lesson 21: curve sketching (section 041 slides)

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

. . . . . .

Graphing functions is likedissection

… or diagrammingsentencesYou can really know a lot abouta function when you know all ofits anatomy.

. . . . . .

Graphing functions is likedissection … or diagrammingsentences

You can really know a lot abouta function when you know all ofits anatomy.

. . . . . .

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), then fis decreasing on (a,b).

Example

Here f(x) = x3 + x2, and f′(x) = 3x2 + 2x.

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)

. . . . . .

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!

. . . . . .

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 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)

3±√105

4It’s OK to skip this step for now since the roots are so complicated.

. . . . . .

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)

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

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

f(x) = 2x3 − 3x2 − 12x

=⇒ f′(x) = 6x2 − 6x− 12 = 6(x+ 1)(x− 2)

.. x− 2..2

.− . −. +.

f′(x)

. . . . . .

Step 2: Concavity

f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)

Another sign chart: .

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)

Another sign chart: ..

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 6x2 − 6x− 12=⇒ f′′(x) = 12x− 6 = 6(2x− 1)

f′′(x)

−−

. . . . . .

Step 3: One sign chart to rule them all

Remember, f(x) = 2x3 − 3x2 − 12x.

. f′(x).monotonicity

..−1

. +.↗

.− .↘

. −.↘

.+ .↗

f′′(x)

concavity

−−

shape of f

−61/2

. . . . . .

Remember, f(x) = 2x3 − 3x2 − 12x.

.. f′(x).monotonicity

..−1

. +.↗

.− .↘

. −.↘

.+ .↗

f′′(x)

concavity

−−

shape of f

−61/2

. . . . . .

Remember, f(x) = 2x3 − 3x2 − 12x.

..−1

. +.↗

.− .↘

. −.↘

.+ .↗

f′′(x)

concavity

−−

shape of f

−61/2

. . . . . .

Remember, f(x) = 2x3 − 3x2 − 12x.

..−1

. +.↗

.− .↘

. −.↘

.+ .↗

f′′(x)

concavity

−−

shape of f

−61/2

. . . . . .

Combinations of monotonicity and concavity

decreasing,concavedown

increasing,concavedown

decreasing,concave up

increasing,concave up

. . . . . .

Remember, f(x) = 2x3 − 3x2 − 12x.

..−1

. +.↗

.− .↘

. −.↘

.+ .↗

f′′(x)

concavity

−−

shape of f

−61/2

. . . . . .

Remember, f(x) = 2x3 − 3x2 − 12x.

..−1

. +.↗

.− .↘

. −.↘

.+ .↗

f′′(x)

concavity

−−

shape of f

−61/2

. . . . . .

Remember, f(x) = 2x3 − 3x2 − 12x.

..−1

. +.↗

.− .↘

. −.↘

.+ .↗

f′′(x)

concavity

−−

shape of f

−61/2

. . . . . .

Remember, f(x) = 2x3 − 3x2 − 12x.

..−1

. +.↗

.− .↘

. −.↘

.+ .↗

f′′(x)

concavity

−−

shape of f

−61/2

. . . . . .

Step 4: Graph

f(x) = 2x3 − 3x2 − 12x

shape of f

−61/2

√105

(−1,7)

..(0,0)..

(1/2,−61/2)..

(2,−20)

.. (3+

√105

. . . . . .

Step 4: Graph

f(x) = 2x3 − 3x2 − 12x

shape of f

−61/2

√105

(−1,7)

..(0,0)..

(1/2,−61/2)..

(2,−20)

.. (3+

√105

. . . . . .

Step 4: Graph

f(x) = 2x3 − 3x2 − 12x

shape of f

−61/2

√105

(−1,7)

..(0,0)..

(1/2,−61/2)..

(2,−20)

.. (3+

√105

. . . . . .

Step 4: Graph

f(x) = 2x3 − 3x2 − 12x

shape of f

−61/2

√105

(−1,7)

..(0,0)..

(1/2,−61/2)..

(2,−20)

.. (3+

√105

. . . . . .

Step 4: Graph

f(x) = 2x3 − 3x2 − 12x

shape of f

−61/2

√105

(−1,7)

..(0,0)..

(1/2,−61/2)..

(2,−20)

.. (3+

√105

. . . . . .

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.

. . . . . .

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.

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

We make its sign chart.

. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0

.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+

. +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

f(x) = x4 − 4x3 + 10

=⇒ f′(x) = 4x3 − 12x2 = 4x2(x− 3)

.. 4x2..0.0.+ . +. +.

(x− 3)

f′(x)

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

Here is its sign chart:

. 12x..0.0

.− . +. +

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

. 12x..0.0

.− . +. +

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0

.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.−

. +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 2: Concavity

f′(x) = 4x3 − 12x2

=⇒ f′′(x) = 12x2 − 24x = 12x(x− 2)

.. 12x..0.0.− . +. +.

x− 2

f′′(x)

−−

. . . . . .

Step 3: Grand Unified Sign Chart

Remember, f(x) = x4 − 4x3 + 10.

f′(x)

monotonicity

f′′(x)

concavity

−−

. . . . . .

Remember, f(x) = x4 − 4x3 + 10.

f′(x)

monotonicity

f′′(x)

concavity

−−

. . . . . .

Remember, f(x) = x4 − 4x3 + 10.

f′(x)

monotonicity

f′′(x)

concavity

−−

. . . . . .

Remember, f(x) = x4 − 4x3 + 10.

f′(x)

monotonicity

f′′(x)

concavity

−−

. . . . . .

Remember, f(x) = x4 − 4x3 + 10.

f′(x)

monotonicity

f′′(x)

concavity

−−

. . . . . .

Step 4: Graph

f(x) = x4 − 4x3 + 10

..(0,10)

..(2,−6)

(3,−17)

. . . . . .

Step 4: Graph

f(x) = x4 − 4x3 + 10

..(0,10)

..(2,−6)

(3,−17)

. . . . . .

Step 4: Graph

f(x) = x4 − 4x3 + 10

..(0,10)

..(2,−6)

(3,−17)

. . . . . .

Step 4: Graph

f(x) = x4 − 4x3 + 10

..(0,10)

..(2,−6)

(3,−17)

. . . . . .

Step 4: Graph

f(x) = x4 − 4x3 + 10

..(0,10)

..(2,−6)

(3,−17)

. . . . . .

Outline

Simple examplesA cubic functionA quartic function

More ExamplesPoints of nondifferentiabilityHorizontal asymptotesVertical asymptotesTrigonometric and polynomial togetherLogarithmic

. . . . . .

Graphing a function with a cusp

Example

Graph f(x) = x+√

This function looks strange because of the absolute value. Butwhenever we become nervous, we can just take cases.

. . . . . .

Graphing a function with a cusp

Example

Graph f(x) = x+√

This function looks strange because of the absolute value. Butwhenever we 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) > 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.

. . . . . .

f(x) = x+√

x is positive.I Are there negative numbers which are zeroes for f?

x+√−x = 0

√−x = −x

−x = x2

x2 + x = 0

. . . . . .

f(x) = x+√

x is positive.I Are there negative numbers which are zeroes for f?

x+√−x = 0

√−x = −x

−x = x2

x2 + x = 0

. . . . . .

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+

limy→+∞

(−y+√y) = lim

y→∞(√y− y) ·

√y+ y√y+ y

= limy→∞

y− y2√y+ y

= −∞

. . . . . .

f(x) = x+√

|x|I lim

I limx→−∞

y→+∞(−y+

limy→+∞

(−y+√y) = lim

y→∞(√y− y) ·

√y+ y√y+ y

= limy→∞

y− y2√y+ y

= −∞

. . . . . .

f(x) = x+√

|x|I lim

I limx→−∞

y→+∞(−y+

limy→+∞

(−y+√y) = lim

y→∞(√y− y) ·

√y+ y√y+ y

= limy→∞

y− y2√y+ y

= −∞

. . . . . .

Step 1: The derivative

Remember, f(x) = x+√

|x|.To find f′, first assume x > 0. Then

f′(x) =ddx

√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)

. . . . . .

f′(x) =ddx

√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) =ddx

√x)= 1+

12√x

I limx→∞

. . . . . .

f′(x) =ddx

√x)= 1+

12√x

I limx→∞

. . . . . .

|x|.If x is negative, we have

f′(x) =ddx

√−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 =

=⇒ −x =14

=⇒ x = −14

. . . . . .

f′(x) =ddx

√−x

)= 1− 1

2√−x

NoticeI lim

I limx→−∞

I f′(x) = 0 when

1− 12√−x

= 0 =⇒√−x =

=⇒ −x =14

=⇒ x = −14

. . . . . .

f′(x) =ddx

√−x

)= 1− 1

2√−x

NoticeI lim

I limx→−∞

I f′(x) = 0 when

1− 12√−x

= 0 =⇒√−x =

=⇒ −x =14

=⇒ x = −14

. . . . . .

f′(x) =

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

.0 ..0.∓∞.+ .− . +.

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

..0.∓∞.+ .− . +.

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞

.+ .− . +.↗

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞.+

.− . +.↗

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞.+ .−

. +.↗

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞.+ .− . +

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞.+ .− . +.

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞.+ .− . +.

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞.+ .− . +.

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞.+ .− . +.

. . . . . .

f′(x) =

12√x

if x > 0

1− 12√−x

if x < 0

.. f′(x).f(x)

..−1

.0 ..0.∓∞.+ .− . +.

. . . . . .

Step 2: Concavity

I If x > 0, then

f′′(x) =ddx

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.−∞.−− .

⌢... −−.

. . . . . .

Step 2: Concavity

I If x > 0, then

f′′(x) =ddx

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

.. f′′(x).f(x)

..0.−∞.−− .

⌢... −−.

. . . . . .

Step 2: Concavity

I If x > 0, then

f′′(x) =ddx

12x−1/2

)= −1

4x−3/2

f′′(x) =ddx

(1− 1

2(−x)−1/2

)= −1

4(−x)−3/2

.. f′′(x).f(x)

..0.−∞.−− .

⌢... −−.

. . . . . .

Step 2: Concavity

I If x > 0, then

f′′(x) =ddx

12x−1/2

)= −1

4x−3/2

f′′(x) =ddx

(1− 1

2(−x)−1/2

)= −1

4(−x)−3/2

.. f′′(x).f(x)

..0.−∞.−− .

⌢... −−.

. . . . . .

Step 3: Synthesis

Now we can put these things together.

f(x) = x+√

..−1

.0 ..0.∓∞.+1 .

↗.+ .

↗.− .

↘. +.

↗. +1.

f′′(x)

concavity

−∞

−−

−∞

. . . . . .

Step 3: Synthesis

f(x) = x+√

..−1

.0 ..0.∓∞.+1 .

↗.+ .

↗.− .

↘. +.

↗. +1.

f′′(x)

concavity

−∞

−−

−∞

. . . . . .

Step 3: Synthesis

f(x) = x+√

..−1

.0 ..0.∓∞.+1 .

↗.+ .

↗.− .

↘. +.

↗. +1.

f′′(x)

concavity

−∞

−−

−∞

. . . . . .

Step 3: Synthesis

f(x) = x+√

..−1

.0 ..0.∓∞.+1 .

↗.+ .

↗.− .

↘. +.

↗. +1.

f′′(x)

concavity

−∞

−−

−∞

. . . . . .

Step 3: Synthesis

f(x) = x+√

..−1

.0 ..0.∓∞.+1 .

↗.+ .

↗.− .

↘. +.

↗. +1.

f′′(x)

concavity

−∞

−−

−∞

. . . . . .

f(x) = x+√

−∞

..(−1,0)

..(−1

4 ,14)

..(0,0)

. . . . . .

f(x) = x+√

−∞

..(−1,0)

..(−1

4 ,14)

..(0,0)

. . . . . .

f(x) = x+√

−∞

..(−1,0)

..(−1

4 ,14)

..(0,0)

. . . . . .

f(x) = x+√

−∞

..(−1,0)

..(−1

4 ,14)

..(0,0)

. . . . . .

f(x) = x+√

−∞

..(−1,0)

..(−1

4 ,14)

..(0,0)

. . . . . .

f(x) = x+√

−∞

..(−1,0)

..(−1

4 ,14)

..(0,0)

. . . . . .

f(x) = x+√

−∞

..(−1,0)

..(−1

4 ,14)

..(0,0)

. . . . . .

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

. . . . . .

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−x2

=(1−

)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+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

.+. −

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

f′(x) = 1 · e−x2 + xe−x2(−2x) =(1− 2x2

)e−x2

=(1−

)e−x2

.. 1−√2x.. √

. 0.+ .+. −.

1+√2x

−√

f′(x)

−√

√1/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

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−

√3/2

√2x+

−√

f′′(x)

−−

−√

√3/2

. . . . . .

Step 3: Synthesis

f(x) = xe−x2

..−√

.0 .. √1/2

. 0.− .↘

.− .↘

.+ .↗

. +.↗

. −.↘

f′′(x)

concavity

−√

√3/2

−−

−√

− 1√2e

√1/2

1√2e

−√

√3/2

√32e3

. . . . . .

Step 3: Synthesis

f(x) = xe−x2

..−√

.0 .. √1/2

. 0.− .↘

.− .↘

.+ .↗

. +.↗

. −.↘

f′′(x)

concavity

−√

√3/2

−−

−√

− 1√2e

√1/2

1√2e

−√

√3/2

√32e3

. . . . . .

Step 3: Synthesis

f(x) = xe−x2

..−√

.0 .. √1/2

. 0.− .↘

.− .↘

.+ .↗

. +.↗

. −.↘

f′′(x)

concavity

−√

√3/2

−−

−√

− 1√2e

√1/2

1√2e

−√

√3/2

√32e3

. . . . . .

Step 3: Synthesis

f(x) = xe−x2

..−√

.0 .. √1/2

. 0.− .↘

.− .↘

.+ .↗

. +.↗

. −.↘

f′′(x)

concavity

−√

√3/2

−−

−√

− 1√2e

√1/2

1√2e

−√

√3/2

√32e3

. . . . . .

Step 3: Synthesis

f(x) = xe−x2

..−√

.0 .. √1/2

. 0.− .↘

.− .↘

.+ .↗

. +.↗

. −.↘

f′′(x)

concavity

−√

√3/2

−−

−√

− 1√2e

√1/2

1√2e

−√

√3/2

√32e3

. . . . . .

Step 3: Synthesis

f(x) = xe−x2

..−√

.0 .. √1/2

. 0.− .↘

.− .↘

.+ .↗

. +.↗

. −.↘

f′′(x)

concavity

−√

√3/2

−−

−√

− 1√2e

√1/2

1√2e

−√

√3/2

√32e3

. . . . . .

Step 3: Synthesis

f(x) = xe−x2

..−√

.0 .. √1/2

. 0.− .↘

.− .↘

.+ .↗

. +.↗

. −.↘

f′′(x)

concavity

−√

√3/2

−−

−√

− 1√2e

√1/2

1√2e

−√

√3/2

√32e3

. . . . . .

Step 4: Graph

f(x) = xe−x2

(−√

1/2,− 1√2e

(√1/2, 1√

(−√

3/2,−√

(√3/2,

√32e3

. f(x).shape

..−√

.− 1√

.. √1/2

.1√2e.

..−√

.−√

32e3 .

..0.0.

.. √3/2

√32e3.

. . ." . ". .

. . . . . .

Step 4: Graph

f(x) = xe−x2

(−√

1/2,− 1√2e

(√1/2, 1√

(−√

3/2,−√

(√3/2,

√32e3

. f(x).shape

..−√

.− 1√

.. √1/2

.1√2e.

..−√

.−√

32e3 .

..0.0.

.. √3/2

√32e3.

. . ." . ". .

. . . . . .

Step 4: Graph

f(x) = xe−x2

(−√

1/2,− 1√2e

(√1/2, 1√

(−√

3/2,−√

(√3/2,

√32e3

. f(x).shape

..−√

.− 1√

.. √1/2

.1√2e.

..−√

.−√

32e3 .

..0.0.

.. √3/2

√32e3.

. . ." . ". .

. . . . . .

Step 4: Graph

f(x) = xe−x2

(−√

1/2,− 1√2e

(√1/2, 1√

(−√

3/2,−√

(√3/2,

√32e3

. f(x).shape

..−√

.− 1√

.. √1/2

.1√2e.

..−√

.−√

32e3 .

..0.0.

.. √3/2

√32e3.

. . ." . ". .

. . . . . .

Step 4: Graph

f(x) = xe−x2

(−√

1/2,− 1√2e

(√1/2, 1√

(−√

3/2,−√

(√3/2,

√32e3

. f(x).shape

..−√

.− 1√

.. √1/2

.1√2e.

..−√

.−√

32e3 .

..0.0.

.. √3/2

√32e3.

. . ." . ". .

. . . . . .

Step 4: Graph

f(x) = xe−x2

(−√

1/2,− 1√2e

(√1/2, 1√

(−√

3/2,−√

(√3/2,

√32e3

. f(x).shape

..−√

.− 1√

.. √1/2

.1√2e.

..−√

.−√

32e3 .

..0.0.

.. √3/2

√32e3.

. . ." . ". .

. . . . . .

Step 4: Graph

f(x) = xe−x2

(−√

1/2,− 1√2e

(√1/2, 1√

(−√

3/2,−√

(√3/2,

√32e3

. f(x).shape

..−√

.− 1√

.. √1/2

.1√2e.

..−√

.−√

32e3 .

..0.0.

.. √3/2

√32e3.

. . ." . ". .

. . . . . .

Example with Vertical Asymptotes

Example

Graph f(x) =1x+

. . . . . .

Step 0

Find when f is positive, negative, zero, not defined.

We need to factor f:

f(x) =1x+

=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

.− . +.

. . . . . .

Step 0

Find when f is positive, negative, zero, not defined. We need to factor f:

f(x) =1x+

=x+ 1x2

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

.− . +.

. . . . . .

Step 0

Find when f is positive, negative, zero, not defined. We need to factor f:

f(x) =1x+

=x+ 1x2

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

.− . +.

. . . . . .

Step 0, continued

For horizontal asymptotes, notice that

limx→∞

x+ 1x2

so y = 0 is a horizontal asymptote of the graph. The same is true at−∞.

. . . . . .

We havef′(x) = − 1

x2− 2

x3= −x+ 2

The critical points are x = −2 and x = 0. We have the following signchart:

.. −(x+ 2)..0 .−2

.+ . −.

f′(x)

. . . . . .

x2− 2

x3= −x+ 2

.. −(x+ 2)..0 .−2

.+ . −.

f′(x)

. . . . . .

x2− 2

x3= −x+ 2

.. −(x+ 2)..0 .−2

.+ . −.

f′(x)

. . . . . .

x2− 2

x3= −x+ 2

.. −(x+ 2)..0 .−2

.+ . −.

f′(x)

. . . . . .

x2− 2

x3= −x+ 2

.. −(x+ 2)..0 .−2

.+ . −.

f′(x)

. . . . . .

x2− 2

x3= −x+ 2

.. −(x+ 2)..0 .−2

.+ . −.

f′(x)

. . . . . .

x2− 2

x3= −x+ 2

.. −(x+ 2)..0 .−2

.+ . −.

f′(x)

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

The critical points of f′ are −3 and 0. Sign chart:

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 2: Concavity

We havef′′(x) =

=2(x+ 3)

.. (x+ 3)..0 .−3

.− . +.

f′′(x)

−−

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 3: Synthesis

monotonicity

f′′

concavity

−−

shape of f

−1/4

−2/9

−∞

. . . . . .

Step 4: Graph

(−3,−2/9)

(−2,−1/4)

shape of f

−1/4

−2/9

−∞

. . . . . .

Trigonometric and polynomial together

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 negative at all other points.

.. f′(x).f(x)

..−π/2

.0 ..3π/2

. 0..7π/2

. −.↘

. . . . . .

0 ≤ sin x+ 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x+ 1) ≤ 0

.. f′(x).f(x)

..−π/2

.0 ..3π/2

. 0..7π/2

. 0. −

. −.↘

. . . . . .

0 ≤ sin x+ 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x+ 1) ≤ 0

.. f′(x).f(x)

..−π/2

.0 ..3π/2

. 0..7π/2

. 0. −

. . . . . .

0 ≤ sin x+ 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x+ 1) ≤ 0

.. f′(x).f(x)

..−π/2

.0 ..3π/2

. 0..7π/2

. 0. −.↘

. . . . . .

0 ≤ sin x+ 1 ≤ 2 =⇒ −2 ≤ (−1)(sin x+ 1) ≤ 0

.. 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

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−

.⌢. ++.

⌣. −−.

⌢. ++.

..−π/2

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−

. −−.⌢

. ++.⌣

..−π/2

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−

. −−

. ++.⌣

..−π/2

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−

. −−

..−π/2

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++

. −−

..−π/2

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++.

⌣. −−

..−π/2

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++.

⌣. −−.

⌢. ++

..−π/2

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++.

⌣. −−.

⌢. ++.

−π/2.0

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++.

⌣. −−.

⌢. ++.

−π/2.0 .

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++.

⌣. −−.

⌢. ++.

−π/2.0 .

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++.

⌣. −−.

⌢. ++.

−π/2.0 .

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++.

⌣. −−.

⌢. ++.

−π/2.0 .

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 2: Concavity

.. f′′(x).f(x)

.−−.⌢. ++.

⌣. −−.

⌢. ++.

−π/2.0 .

..π/2

..3π/2

..5π/2

..7π/2

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..−π/2

.0 ..3π/2

. 0..7π/2

. 0. −.↘

. −.↘

f′′(x)

−π/2

−−

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..−π/2

.0 ..3π/2

. 0..7π/2

. 0. −.↘

. −.↘

f′′(x)

−π/2

−−

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..−π/2

.0 ..3π/2

. 0..7π/2

. 0. −.↘

. −.↘

f′′(x)

−π/2

−−

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..−π/2

.0 ..3π/2

. 0..7π/2

. 0. −.↘

. −.↘

f′′(x)

−π/2

−−

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..−π/2

.0 ..3π/2

. 0..7π/2

. 0. −.↘

. −.↘

f′′(x)

−π/2

−−

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 4: Graph

f(x) = cos x− x

.y.......

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 4: Graph

f(x) = cos x− x

.y.......

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 4: Graph

f(x) = cos x− x

.y.......

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 4: Graph

f(x) = cos x− x

.y.......

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 4: Graph

f(x) = cos x− x

.y.......

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 4: Graph

f(x) = cos x− x

.y.......

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Step 4: Graph

f(x) = cos x− x

.y.......

−π/2

−3π/2

−5π/2

−7π/2

. . . . . .

Logarithmic

ProblemGraph f(x) = x ln x2

. . . . . .

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

1/xH= lim

(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

. . . . . .

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.

.+ .↗

.− .↘

. −.↘

. +.↗

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

.− .↘

. −.↘

. +.↗

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

. −.↘

. +.↗

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

. +.↗

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.−

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

. . . . . .

f′(x) = ln x2 + x · 1x2

(2x) = ln x2 + 2

ln x2 = −2 =⇒ x2 = e−2 =⇒ x = ±e−1

.. f′(x).f(x)

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

. . . . . .

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

.−− .⌢

. ++.⌣

. . . . . .

Step 2: Concavity

.. f′(x).f(x)

..×.0.−−

. ++.⌣

. . . . . .

Step 2: Concavity

.. f′(x).f(x)

..×.0.−−

. . . . . .

Step 2: Concavity

.. f′(x).f(x)

..×.0.−− .

⌢. ++

. . . . . .

Step 2: Concavity

.. f′(x).f(x)

..×.0.−− .

⌢. ++.

. . . . . .

Step 2: Concavity

.. f′(x).f(x)

..×.0.−− .

⌢. ++.

. . . . . .

Step 2: Concavity

.. f′(x).f(x)

..×.0.−− .

⌢. ++.

. . . . . .

Step 2: Concavity

.. f′(x).f(x)

..×.0.−− .

⌢. ++.

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

f′(x)

−−

f′(x)

−1/e

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

f′(x)

−−

f′(x)

−1/e

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

f′(x)

−−

f′(x)

−1/e

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

f′(x)

−−

f′(x)

−1/e

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

f′(x)

−−

f′(x)

−1/e

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

f′(x)

−−

f′(x)

−1/e

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

f′(x)

−−

f′(x)

−1/e

. . . . . .

Step 3: Synthesis

.. f′(x).mono

..0 .−1/e

..×.0.. 0.

1/e.+ .

↗.− .

↘. −.↘

. +.↗

f′(x)

−−

f′(x)

−1/e

. . . . . .

Step 4: Graph

. . . . . .

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?

lesson 21: curve sketching (section 041 slides)

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