
x + 1
x2
= ∞,
so x = 0 is a vertical asymptote of the graph.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
200. . . . . . .
Step 0
Find when f is positive, negative, zero, not defined. We need to factor f:
f(x) =
1
x
+
1
x2
=
x + 1
x2
.
This means f is 0 at −1 and has trouble at 0. In fact,
lim
x→0
x + 1
x2
= ∞,
so x = 0 is a vertical asymptote of the graph. We can make a sign
chart as follows:
.. x + 1..0 .
−1
.− . +.
x2
..
0
.
0
.
+
.
+
.
f(x)
..
∞
.
0
..
0
.
−1
.
−
.
+
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 37 / 55
201. . . . . . .
Step 0, continued
For horizontal asymptotes, notice that
lim
x→∞
x + 1
x2
= 0,
so y = 0 is a horizontal asymptote of the graph. The same is true at
−∞.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 38 / 55
202. . . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
203. . . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
204. . . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
205. . . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
206. . . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
207. . . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
208. . . . . . .
Step 1: Monotonicity
We have
f′
(x) = −
1
x2
−
2
x3
= −
x + 2
x3
.
The critical points are x = −2 and x = 0. We have the following sign
chart:
.. −(x + 2)..0 .
−2
.+ . −.
x3
..
0
.
0
.
−
.
+
.
f′
(x)
.
f(x)
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
min
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 39 / 55
230. . . . . . .
Step 4: Graph
.. x.
y
..
(−3, −2/9)
..
(−2, −1/4)
.
f
.
shape of f
..
∞
.
0
..
0
.
−1
..
−2
.
−1/4
..
−3
.
−2/9
.
−∞
.
0
.
∞
.
0
.
−
.
+
.
+
.
HA
..
IP
..
min
..
0
..
VA
..
HA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 42 / 55
231. . . . . . .
Trigonometric and polynomial together
Problem
Graph f(x) = cos x − x
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 43 / 55
232. . . . . . .
Step 0: intercepts and asymptotes
f(0) = 1 and f(−π/2) = −π/2. So by the Intermediate Value
Theorem there is a zero in between. We don’t know it’s precise
value, though.
Since −1 ≤ cos x ≤ 1 for all x, we have
−1 − x ≤ cos x − x ≤ 1 − x
for all x. This means that lim
x→∞
f(x) = −∞ and lim
x→−∞
f(x) = ∞.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 44 / 55
233. . . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
234. . . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
235. . . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −. −
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
236. . . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
237. . . . . . .
Step 1: Monotonicity
If f(x) = cos x − x, then f′
(x) = − sin x − 1 = (−1)(sin x + 1).
f′
(x) = 0 if x = 3π/2 + 2πk, where k is any integer
f′
(x) is periodic with period 2π
Since −1 ≤ sin x ≤ 1 for all x, we have
0 ≤ sin x + 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x + 1) ≤ 0
for all x. This means f′
(x) is negative at all other points.
..
f′
(x)
.
f(x)
..
−π/2
.0 ..
3π/2
. 0..
7π/2
. 0. −.
↘
. −.
↘
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 45 / 55
238. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
239. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
240. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
241. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−. ++. −−..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
242. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−. ++. −−. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
243. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++. −−. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
244. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
245. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
246. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 ..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
247. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
248. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0.
IP
..
3π/2
. 0..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
249. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0.
IP
..
3π/2
. 0.
IP
..
5π/2
. 0..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
250. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0.
IP
..
3π/2
. 0.
IP
..
5π/2
. 0.
IP
..
7π/2
. 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55
251. . . . . . .
Step 2: Concavity
If f′
(x) = − sin x − 1, then f′′
(x) = − cos x.
This is 0 when x = π/2 + πk, where k is any integer.
This is periodic with period 2π
..
f′′
(x)
.
f(x)
.−−.
⌢
. ++.
⌣
. −−.
⌢
. ++.
⌣
..
−π/2
.0 .
IP
..
π/2
. 0.
IP
..
3π/2
. 0.
IP
..
5π/2
. 0.
IP
..
7π/2
. 0.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 46 / 55