1. Summary of curve sketching Such that f’(x)=0 or f”(x) does not exist.
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Transcript of 1. Summary of curve sketching Such that f’(x)=0 or f”(x) does not exist.
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: