lesson 21: curve sketching (section 041 slides)
Post on 04-Jul-2015
669 Views
Preview:
DESCRIPTION
TRANSCRIPT
..
Section 4.4Curve Sketching
V63.0121.041, Calculus I
New York University
November 17, 2010
AnnouncementsI Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7I There is class on November 24
. . . . . .
. . . . . .
Announcements
I Quiz 4 this week inrecitation on 3.3, 3.4, 3.5,3.7
I There is class onNovember 24
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 2 / 55
. . . . . .
Objectives
I given a function, graph itcompletely, indicating
I zeroes (if easy)I asymptotes if applicableI critical pointsI local/global max/minI inflection points
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 3 / 55
. . . . . .
Why?
Graphing functions is likedissection
… or diagrammingsentencesYou can really know a lot abouta function when you know all ofits anatomy.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
. . . . . .
Why?
Graphing functions is likedissection … or diagrammingsentences
You can really know a lot abouta function when you know all ofits anatomy.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
. . . . . .
Why?
Graphing functions is likedissection … or diagrammingsentencesYou can really know a lot abouta function when you know all ofits anatomy.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 4 / 55
. . . . . .
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 fis 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
. . . . . .
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 concavedownward 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
. . . . . .
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/decreasingand local max/min.
2. Find f′′ and form its sign chart. Concludeconcave up/concave down and inflection.
3. Put together a big chart to assemblemonotonicity and concavity data
4. Graph!
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 7 / 55
. . . . . .
Outline
Simple examplesA cubic functionA quartic function
More ExamplesPoints of nondifferentiabilityHorizontal asymptotesVertical asymptotesTrigonometric and polynomial togetherLogarithmic
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 8 / 55
. . . . . .
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 powerof x:
f(x) = x(2x2 − 3x− 12)
so f(0) = 0. The other factor is a quadratic, so we the other two rootsare
x =3±
√32 − 4(2)(−12)
4=
3±√105
4It’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
. . . . . .
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 powerof x:
f(x) = x(2x2 − 3x− 12)
so f(0) = 0. The other factor is a quadratic, so we the other two rootsare
x =3±
√32 − 4(2)(−12)
4=
3±√105
4It’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
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
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
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 10 / 55
. . . . . .
Step 2: Concavity
f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)
Another sign chart: .
.
f′′(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)
Another sign chart: ..
f′′(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)
Another sign chart: ..
f′′(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)
Another sign chart: ..
f′′(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)
Another sign chart: ..
f′′(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)
Another sign chart: ..
f′′(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
Step 2: Concavity
f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)
Another sign chart: ..
f′′(x)
.
f(x)
..
1/2
.
−−
.
++
.
⌢
.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 11 / 55
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,concavedown
.
increasing,concavedown
.
decreasing,concave up
.
increasing,concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,concavedown
.
increasing,concavedown
.
decreasing,concave up
.
increasing,concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,concavedown
.
increasing,concavedown
.
decreasing,concave up
.
increasing,concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,concavedown
.
increasing,concavedown
.
decreasing,concave up
.
increasing,concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
Combinations of monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,concavedown
.
increasing,concavedown
.
decreasing,concave up
.
increasing,concave up
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 13 / 55
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
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
. . . . . .
Step 4: Graph
..
f(x) = 2x3 − 3x2 − 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
.
"
.
.
.
"
..
(3−
√105
4 ,0)
..
(−1,7)
..(0,0)..
(1/2,−61/2)..
(2,−20)
.. (3+
√105
4 ,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3 − 3x2 − 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
.
"
.
.
.
"
..
(3−
√105
4 ,0)
..
(−1,7)
..(0,0)..
(1/2,−61/2)..
(2,−20)
.. (3+
√105
4 ,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3 − 3x2 − 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
.
"
.
.
.
"
..
(3−
√105
4 ,0)
..
(−1,7)
..(0,0)..
(1/2,−61/2)..
(2,−20)
.. (3+
√105
4 ,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3 − 3x2 − 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
.
"
.
.
.
"
..
(3−
√105
4 ,0)
..
(−1,7)
..(0,0)..
(1/2,−61/2)..
(2,−20)
.. (3+
√105
4 ,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Step 4: Graph
..
f(x) = 2x3 − 3x2 − 12x
. x.
f(x)
.
f(x)
.
shape of f
..
−1
.
7
.
max
..
2
.
−20
.
min
..
1/2
.
−61/2
.
IP
.
"
.
.
.
"
..
(3−
√105
4 ,0)
..
(−1,7)
..(0,0)..
(1/2,−61/2)..
(2,−20)
.. (3+
√105
4 ,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 15 / 55
. . . . . .
Graphing a quartic
Example
Graph f(x) = x4 − 4x3 + 10
(Step 0) We know f(0) = 10 and limx→±∞
f(x) = +∞. Not too many otherpoints on the graph are evident.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 16 / 55
. . . . . .
Graphing a quartic
Example
Graph f(x) = x4 − 4x3 + 10
(Step 0) We know f(0) = 10 and limx→±∞
f(x) = +∞. Not too many otherpoints on the graph are evident.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 16 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.
. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.
. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0
.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+
. +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +
. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +
.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 1: Monotonicity
f(x) = x4 − 4x3 + 10
=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)
We make its sign chart.
.. 4x2..0.0.+ . +. +.
(x− 3)
..
3
.
0
.
−
.
−
.
+
.
f′(x)
.
f(x)
..
3
.
0
..
0
.
0
.
−
.
−
.
+
.
↘
.
↘
.
↗
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 17 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.
. 12x..0.0
.− . +. +
.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.
. 12x..0.0
.− . +. +
.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0
.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.−
. +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +
. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +
.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 2: Concavity
f′(x) = 4x3 − 12x2
=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)
Here is its sign chart:
.. 12x..0.0.− . +. +.
x− 2
..
2
.
0
.
−
.
−
.
+
.
f′′(x)
.
f(x)
..
0
.
0
..
2
.
0
.
++
.
−−
.
++
.
⌣
.
⌢
.
⌣
.
IP
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 18 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4 − 4x3 + 10.
..
f′(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4 − 4x3 + 10.
..
f′(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4 − 4x3 + 10.
..
f′(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4 − 4x3 + 10.
..
f′(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 3: Grand Unified Sign Chart
Remember, f(x) = x4 − 4x3 + 10.
..
f′(x)
.
monotonicity
..
3
.
0
..
0
.
0
.
−
.
↘
.
−
.
↘
.
−
.
↘
.
+
.
↗
.
f′′(x)
.
concavity
..
0
.
0
..
2
.
0
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 19 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4 − 4x3 + 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
..(0,10)
..(2,−6)
..
(3,−17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4 − 4x3 + 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
..(0,10)
..(2,−6)
..
(3,−17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4 − 4x3 + 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
..(0,10)
..(2,−6)
..
(3,−17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4 − 4x3 + 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
..(0,10)
..(2,−6)
..
(3,−17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Step 4: Graph
..
f(x) = x4 − 4x3 + 10
. x.
y
.
f(x)
.
shape
..
0
.
10
.
IP
..
2
.
−6
.
IP
..
3
.
−17
.
min
.
.
.
.
"
..(0,10)
..(2,−6)
..
(3,−17)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 20 / 55
. . . . . .
Outline
Simple examplesA cubic functionA quartic function
More ExamplesPoints of nondifferentiabilityHorizontal asymptotesVertical asymptotesTrigonometric and polynomial togetherLogarithmic
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 21 / 55
. . . . . .
Graphing a function with a cusp
Example
Graph f(x) = x+√
|x|
This function looks strange because of the absolute value. Butwhenever we become nervous, we can just take cases.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 22 / 55
. . . . . .
Graphing a function with a cusp
Example
Graph f(x) = x+√
|x|
This function looks strange because of the absolute value. Butwhenever we become nervous, we can just take cases.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 22 / 55
. . . . . .
Step 0: Finding Zeroes
f(x) = x+√
|x|I First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if
x is positive.
I 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
. . . . . .
Step 0: Finding Zeroes
f(x) = x+√
|x|I First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if
x is positive.I 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
. . . . . .
Step 0: Finding Zeroes
f(x) = x+√
|x|I First, look at f by itself. We can tell that f(0) = 0 and that f(x) > 0 if
x is positive.I 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
. . . . . .
Step 0: Asymptotic behavior
f(x) = x+√
|x|I lim
x→∞f(x) = ∞, because both terms tend to ∞.
I limx→−∞
f(x) is indeterminate of the form −∞+∞. It’s the same aslim
y→+∞(−y+
√y)
limy→+∞
(−y+√y) = lim
y→∞(√y− y) ·
√y+ y√y+ y
= limy→∞
y− y2√y+ y
= −∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
. . . . . .
Step 0: Asymptotic behavior
f(x) = x+√
|x|I lim
x→∞f(x) = ∞, because both terms tend to ∞.
I limx→−∞
f(x) is indeterminate of the form −∞+∞. It’s the same aslim
y→+∞(−y+
√y)
limy→+∞
(−y+√y) = lim
y→∞(√y− y) ·
√y+ y√y+ y
= limy→∞
y− y2√y+ y
= −∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
. . . . . .
Step 0: Asymptotic behavior
f(x) = x+√
|x|I lim
x→∞f(x) = ∞, because both terms tend to ∞.
I limx→−∞
f(x) is indeterminate of the form −∞+∞. It’s the same aslim
y→+∞(−y+
√y)
limy→+∞
(−y+√y) = lim
y→∞(√y− y) ·
√y+ y√y+ y
= limy→∞
y− y2√y+ y
= −∞
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 24 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x+√
|x|.To find f′, first assume x > 0. Then
f′(x) =ddx
(x+
√x)= 1+
12√x
NoticeI f′(x) > 0 when x > 0 (so no critical points here)I lim
x→0+f′(x) = ∞ (so 0 is a critical point)
I limx→∞
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
. . . . . .
Step 1: The derivative
Remember, f(x) = x+√
|x|.To find f′, first assume x > 0. Then
f′(x) =ddx
(x+
√x)= 1+
12√x
NoticeI f′(x) > 0 when x > 0 (so no critical points here)
I limx→0+
f′(x) = ∞ (so 0 is a critical point)
I limx→∞
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
. . . . . .
Step 1: The derivative
Remember, f(x) = x+√
|x|.To find f′, first assume x > 0. Then
f′(x) =ddx
(x+
√x)= 1+
12√x
NoticeI f′(x) > 0 when x > 0 (so no critical points here)I lim
x→0+f′(x) = ∞ (so 0 is a critical point)
I limx→∞
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
. . . . . .
Step 1: The derivative
Remember, f(x) = x+√
|x|.To find f′, first assume x > 0. Then
f′(x) =ddx
(x+
√x)= 1+
12√x
NoticeI f′(x) > 0 when x > 0 (so no critical points here)I lim
x→0+f′(x) = ∞ (so 0 is a critical point)
I limx→∞
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
. . . . . .
Step 1: The derivative
Remember, f(x) = x+√
|x|.If x is negative, we have
f′(x) =ddx
(x+
√−x
)= 1− 1
2√−x
NoticeI lim
x→0−f′(x) = −∞ (other side of the critical point)
I limx→−∞
f′(x) = 1 (asymptotic to a line of slope 1)
I f′(x) = 0 when
1− 12√−x
= 0 =⇒√−x =
12
=⇒ −x =14
=⇒ x = −14
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x+√
|x|.If x is negative, we have
f′(x) =ddx
(x+
√−x
)= 1− 1
2√−x
NoticeI lim
x→0−f′(x) = −∞ (other side of the critical point)
I limx→−∞
f′(x) = 1 (asymptotic to a line of slope 1)
I f′(x) = 0 when
1− 12√−x
= 0 =⇒√−x =
12
=⇒ −x =14
=⇒ x = −14
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
. . . . . .
Step 1: The derivative
Remember, f(x) = x+√
|x|.If x is negative, we have
f′(x) =ddx
(x+
√−x
)= 1− 1
2√−x
NoticeI lim
x→0−f′(x) = −∞ (other side of the critical point)
I limx→−∞
f′(x) = 1 (asymptotic to a line of slope 1)
I f′(x) = 0 when
1− 12√−x
= 0 =⇒√−x =
12
=⇒ −x =14
=⇒ x = −14
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 26 / 55
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 1: Monotonicity
f′(x) =
1+
12√x
if x > 0
1− 12√−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
. . . . . .
Step 2: Concavity
I If x > 0, then
f′′(x) =ddx
(1+
12x−1/2
)= −1
4x−3/2
This is negative whenever x > 0.
I If x < 0, then
f′′(x) =ddx
(1− 1
2(−x)−1/2
)= −1
4(−x)−3/2
which is also always negative for negative x.
I In other words, f′′(x) = −14|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
. . . . . .
Step 2: Concavity
I If x > 0, then
f′′(x) =ddx
(1+
12x−1/2
)= −1
4x−3/2
This is negative whenever x > 0.I If x < 0, then
f′′(x) =ddx
(1− 1
2(−x)−1/2
)= −1
4(−x)−3/2
which is also always negative for negative x.
I In other words, f′′(x) = −14|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
. . . . . .
Step 2: Concavity
I If x > 0, then
f′′(x) =ddx
(1+
12x−1/2
)= −1
4x−3/2
This is negative whenever x > 0.I If x < 0, then
f′′(x) =ddx
(1− 1
2(−x)−1/2
)= −1
4(−x)−3/2
which is also always negative for negative x.
I In other words, f′′(x) = −14|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
. . . . . .
Step 2: Concavity
I If x > 0, then
f′′(x) =ddx
(1+
12x−1/2
)= −1
4x−3/2
This is negative whenever x > 0.I If x < 0, then
f′′(x) =ddx
(1− 1
2(−x)−1/2
)= −1
4(−x)−3/2
which is also always negative for negative x.
I In other words, f′′(x) = −14|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
. . . . . .
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
..
−14
.
14
.
max
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
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
..
−14
.
14
.
max
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
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
..
−14
.
14
.
max
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
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
..
−14
.
14
.
max
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
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
..
−14
.
14
.
max
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 29 / 55
. . . . . .
Graph
f(x) = x+√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−14
.
14
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
. x.
f(x)
..(−1,0)
..(−1
4 ,14)
..(0,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x+√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−14
.
14
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
. x.
f(x)
..(−1,0)
..(−1
4 ,14)
..(0,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x+√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−14
.
14
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
. x.
f(x)
..(−1,0)
..(−1
4 ,14)
..(0,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x+√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−14
.
14
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
. x.
f(x)
..(−1,0)
..(−1
4 ,14)
..(0,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x+√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−14
.
14
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
. x.
f(x)
..(−1,0)
..(−1
4 ,14)
..(0,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x+√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−14
.
14
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
. x.
f(x)
..(−1,0)
..(−1
4 ,14)
..(0,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Graph
f(x) = x+√
|x|
..
f(x)
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−14
.
14
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
.
"
.
"
.
.
"
. x.
f(x)
..(−1,0)
..(−1
4 ,14)
..(0,0)
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 30 / 55
. . . . . .
Example with Horizontal Asymptotes
Example
Graph f(x) = xe−x2
Before taking derivatives, we notice that f is odd, that f(0) = 0, andlimx→∞
f(x) = 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 31 / 55
. . . . . .
Example with Horizontal Asymptotes
Example
Graph f(x) = xe−x2
Before taking derivatives, we notice that f is odd, that f(0) = 0, andlimx→∞
f(x) = 0
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 31 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
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 off′(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
.
min
..
√1/2
.
0
.
max
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 32 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0
.− .− . +. +
.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.−
.− . +. +
.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .−
. +. +
.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +
. +
.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 2: Concavity
If f′(x) = (1− 2x2)e−x2 , we know
f′′(x) = (−4x)e−x2 + (1− 2x2)e−x2(−2x) =(4x3 − 6x
)e−x2
= 2x(2x2 − 3)e−x2
.. 2x..0.0.− .− . +. +.
√2x−
√3
..
√3/2
.
0
.
−
.
−
.
−
.
+
.
√2x+
√3
..
−√
3/2
.
0
.
−
.
+
.
+
.
+
.
f′′(x)
.
f(x)
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
..
−√
3/2
.
0
.
IP
..
0
.
0
.
IP
..
√3/2
.
0
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 33 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
.. f′(x).monotonicity
..−√
1/2
.0 .. √1/2
. 0.− .↘
.− .↘
.+ .↗
. +.↗
. −.↘
. −.↘
.
f′′(x)
.
concavity
..
−√
3/2
.
0
..
0
.
0
..
√3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−√
1/2
.
− 1√2e
.
min
..
√1/2
.
1√2e
.
max
..
−√
3/2
.
−√
32e3
.
IP
..
0
.
0
.
IP
..
√3/2
.
√32e3
.
IP
.
.
.
"
.
"
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
.. f′(x).monotonicity
..−√
1/2
.0 .. √1/2
. 0.− .↘
.− .↘
.+ .↗
. +.↗
. −.↘
. −.↘
.
f′′(x)
.
concavity
..
−√
3/2
.
0
..
0
.
0
..
√3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−√
1/2
.
− 1√2e
.
min
..
√1/2
.
1√2e
.
max
..
−√
3/2
.
−√
32e3
.
IP
..
0
.
0
.
IP
..
√3/2
.
√32e3
.
IP
.
.
.
"
.
"
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
.. f′(x).monotonicity
..−√
1/2
.0 .. √1/2
. 0.− .↘
.− .↘
.+ .↗
. +.↗
. −.↘
. −.↘
.
f′′(x)
.
concavity
..
−√
3/2
.
0
..
0
.
0
..
√3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−√
1/2
.
− 1√2e
.
min
..
√1/2
.
1√2e
.
max
..
−√
3/2
.
−√
32e3
.
IP
..
0
.
0
.
IP
..
√3/2
.
√32e3
.
IP
.
.
.
"
.
"
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
.. f′(x).monotonicity
..−√
1/2
.0 .. √1/2
. 0.− .↘
.− .↘
.+ .↗
. +.↗
. −.↘
. −.↘
.
f′′(x)
.
concavity
..
−√
3/2
.
0
..
0
.
0
..
√3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−√
1/2
.
− 1√2e
.
min
..
√1/2
.
1√2e
.
max
..
−√
3/2
.
−√
32e3
.
IP
..
0
.
0
.
IP
..
√3/2
.
√32e3
.
IP
.
.
.
"
.
"
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
.. f′(x).monotonicity
..−√
1/2
.0 .. √1/2
. 0.− .↘
.− .↘
.+ .↗
. +.↗
. −.↘
. −.↘
.
f′′(x)
.
concavity
..
−√
3/2
.
0
..
0
.
0
..
√3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−√
1/2
.
− 1√2e
.
min
..
√1/2
.
1√2e
.
max
..
−√
3/2
.
−√
32e3
.
IP
..
0
.
0
.
IP
..
√3/2
.
√32e3
.
IP
.
.
.
".
"
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
.. f′(x).monotonicity
..−√
1/2
.0 .. √1/2
. 0.− .↘
.− .↘
.+ .↗
. +.↗
. −.↘
. −.↘
.
f′′(x)
.
concavity
..
−√
3/2
.
0
..
0
.
0
..
√3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−√
1/2
.
− 1√2e
.
min
..
√1/2
.
1√2e
.
max
..
−√
3/2
.
−√
32e3
.
IP
..
0
.
0
.
IP
..
√3/2
.
√32e3
.
IP
.
.
.
".
"
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 3: Synthesis
f(x) = xe−x2
.. f′(x).monotonicity
..−√
1/2
.0 .. √1/2
. 0.− .↘
.− .↘
.+ .↗
. +.↗
. −.↘
. −.↘
.
f′′(x)
.
concavity
..
−√
3/2
.
0
..
0
.
0
..
√3/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
++
.
⌣
.
−−
.
⌢
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−√
1/2
.
− 1√2e
.
min
..
√1/2
.
1√2e
.
max
..
−√
3/2
.
−√
32e3
.
IP
..
0
.
0
.
IP
..
√3/2
.
√32e3
.
IP
.
.
.
".
"
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 34 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(−√
1/2,− 1√2e
)..
(√1/2, 1√
2e
)
..
(−√
3/2,−√
32e3
)
..
(0,0)
..
(√3/2,
√32e3
)
. f(x).shape
..−√
1/2
.− 1√
2e .
min
.. √1/2
.1√2e.
max
..−√
3/2
.−√
32e3 .
IP
..0.0.
IP
.. √3/2
.
√32e3.
IP
. . ." . ". .
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(−√
1/2,− 1√2e
)..
(√1/2, 1√
2e
)
..
(−√
3/2,−√
32e3
)
..
(0,0)
..
(√3/2,
√32e3
)
. f(x).shape
..−√
1/2
.− 1√
2e .
min
.. √1/2
.1√2e.
max
..−√
3/2
.−√
32e3 .
IP
..0.0.
IP
.. √3/2
.
√32e3.
IP
. . ." . ". .
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(−√
1/2,− 1√2e
)..
(√1/2, 1√
2e
)
..
(−√
3/2,−√
32e3
)
..
(0,0)
..
(√3/2,
√32e3
)
. f(x).shape
..−√
1/2
.− 1√
2e .
min
.. √1/2
.1√2e.
max
..−√
3/2
.−√
32e3 .
IP
..0.0.
IP
.. √3/2
.
√32e3.
IP
. . ." . ". .
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(−√
1/2,− 1√2e
)..
(√1/2, 1√
2e
)
..
(−√
3/2,−√
32e3
)
..
(0,0)
..
(√3/2,
√32e3
)
. f(x).shape
..−√
1/2
.− 1√
2e .
min
.. √1/2
.1√2e.
max
..−√
3/2
.−√
32e3 .
IP
..0.0.
IP
.. √3/2
.
√32e3.
IP
. . ." . ". .
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(−√
1/2,− 1√2e
)..
(√1/2, 1√
2e
)
..
(−√
3/2,−√
32e3
)
..
(0,0)
..
(√3/2,
√32e3
)
. f(x).shape
..−√
1/2
.− 1√
2e .
min
.. √1/2
.1√2e.
max
..−√
3/2
.−√
32e3 .
IP
..0.0.
IP
.. √3/2
.
√32e3.
IP
. . ." . ". .
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(−√
1/2,− 1√2e
)..
(√1/2, 1√
2e
)
..
(−√
3/2,−√
32e3
)
..
(0,0)
..
(√3/2,
√32e3
)
. f(x).shape
..−√
1/2
.− 1√
2e .
min
.. √1/2
.1√2e.
max
..−√
3/2
.−√
32e3 .
IP
..0.0.
IP
.. √3/2
.
√32e3.
IP
. . ." . ". .
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Step 4: Graph
..
x
.
f(x)
.
f(x) = xe−x2
..
(−√
1/2,− 1√2e
)..
(√1/2, 1√
2e
)
..
(−√
3/2,−√
32e3
)
..
(0,0)
..
(√3/2,
√32e3
)
. f(x).shape
..−√
1/2
.− 1√
2e .
min
.. √1/2
.1√2e.
max
..−√
3/2
.−√
32e3 .
IP
..0.0.
IP
.. √3/2
.
√32e3.
IP
. . ." . ". .
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 35 / 55
. . . . . .
Example with Vertical Asymptotes
Example
Graph f(x) =1x+
1x2
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 36 / 55
. . . . . .
Step 0
Find when f is positive, negative, zero, not defined.
We need to factor f:
f(x) =1x+
1x2
=x+ 1x2
.
This means f is 0 at −1 and has trouble at 0. In fact,
limx→0
x+ 1x2
= ∞,
so x = 0 is a vertical asymptote of the graph. We can make a signchart 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
. . . . . .
Step 0
Find when f is positive, negative, zero, not defined. We need to factor f:
f(x) =1x+
1x2
=x+ 1x2
.
This means f is 0 at −1 and has trouble at 0. In fact,
limx→0
x+ 1x2
= ∞,
so x = 0 is a vertical asymptote of the graph.
We can make a signchart 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
. . . . . .
Step 0
Find when f is positive, negative, zero, not defined. We need to factor f:
f(x) =1x+
1x2
=x+ 1x2
.
This means f is 0 at −1 and has trouble at 0. In fact,
limx→0
x+ 1x2
= ∞,
so x = 0 is a vertical asymptote of the graph. We can make a signchart 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
. . . . . .
Step 0, continued
For horizontal asymptotes, notice that
limx→∞
x+ 1x2
= 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
. . . . . .
Step 1: Monotonicity
We havef′(x) = − 1
x2− 2
x3= −x+ 2
x3.
The critical points are x = −2 and x = 0. We have the following signchart:
.. −(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
. . . . . .
Step 1: Monotonicity
We havef′(x) = − 1
x2− 2
x3= −x+ 2
x3.
The critical points are x = −2 and x = 0. We have the following signchart:
.. −(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
. . . . . .
Step 1: Monotonicity
We havef′(x) = − 1
x2− 2
x3= −x+ 2
x3.
The critical points are x = −2 and x = 0. We have the following signchart:
.. −(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
. . . . . .
Step 1: Monotonicity
We havef′(x) = − 1
x2− 2
x3= −x+ 2
x3.
The critical points are x = −2 and x = 0. We have the following signchart:
.. −(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
. . . . . .
Step 1: Monotonicity
We havef′(x) = − 1
x2− 2
x3= −x+ 2
x3.
The critical points are x = −2 and x = 0. We have the following signchart:
.. −(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
. . . . . .
Step 1: Monotonicity
We havef′(x) = − 1
x2− 2
x3= −x+ 2
x3.
The critical points are x = −2 and x = 0. We have the following signchart:
.. −(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
. . . . . .
Step 1: Monotonicity
We havef′(x) = − 1
x2− 2
x3= −x+ 2
x3.
The critical points are x = −2 and x = 0. We have the following signchart:
.. −(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
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The critical points of f′ are −3 and 0. Sign chart:
.. (x+ 3)..0 .−3
.− . +.
x4
..
0
.
0
.
+
.
+
.
f′′(x)
.
f(x)
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
IP
.
VA
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 40 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
Step 3: Synthesis
..
f′
.
monotonicity
..
∞
.
0
..
0
.
−2
.
−
.
+
.
−
.
↘
.
↗
.
↘
.
f′′
.
concavity
..
∞
.
0
..
0
.
−3
.
−−
.
++
.
++
.
⌢
.
⌣
.
⌣
.
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 41 / 55
. . . . . .
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
. . . . . .
Trigonometric and polynomial together
ProblemGraph f(x) = cos x− x
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 43 / 55
. . . . . .
Step 0: intercepts and asymptotes
I f(0) = 1 and f(−π/2) = −π/2. So by the Intermediate ValueTheorem there is a zero in between. We don’t know it’s precisevalue, though.
I Since −1 ≤ cos x ≤ 1 for all x, we have
−1− x ≤ cos x− x ≤ 1− x
for all x. This means that limx→∞
f(x) = −∞ and limx→−∞
f(x) = ∞.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 44 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = cos x− x, then f′(x) = − sin x− 1 = (−1)(sin x+ 1).I f′(x) = 0 if x = 3π/2+ 2πk, where k is any integerI f′(x) is periodic with period 2πI 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
. . . . . .
Step 1: Monotonicity
If f(x) = cos x− x, then f′(x) = − sin x− 1 = (−1)(sin x+ 1).I f′(x) = 0 if x = 3π/2+ 2πk, where k is any integerI f′(x) is periodic with period 2πI 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
. . . . . .
Step 1: Monotonicity
If f(x) = cos x− x, then f′(x) = − sin x− 1 = (−1)(sin x+ 1).I f′(x) = 0 if x = 3π/2+ 2πk, where k is any integerI f′(x) is periodic with period 2πI 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
. . . . . .
Step 1: Monotonicity
If f(x) = cos x− x, then f′(x) = − sin x− 1 = (−1)(sin x+ 1).I f′(x) = 0 if x = 3π/2+ 2πk, where k is any integerI f′(x) is periodic with period 2πI 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
. . . . . .
Step 1: Monotonicity
If f(x) = cos x− x, then f′(x) = − sin x− 1 = (−1)(sin x+ 1).I f′(x) = 0 if x = 3π/2+ 2πk, where k is any integerI f′(x) is periodic with period 2πI 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 2: Concavity
If f′(x) = − sin x− 1, then f′′(x) = − cos x.I This is 0 when x = π/2+ πk, where k is any integer.I 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
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..−π/2
.0 ..3π/2
. 0..7π/2
. 0. −.↘
. −.↘
.
f′′(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..−π/2
.0 ..3π/2
. 0..7π/2
. 0. −.↘
. −.↘
.
f′′(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..−π/2
.0 ..3π/2
. 0..7π/2
. 0. −.↘
. −.↘
.
f′′(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..−π/2
.0 ..3π/2
. 0..7π/2
. 0. −.↘
. −.↘
.
f′′(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..−π/2
.0 ..3π/2
. 0..7π/2
. 0. −.↘
. −.↘
.
f′′(x)
.
conc
..
−π/2
.
0
..
π/2
.
0
..
3π/2
.
0
..
5π/2
.
0
..
7π/2
.
0
.
−−
.
⌢
.
++
.
⌣
.
−−
.
⌢
.
++
.
⌣
.
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 47 / 55
. . . . . .
Step 4: Graph
f(x) = cos x− x
..x
.y.......
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 48 / 55
. . . . . .
Step 4: Graph
f(x) = cos x− x
..x
.y.......
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 48 / 55
. . . . . .
Step 4: Graph
f(x) = cos x− x
..x
.y.......
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 48 / 55
. . . . . .
Step 4: Graph
f(x) = cos x− x
..x
.y.......
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 48 / 55
. . . . . .
Step 4: Graph
f(x) = cos x− x
..x
.y.......
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 48 / 55
. . . . . .
Step 4: Graph
f(x) = cos x− x
..x
.y.......
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 48 / 55
. . . . . .
Step 4: Graph
f(x) = cos x− x
..x
.y.......
f(x)
.
shape
..
−π/2
.
π/2
.
IP
..
π/2
.
−π/2
.
IP
..
3π/2
.
−3π/2
.
IP
..
5π/2
.
−5π/2
.
IP
..
7π/2
.
−7π/2
.
IP
.
.
.
.
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 48 / 55
. . . . . .
Logarithmic
ProblemGraph f(x) = x ln x2
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 49 / 55
. . . . . .
Step 0: Intercepts and Asymptotes
I limx→∞
f(x) = ∞, limx→−∞
f(x) = −∞.
I f is not originally defined at 0 because limx→0
ln x2 = −∞. But
limx→0
x ln x2 = limx→0
ln x2
1/xH= lim
x→0
(1/x2)(2x)−1/x2
= limx→0
2x = 0.
So we can define f(0) = 0 to make it a continuous function on(−∞,∞).
I Other zeroes?
ln x2 = 0 =⇒ x2 = 1 =⇒ x = ±1
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 50 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e
.+ .↗
.− .↘
. −.↘
. +.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+
.↗
.− .↘
. −.↘
. +.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+
.↗
.−
.↘
. −.↘
. +.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+
.↗
.−
.↘
. −
.↘
. +.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+
.↗
.−
.↘
. −
.↘
. +
.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.−
.↘
. −
.↘
. +
.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −
.↘
. +
.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +
.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 1: Monotonicity
If f(x) = x ln x2, then
f′(x) = ln x2 + x · 1x2
(2x) = ln x2 + 2
This is not defined at 0 and is 0 when
ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1
Use test points ±1, ±e−2. Here is the sign chart:
.. f′(x).f(x)
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 51 / 55
. . . . . .
Step 2: Concavity
If f′(x) = ln x2+ 2, then f′′(x) = 1/x2 · (2x) = 2/x. Here is the sign chart:
.. f′(x).f(x)
..×.0
.−− .⌢
. ++.⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 52 / 55
. . . . . .
Step 2: Concavity
If f′(x) = ln x2+ 2, then f′′(x) = 1/x2 · (2x) = 2/x. Here is the sign chart:
.. f′(x).f(x)
..×.0.−−
.⌢
. ++.⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 52 / 55
. . . . . .
Step 2: Concavity
If f′(x) = ln x2+ 2, then f′′(x) = 1/x2 · (2x) = 2/x. Here is the sign chart:
.. f′(x).f(x)
..×.0.−−
.⌢
. ++
.⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 52 / 55
. . . . . .
Step 2: Concavity
If f′(x) = ln x2+ 2, then f′′(x) = 1/x2 · (2x) = 2/x. Here is the sign chart:
.. f′(x).f(x)
..×.0.−− .
⌢. ++
.⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 52 / 55
. . . . . .
Step 2: Concavity
If f′(x) = ln x2+ 2, then f′′(x) = 1/x2 · (2x) = 2/x. Here is the sign chart:
.. f′(x).f(x)
..×.0.−− .
⌢. ++.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 52 / 55
. . . . . .
Step 2: Concavity
If f′(x) = ln x2+ 2, then f′′(x) = 1/x2 · (2x) = 2/x. Here is the sign chart:
.. f′(x).f(x)
..×.0.−− .
⌢. ++.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 52 / 55
. . . . . .
Step 2: Concavity
If f′(x) = ln x2+ 2, then f′′(x) = 1/x2 · (2x) = 2/x. Here is the sign chart:
.. f′(x).f(x)
..×.0.−− .
⌢. ++.
⌣
.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 52 / 55
. . . . . .
Step 2: Concavity
If f′(x) = ln x2+ 2, then f′′(x) = 1/x2 · (2x) = 2/x. Here is the sign chart:
.. f′(x).f(x)
..×.0.−− .
⌢. ++.
⌣.
IP
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 52 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
0
.
−1/e
..
×
.
0
..
0
.
1/e
.
"
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 53 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
0
.
−1/e
..
×
.
0
..
0
.
1/e
.
"
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 53 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
0
.
−1/e
..
×
.
0
..
0
.
1/e
.
"
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 53 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
0
.
−1/e
..
×
.
0
..
0
.
1/e
.
"
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 53 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
0
.
−1/e
..
×
.
0
..
0
.
1/e
.
"
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 53 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
0
.
−1/e
..
×
.
0
..
0
.
1/e
.
"
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 53 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
0
.
−1/e
..
×
.
0
..
0
.
1/e
.
"
.
.
.
"
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 53 / 55
. . . . . .
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
0
.
−1/e
..
×
.
0
..
0
.
1/e
.
"
.
.
."
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 53 / 55
. . . . . .
Step 4: Graph
.. x.
y
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 54 / 55
. . . . . .
Summary
I Graphing is a procedure that gets easier with practice.I Remember to follow the checklist.I Graphing is like dissection—or is it vivisection?
V63.0121.041, Calculus I (NYU) Section 4.4 Curve Sketching November 17, 2010 55 / 55
top related