Lesson 21: Curve Sketching (handout)

.

...

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.


.

.

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.


.

.

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.


.

.

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.


..−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.


.

.

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.

.


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.


.

.

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.


.

.

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.


.

.

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.


.

.


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|


..−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.


.

.

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

.


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.


.

.

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


..−√

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.


.

.

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

.

−

.

+

.

+

.


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.


.

.

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.


.

.

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) = ∞.

.


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.


.

.

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.


.

.

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

.


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.


.

.

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.


.

.

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.


Documents

Lesson 21: Curve Sketching (handout)