lesson 21: curve sketching (section 021 handout)

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


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.


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)


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)


Notes

Notes

Notes

2


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!


Outline

Simple examplesA cubic functionA quartic function

More ExamplesPoints of nondifferentiabilityHorizontal asymptotesVertical asymptotesTrigonometric and polynomial togetherLogarithmic


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.


Notes

Notes

Notes

3


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


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


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

"

"


Notes

Notes

Notes

4


Combinations of monotonicity and concavity

III

III IV

decreasing,concavedown

increasing,concavedown

decreasing,concave up

increasing,concave up


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

"

"


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)


Notes

Notes

Notes

5


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.



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


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


Notes

Notes

Notes

6


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

"


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)


Outline

Simple examplesA cubic functionA quartic function

More ExamplesPoints of nondifferentiabilityHorizontal asymptotesVertical asymptotesTrigonometric and polynomial togetherLogarithmic


Notes

Notes

Notes

7


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.


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.


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

= −∞


Notes

Notes

Notes

8


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)


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



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


Notes

Notes

Notes

9


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

−∞−−_

−−_


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

−∞ +∞" " "


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)


Notes

Notes

Notes

10


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



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


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


Notes

Notes

Notes

11


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

"

"


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

"

"


Example with Vertical Asymptotes

Example

Graph f (x) =1

x+

1

x2


Notes

Notes

Notes

12


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

− + +



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


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


Notes

Notes

Notes

13


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


Trigonometric and polynomial together

Problem

Graph f (x) = cos x − x


Notes

Notes

Notes

14


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) =∞.



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−↘

−↘


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


Notes

Notes

Notes

15


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


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


Logarithmic

Problem

Graph f (x) = x ln x2


Notes

Notes

Notes

16


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



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


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


Notes

Notes

Notes

17


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"

"


Step 4: Graph

x

y


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?


Notes

Notes

Notes

18


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