lesson 21: curve sketching (handout)
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Sec on 4.4Curve Sketching
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
April 13, 2011
.
AnnouncementsI Quiz 4 on Sec ons 3.3,3.4, 3.5, and 3.7 thisweek (April 14/15)
I Quiz 5 on Sec ons4.1–4.4 April 28/29
I Final Exam Thursday May12, 2:00–3:50pm
I I am teaching Calc II MW2:00pm and Calc III TR2:00pm both Fall ’11 andSpring ’12
.
Objectives
I given a func on, graph itcompletely, indica ng
I zeroes (if easy)I asymptotes if applicableI cri cal pointsI local/global max/minI inflec on points
.
Notes
.
Notes
.
Notes
. 1.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Why?
Graphing func ons is likedissec on … or diagrammingsentencesYou can really know a lotabout a func on when youknow all of its anatomy.
.
The Increasing/Decreasing TestTheorem (The Increasing/Decreasing Test)
If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b),then f is decreasing on (a, b).
Example
f(x) = x3 + x2
f′(x) = 3x2 + 2x ..
f(x)
.
f′(x)
.
Testing for ConcavityTheorem (Concavity Test)
If f′′(x) > 0 for all x in (a, b), then the graph of f is concave upwardon (a, b) If f′′(x) < 0 for all x in (a, b), then the graph of f is concavedownward on (a, b).
Example
f(x) = x3 + x2
f′(x) = 3x2 + 2xf′′(x) = 6x+ 2
..
f(x)
.
f′(x)
.
f′′(x)
.
Notes
.
Notes
.
Notes
. 2.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Graphing ChecklistTo graph a func on f, follow this plan:
0. Find when f is posi ve, nega ve, zero,not defined.
1. Find f′ and form its sign chart.Conclude informa on aboutincreasing/decreasing and localmax/min.
2. Find f′′ and form its sign chart.Conclude concave up/concave downand inflec on.
.
Graphing ChecklistTo graph a func on f, follow this plan:
3. Put together a big chart to assemblemonotonicity and concavity data
4. Graph!
.
OutlineSimple examples
A cubic func onA quar c func on
More ExamplesPoints of nondifferen abilityHorizontal asymptotesVer cal asymptotesTrigonometric and polynomial togetherLogarithmic
.
Notes
.
Notes
.
Notes
. 3.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Graphing a cubicExample
Graph f(x) = 2x3 − 3x2 − 12x.
(Step 0) First, let’s find the zeros. We can at least factor out onepower of x:
f(x) = x(2x2 − 3x− 12)
so f(0) = 0. The other factor is a quadra c, so we the other tworoots are
x =3±
√32 − 4(2)(−12)
4=
3±√105
4It’s OK to skip this step for now since the roots are so complicated.
.
Step 1: Monotonicity
f(x) = 2x3 − 3x2 − 12x=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)
We can form a sign chart from this:
.. x− 2..2
.− . −. +.
x+ 1
..
−1
.+
.+
.−
.
f′(x)
.
f(x)
..
2
..
−1
.
+
.
−
.
+
.
↗
.
↘
.
↗
.
max
.
min
.
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
.
Notes
.
Notes
.
Notes
. 4.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Step 3: One sign chart to rule them allRemember, 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
.
"
.
.
.
"
.
monotonicity and concavity
..
I
.
II
.
III
.
IV
.
decreasing,concavedown
.
increasing,concavedown
.
decreasing,concaveup
.
increasing,concaveup
.
Step 3: One sign chart to rule them allRemember, 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
.
"
.
.
.
"
.
Notes
.
Notes
.
Notes
. 5.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
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)
.
Graphing a quartic
Example
Graph f(x) = x4 − 4x3 + 10
(Step 0) We know f(0) = 10 and limx→±∞
f(x) = +∞. Not too manyother points on the graph are evident.
.
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
.
Notes
.
Notes
.
Notes
. 6.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
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
.
Step 3: Grand Unified Sign ChartRemember, 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
.
.
.
.
"
.
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)
.
Notes
.
Notes
.
Notes
. 7.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
OutlineSimple examples
A cubic func onA quar c func on
More ExamplesPoints of nondifferen abilityHorizontal asymptotesVer cal asymptotesTrigonometric and polynomial togetherLogarithmic
.
Graphing a function with a cusp
Example
Graph f(x) = x+√
|x|
This func on looks strange because of the absolute value. Butwhenever we become nervous, we can just take cases.
.
Step 0: Finding Zeroesf(x) = x+
√|x|
I First, look at f by itself. We can tell that f(0) = 0 and thatf(x) > 0 if x is posi ve.
I Are there nega ve numbers which are zeroes for f?
x+√−x = 0 =⇒
√−x = −x
−x = x2 =⇒ x2 + x = 0
The only solu ons are x = 0 and x = −1.
.
Notes
.
Notes
.
Notes
. 8.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Step 0: Asymptotic behaviorf(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= −∞
.
Step 1: The derivativeRemember, f(x) = x+
√|x|.
To find f′, first assume x > 0. Then
f′(x) =ddx
(x+
√x)= 1+
12√
x
No ceI f′(x) > 0 when x > 0 (so no cri cal points here)I lim
x→0+f′(x) = ∞ (so 0 is a cri cal point)
I limx→∞
f′(x) = 1 (so the graph is asympto c to a line of slope 1)
.
Step 1: The derivativeRemember, f(x) = x+
√|x|.
If x is nega ve, we have
f′(x) =ddx
(x+
√−x
)= 1− 1
2√−x
No ceI lim
x→0−f′(x) = −∞ (other side of the cri cal point)
I limx→−∞
f′(x) = 1 (asympto c to a line of slope 1)I f′(x) = 0 when
1− 12√−x
= 0 =⇒√−x =
12
=⇒ −x =14
=⇒ x = −14
.
Notes
.
Notes
.
Notes
. 9.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Step 1: Monotonicity
f′(x) =
1+
12√
xif x > 0
1− 12√−x
if x < 0
We can’t make a mul -factor sign chart because of the absolutevalue, but we can test points in between cri cal points.
.. f′(x).f(x)
..−1
4
.0 ..0.∓∞.+ .− . +.
↗.
↘.
↗.
max
.
min
.
Step 2: ConcavityI If x > 0, then f′′(x) =
ddx
(1+
12x−1/2
)= −1
4x−3/2 This is
nega ve 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 nega ve for nega ve x.I In other words, f′′(x) = −1
4|x|−3/2.
Here is the sign chart:
.. f′′(x).f(x)
..0.−∞.−− .
⌢... −−.
⌢
.
Step 3: SynthesisNow 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
.
−∞
.
+∞
.
".
".
.
"
.
Notes
.
Notes
.
Notes
. 10.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Graph
..
x
.
shape
..
−1
.
0
.
zero
.
−∞
.
+∞
..
−14
.
14
.
max
.
−∞
.
+∞
..
0
.
0
.
min
.
−∞
.
+∞
.
".
".
.
". x.
f(x) = x+√
|x|
..(−1, 0)
..(−1
4 ,14)
..(0, 0)
.
Example with HorizontalAsymptotes
Example
Graph f(x) = xe−x2
Before taking deriva ves, we no ce that f is odd, that f(0) = 0, andlimx→∞
f(x) = 0
.
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
.. 1−√2x.. √
1/2
. 0.+ .+. −.
1+√2x
..
−√
1/2
.
0
.−
.+
.+
.
f′(x)
.
f(x)
.
−
.
↘
.
+
.
↗
.
−
.
↘
..
−√
1/2
.
0
.
min
..
√1/2
.
0
.
max
.
Notes
.
Notes
.
Notes
. 11.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
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
.
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
.
.
.
"
.
"
.
.
.
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
. . ." . ". .
.
Notes
.
Notes
.
Notes
. 12.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Example with Vertical Asymptotes
Example
Graph f(x) =1x+
1x2
.
Step 0
Find when f is posi ve, nega ve, zero, not defined. We need tofactor 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 ver cal asymptote of the graph. We can make a signchart as follows:
.. x+ 1..0 .−1
.− . +.
x2
..
0
.
0
.+
.+
.
f(x)
..
∞
.
0
..
0
.
−1
.
−
.
+
.
+
.
Step 1: Monotonicity
We havef′(x) = − 1
x2− 2
x3= −x+ 2
x3.
The cri cal 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
.
Notes
.
Notes
.
Notes
. 13.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Step 2: Concavity
We havef′′(x) =
2x3
+6x4
=2(x+ 3)
x4.
The cri cal 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
.
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
.
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
.
Notes
.
Notes
.
Notes
. 14.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Trigonometric and polynomialtogether
ProblemGraph f(x) = cos x− x
.
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) = ∞.
.
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 nega ve at all other points.
.. f′(x).f(x)
..−π/2
.0 ..3π/2
. 0..7π/2
. 0. −.↘
. −.↘
.
Notes
.
Notes
.
Notes
. 15.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
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
.
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
.
.
.
.
.
Step 4: Graphf(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
.
.
.
.
.
Notes
.
Notes
.
Notes
. 16.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Logarithmic
ProblemGraph f(x) = x ln x2
.
Step 0: Intercepts andAsymptotes
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 con nuous func on on(−∞,∞).
I Other zeroes?
ln x2 = 0 =⇒ x2 = 1 =⇒ x = ±1
.
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
.
Notes
.
Notes
.
Notes
. 17.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
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
.
Step 3: Synthesis
.. f′(x).mono
..0 .−1/e
..×.0.. 0.
1/e.+ .
↗.− .
↘. −.↘
. +.↗
.
max
.
min
.
f′(x)
.
conc
..
×
.
0
.
−−
.
⌢
.
++
.
⌣
.
IP
.
f′(x)
.
shape
..
2
.
−1/e
..
0
.
0
..
−2
.
1/e
.
"
.
.
.
"
.
Step 4: Graph
.. x.
y
......
f(x)
.
shape
..
0
.
−1
..
2
.
−1/e
..
0
.
0
..
2
.
1/e
..
0
.
1
.
−∞
.
∞
.
"
.
"
.
.
.
"
.
"
.
max
.
IP
.
min
.
Notes
.
Notes
.
Notes
. 18.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
.
.
Summary
I Graphing is a procedure that gets easier with prac ce.I Remember to follow the checklist.
I Graphing is like dissec on—or is it vivisec on?
.
.
.
Notes
.
Notes
.
Notes
. 19.
. Sec on 4.4: Curve Sketching. V63.0121.001: Calculus I . April 13, 2011
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