lesson 21: curve sketching (section 021 handout)
DESCRIPTION
We can put all of our graph-description techniques into a single picture.TRANSCRIPT
Section 4.4Curve Sketching
V63.0121.021, Calculus I
New York University
November 18, 2010
Announcements
I There is class on November 23. The homework is due onNovember 24. Turn in homework to my mailbox or bring to class onNovember 23.
Announcements
I There is class onNovember 23. Thehomework is due onNovember 24. Turn inhomework to my mailbox orbring to class onNovember 23.
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 3 / 55
Notes
Notes
Notes
1
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Why?
Graphing functions is likedissection . . . or diagrammingsentencesYou can really know a lot abouta function when you know all ofits anatomy.
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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), thenf is decreasing on (a, b).
Example
Here f (x) = x3 + x2, and f ′(x) = 3x2 + 2x .
f (x)f ′(x)
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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)
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 6 / 55
Notes
Notes
Notes
2
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Graphing Checklist
To graph a function f , follow this plan:
0. Find when f is positive, negative, zero, notdefined.
1. Find f ′ and form its sign chart. Concludeinformation 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!
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 7 / 55
Outline
Simple examplesA cubic functionA quartic function
More ExamplesPoints of nondifferentiabilityHorizontal asymptotesVertical asymptotesTrigonometric and polynomial togetherLogarithmic
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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 power ofx :
f (x) = x(2x2 − 3x − 12)
so f (0) = 0. The other factor is a quadratic, so we the other two roots are
x =3±
√32 − 4(2)(−12)
4=
3±√
105
4
It’s OK to skip this step for now since the roots are so complicated.
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 9 / 55
Notes
Notes
Notes
3
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
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 − 22
− − +
x + 1−1
++−
f ′(x)
f (x)2−1
+ − +
↗ ↘ ↗max min
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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)
concavity1/2
−−_
−−_
++^
++^
f (x)
shape of f−1
7
max2
−20
min1/2
−61/2
IP
"
"
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 12 / 55
Notes
Notes
Notes
4
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Combinations of monotonicity and concavity
III
III IV
decreasing,concavedown
increasing,concavedown
decreasing,concave up
increasing,concave up
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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)
concavity1/2
−−_
−−_
++^
++^
f (x)
shape of f−1
7
max2
−20
min1/2
−61/2
IP
"
"
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 14 / 55
Step 4: Graph
f (x) = 2x3 − 3x2 − 12x
x
f (x)
f (x)
shape of f−1
7
max2
−20
min1/2
−61/2
IP
"
"
(3−√105
4 , 0) (−1, 7)
(0, 0)
(1/2,−61/2)
(2,−20)
(3+√105
4 , 0)
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 15 / 55
Notes
Notes
Notes
5
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Graphing a quartic
Example
Graph f (x) = x4 − 4x3 + 10
(Step 0) We know f (0) = 10 and limx→±∞
f (x) = +∞. Not too many other
points on the graph are evident.
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 17 / 55
Step 2: Concavity
f ′(x) = 4x3 − 12x2
=⇒ f ′′(x) = 12x2 − 24x = 12x(x − 2)
Here is its sign chart:
12x0
0− + +
x − 22
0− − +
f ′′(x)
f (x)0
0
2
0++ −− ++^ _ ^
IP IP
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 18 / 55
Notes
Notes
Notes
6
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Step 3: Grand Unified Sign Chart
Remember, f (x) = x4 − 4x3 + 10.
f ′(x)
monotonicity3
0
0
0−↘
−↘
−↘
+
↗f ′′(x)
concavity0
0
2
0++^
−−_
++^
++^
f (x)
shape0
10
IP2
−6
IP3
−17
min
"
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 19 / 55
Step 4: Graph
f (x) = x4 − 4x3 + 10
x
y
f (x)
shape0
10
IP2
−6
IP3
−17
min
"
(0, 10)
(2,−6)(3,−17)
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 20 / 55
Outline
Simple examplesA cubic functionA quartic function
More ExamplesPoints of nondifferentiabilityHorizontal asymptotesVertical asymptotesTrigonometric and polynomial togetherLogarithmic
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 21 / 55
Notes
Notes
Notes
7
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Graphing a function with a cusp
Example
Graph f (x) = x +√|x |
This function looks strange because of the absolute value. But wheneverwe become nervous, we can just take cases.
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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) > 0if 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.
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 23 / 55
Step 0: Asymptotic behavior
f (x) = x +√|x |
I limx→∞
f (x) =∞, because both terms tend to ∞.
I limx→−∞
f (x) is indeterminate of the form −∞+∞. It’s the same as
limy→+∞
(−y +√
y)
limy→+∞
(−y +√
y) = limy→∞
(√
y − y) ·√
y + y√
y + y
= limy→∞
y − y2
√y + y
= −∞
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 24 / 55
Notes
Notes
Notes
8
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Step 1: The derivative
Remember, f (x) = x +√|x |.
To find f ′, first assume x > 0. Then
f ′(x) =d
dx
(x +√
x)
= 1 +1
2√
x
Notice
I 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)
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 25 / 55
Step 1: The derivative
Remember, f (x) = x +√|x |.
If x is negative, we have
f ′(x) =d
dx
(x +√−x)
= 1− 1
2√−x
Notice
I limx→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− 1
2√−x
= 0 =⇒√−x =
1
2=⇒ −x =
1
4=⇒ x = −1
4
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 26 / 55
Step 1: Monotonicity
f ′(x) =
1 +
1
2√
xif x > 0
1− 1
2√−x
if x < 0
We can’t make a multi-factor sign chart because of the absolute value,but we can test points in between critical points.
f ′(x)
f (x)−14
0
0
∓∞+ − +
↗ ↘ ↗max min
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 27 / 55
Notes
Notes
Notes
9
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Step 2: Concavity
I If x > 0, then
f ′′(x) =d
dx
(1 +
1
2x−1/2
)= −1
4x−3/2
This is negative whenever x > 0.I If x < 0, then
f ′′(x) =d
dx
(1− 1
2(−x)−1/2
)= −1
4(−x)−3/2
which is also always negative for negative x .
I In other words, f ′′(x) = −1
4|x |−3/2.
Here is the sign chart:
f ′′(x)
f (x)0
−∞−−_
−−_
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 28 / 55
Step 3: Synthesis
Now we can put these things together.
f (x) = x +√|x |
f ′(x)
monotonicity−14
0
0
∓∞+1
↗+
↗−↘
+
↗+1
↗f ′′(x)
concavity0
−∞−−_
−−_
−−_
−∞_
−∞_
f (x)
shape−1
0
zero−1
4
14
max0
0
min
−∞ +∞" " "
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 29 / 55
Graph
f (x) = x +√|x |
f (x)
shape−1
0
zero
−∞ +∞−1
4
14
max
−∞ +∞0
0
min
−∞ +∞" " "
x
f (x)
(−1, 0)(−1
4 ,14)
(0, 0)
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 30 / 55
Notes
Notes
Notes
10
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 31 / 55
Step 1: Monotonicity
If f (x) = xe−x2, then
f ′(x) = 1 · e−x2 + xe−x2(−2x) =
(1− 2x2
)e−x
2
=(
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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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−x
2
= 2x(2x2 − 3)e−x2
2x0
0− − + +
√2x −
√3√
3/2
0− − − +
√2x +
√3
−√
3/2
0− + + +
f ′′(x)
f (x)
−−_
++^
−−_
++^−
√3/2
0
IP0
0
IP
√3/2
0
IP
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 33 / 55
Notes
Notes
Notes
11
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
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
IP0
0
IP
√3/2
√3
2e3
IP
"
"
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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
IP0
0
IP
√3/2
√3
2e3
IP
"
"
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 35 / 55
Example with Vertical Asymptotes
Example
Graph f (x) =1
x+
1
x2
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 36 / 55
Notes
Notes
Notes
12
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Step 0
Find when f is positive, negative, zero, not defined. We need to factor f :
f (x) =1
x+
1
x2=
x + 1
x2.
This means f is 0 at −1 and has trouble at 0. In fact,
limx→0
x + 1
x2=∞,
so x = 0 is a vertical asymptote of the graph. We can make a sign chartas follows:
x + 10
−1
− +
x20
0
+ +
f (x)∞0
0
−1
− + +
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 37 / 55
Step 1: Monotonicity
We have
f ′(x) = − 1
x2− 2
x3= −x + 2
x3.
The critical points are x = −2 and x = 0. We have the following signchart:
−(x + 2)0
−2
+ −
x30
0
− +
f ′(x)
f (x)
∞0
0
−2
− + −↘ ↗ ↘
min VA
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 39 / 55
Step 2: Concavity
We have
f ′′(x) =2
x3+
6
x4=
2(x + 3)
x4.
The critical points of f ′ are −3 and 0. Sign chart:
(x + 3)0
−3
− +
x40
0
+ +
f ′′(x)
f (x)
∞0
0
−3
−− ++ ++_ ^ ^
IP VA
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 40 / 55
Notes
Notes
Notes
13
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 42 / 55
Trigonometric and polynomial together
Problem
Graph f (x) = cos x − x
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 43 / 55
Notes
Notes
Notes
14
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
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 precise value,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) =∞.
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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 integer
I f ′(x) is periodic with period 2π
I 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−↘
−↘
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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
IPπ/2
0
IP3π/2
0
IP5π/2
0
IP7π/2
0
IP
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 46 / 55
Notes
Notes
Notes
15
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
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
IPπ/2
−π/2
IP3π/2
−3π/2
IP5π/2
−5π/2
IP7π/2
−7π/2
IP
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 47 / 55
Step 4: Graph
f (x) = cos x − x
x
y
f (x)
shape−π/2
π/2
IPπ/2
−π/2
IP3π/2
−3π/2
IP5π/2
−5π/2
IP7π/2
−7π/2
IP
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 48 / 55
Logarithmic
Problem
Graph f (x) = x ln x2
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 49 / 55
Notes
Notes
Notes
16
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
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= lim
x→02x = 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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 50 / 55
Step 1: Monotonicity
If f (x) = x ln x2, then
f ′(x) = ln x2 + x · 1
x2(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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 52 / 55
Notes
Notes
Notes
17
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010
Step 3: Synthesis
f ′(x)
mono0
−1/e
×0
0
1/e
+
↗−↘
−↘
+
↗max min
f ′(x)
conc×0
−−_
++^
IP
f ′(x)
shape
0
−1/e
×0
0
1/e"
"
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 53 / 55
Step 4: Graph
x
y
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 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?
V63.0121.021, Calculus I (NYU) Section 4.4 Curve Sketching November 18, 2010 55 / 55
Notes
Notes
Notes
18
Section 4.4 : Curve SketchingV63.0121.021, Calculus I November 18, 2010