1. Summary of curve sketching a Domain of .f x

b Critical numbers of .f xFind all the numbers in the domain of c f

c Intervals of increase or decrease.Such that f’(x)=0 or f”(x) does not exist

d Local maximum and Local minimum.Determine if a critical number is a localmaximum or local minimum.

e Concavity

f Points of inflection

g Intercepts

i -intercept; find such that 0 .y y y f

for x.

h Asymptotes

(ii) x-intercept: if possible solve f(x)=0

(i) Horizontal Asymptotes

ii Vertical Asymptote : is a V.A. ifx a lim or lim .

x a x af x f x

If f(-x)=f(x), then f(x) is symmetric w.r.t. the y- axis. If f(-x)=-f(x), then f(x) is symmetric w.r.t. the origin.

(i) Symmetry

domain and some fixed p 0,

j Sketch the graph of using a - i .f

(i) Period: If f(x + p)=f(x) for all x in its

Then f(x) is periodic with period p.

Example

Solution:

a Domain: of .f x All real numbers.

b Critical numbers :

x 2.1

or3

x

2(i): f(x)=2x3 +5x2 - 4x

f′(x)= 6x2 +10x -4

=2(3x2 +5x-2)=0

f 2 1 3/

2 1 3/f

-local maximum

-local minimum

c Intervals of increase or decrease.

d Local maximum and Local mimimum.

f′(-2)=12

f′(1/3)=-19/27

12 10x 0 x 5 6/

f 5 6/

f 5 6/

f Points of inflection :

CD CU

(e) Concavity :

f″(x)=

(-5/6, f(-5/6))=(-5/6, 5.65)

g Intercepts :

5 157,

4 4

5 157.

4 4

h Asymptotes :

y-intercept: f(0)=0

x-intercept: f(x)=0, when x=0,

as x→∞, f(x)→∞as x→-∞, f(x)→-∞

No Horizontal Asymptotes or Vertical Asymptotes

(i) Symmetry

f(-x)=2(-x)3 +5(-x)2 – 4(-x)

=-2x3 +5x2 + 4x

No symmetry

j Graph

2 1 3/f′(x)

j Graph

2 1 3/f′(x)

j Graph

2 1 3/

5 6/CD CU

f′(x)

f″(x)

j Graph

2 1 3/

5 6/CD CU

j Graph

2 1 3/

5 6/CD CU

j Graph

2 1 3/

5 6/CD CU

j Graph

2 1 3/

5 6/CD CU

2

2

12 ii . ( ) .

1

xf x

x

Example:

Solution:

a Domain: x 1

b Critical numbers:

22

4

1

x

x

x 0

The only critical number is 0.x

f′(x)=

f′(x)=0 ˂═˃

c Intervals of increase or decrease.

f0

f0

d Local maximum and Local minimum.

Local min. at 0, 0 1.x f No local maximum.

e Concavity:

2

3 3

4 1 3

1 1

x

x x

1 x 1 x 1 3 2 x

x 1 1 1x

1 x

CD CU CD

1 1f

f″(x) f(x)

f″(x)=

1 1CD CDCU

e Concavity:

f(x)

f″(x)_

+ _

f Points of inflection:There are no points of inflection since

g Intercepts:

1 02 x

No - intercepts.x

1 01

1 0

2

2

1( )

1

xf x

x

X=±1 are not in the domain of f(x).

y-intercept: f(0)=

x-intercept:

h Asymptotes:

22

22

11

lim1

1x

xx

xx

0 11

0 1

is an H.A.y 1

Horizontal

ii Vertical Asymptotes:

2

21

1lim

1x

x

x

2

21

1lim

1x

x

x

2

21

1lim

1x

x

x

2

21

1lim

1x

x

x

and are V.A.x x1 1

2(-∞)=

2(∞)=

2(∞)=

2(-∞)=

i Symmetry:

2

2

1

1

x

x

2

2

1

1

x

x

is symmetry w.r.t. the - axis.f y

f(-x)= f(x)

j Graph:

f 0

j Graph:

f 0

j Graph:

1 1f CD CDCU

f 0

j Graph:

1 1f CD CDCU

f 0

j Graph:

1 1f CD CDCU

f 0

j Graph:

1 1f CD CDCU

f 0

j Graph:

1 1f CD CDCU

f 0

j Graph:

1 1f CD CDCU

f 0

is periodic with period 2 . Firstf x draw the graph on Then extend0 2, .it to all of R.

Period:

Example 2(iii) f(x)=2sin(x)+sin2(x)

f(x+2π)

=2sin(x+2π)+sin2(x+2π)

=2sin(x)+sin2(x)=f(x)

a Domain: All real numbers.

b Critical numbers:

or2

x

3

.2

f′(x)= 2 cos(x)+2 sin(x) cos(x)

=2 cos(x)(1+sin(x))

f′(x)=0 ˂=˃ 2cos(x)=0 or sin(x)=-1

c Intervals of increase or decrease.f

/ 2 3 2 /

/ 2 3 2 /

f

d Local maximum and Local minimum.

- local max.

- local min.

f(π/2)=2+1

f(3π/2)=-2+1=-1

e Concavity:

2 sin 1 2sin 1x x

3 or

2x

or

6x

5.

6

f 0 / 6 5 6 / 3 2 / 2

f0 / 6 5 6 / 3 2 / 2CU CD CU CU

f″(x)=

f″(x)=0 ˂=˃ sin(x)=-1, or sin(x)=1/2

f Points of inflection:

21 1

22 2

5

4

21 1

22 2

5

4

5 5 5, and , are inflection points.

6 4 6 4

f(π/6)

f(π/6)

g Intercepts:

0

x 0, , 2 .

No asymptotes.(h) Asymptotes:

y-intercept: f(0)=2 sin(0)+ sin 2(0)=0

(0,0) is the y-intercept

x-intercept: f(x)=

2 sin(x)+sin2 (x)=0

sin(x)(2+sin(x))=0

sin (x)=0 ═˃

j Graph:

/ 2 3 2 /f

/ 2 3 2 /f

j Graph:

/ 2 3 2 /f

j Graph:

f0 / 6 5 6 / 3 2 / 2CU CD CU CU

f0 / 6 5 6 / 3 2 / 2CU CD CU CU

j Graph:

/ 2 3 2 /f

j Graph:

/ 2 3 2 /

f0 / 6 5 6 / 3 2 / 2CU CD CU CU

f

j Graph:

/ 2 3 2 /

f0 / 6 5 6 / 3 2 / 2CU CD CU CU

f

2 / 31/ 32 iv . 3 .f x x x Example:

a Domain: All real numbers.

b Critical numbers:

1/ 32 / 3

1

3

x

x x

x 1.

critical numbers are x 1,

x 3

0, and 3.

f′(x)=

f′(x)=o =˃

f′(x) does not exist if x=0 ,

c Intervals of increase or decrease.

3 1 0

d Local maximum and Local minimum. 1/ 3 2 / 33 ( 3) ( 3 3)f local max.

local minimum.0.

f

f(-1)=-(2)2/3

e Concavity:

4 / 35/ 3

2

3 3x x

f0

0f

f Points of inflection:

0, 0f 0,0 - inflection point.

CU CD

f″(x)=

g Intercepts:

orx 0 x 3.

No vertical or horizontal asymptotes.

h Asymptotes:

y-intercept: f(0)=0x-intercept: x1/3 (x+3)2/3 =0

Limx→∞ x1/3 (x+3)2/3 =∞

Limx→-∞ x1/3 (x+3)2/3 =-∞

i Symmetry:

No symmetry.

f(-x)≠ f(x)

f(-x)≠- f(x)

j Graph:

3 1 0f

j Graph:

3 1 0f

j Graph:

3 1 0f

f0

CU CD

j Graph:

3 1 0f

f0

CU CD

j Graph:

3 1 0f

f0

CU CD

j Graph:

3 1 0f

f0

CU CD

3 1 0f

f0

CU CD

j Graph: