Upload
paolo-gochingco
View
219
Download
0
Embed Size (px)
Citation preview
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 1/176
LIMITS
of FUNCTIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 2/176
LIMITS OF FUNCTIONS
OBJECTIVES:
define limits;
illustrate limits and its theorems; and
evaluate limits applying the given theorems.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 3/176
DEFINITION: Limits
The most basic use of limits is to describe how a
function behaves as the independent variableapproaches a given value. For example let us
examine the behavior of the function
for x-values closer and closer to 2. It is evident from
the graph and the table in the next slide that the
values of f(x) get closer and closer to 3 as the values
of x are selected closer and closer to 2 on either the
left or right side of 2. We describe this by sayingthat the limit of is 3 as x
approaches 2 from either side, we write
1 x x ) x( f 2 !
1 x x ) x( f 2 !
31 x xlim
2
2 x !p
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 4/176
2
3
f(x)
f(x)
x
y
1 x x y 2 !
x 1.9 1.95 1.99 1.995 1.999 2 2.001 2.005 2.01 2.05 2.1
F(x) 2.71 2.852 2.97 2.985 2.997 3.003 3.015 3.031 3.152 3.31
left side r ight side
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 5/176
1.1.1 (p. 70)Limits (An Informal View)
This leads us to the following general idea.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 6/176
EXAMPLE
Use numerical evidence to make a conjecture about
the value of
1 x
1 xlim
1 x
p
Although the function , this has no
bearing on the limit.
The table shows sample x-values approaching 1 from
the left side and from the right side. In both cases the
corresponding values of f(x) appear to get closer andcloser to 2, and hence we conjecture that
and is consistent with the graph of f.
1 x
1 x ) x( f
!
2
1 x
1 xlim
1 x!
p
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 7/176
Figure 1.1.9 (p. 71)
x .99 .999 .9999 .99999 1 1.00001 1.0001 1.001 1.01
F(x) 1.9949 1.9995 1.99995 1.999995 2.000005 2.00005 2.0005 2.004915
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 8/176
THEOREMS ON LIMITS
Our strategy for finding limits algebraically has two parts:
First we will obtain the limits of some simpler function
Then we will develop a list of theorems that will enable us
to use the limits of simple functions as building blocks for
finding limits of more complicated functions.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 9/176
We start with the following basic theorems, which are
illustrated in Fig 1.2.1
Theorem 1.2.1 (p. 80)
a xk k !!pp axax
lim b lim a
numbers.real bek andaLetTheorem 1.2.1
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 10/176
Figure 1.2.1 (p. 80)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 11/176
33lim 33lim 33lim
example, For
a.of valuesall for a xa sk f(x)
whyexpl ainswhichvaries, xa sk at fixed remain
f(x)of valuest het hen funct ion,constant aisk x f If
x0 x-25 x!!!
pp
!
ppp T
Example 1.
TT
!!!
pp!
ppp x x x
If
x-2x0xlim 2lim 0lim
example,For
.axf thattrue bealsomustitathen xx,xf
Example 2.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 12/176
Theorem 1.2.2 (p. 81)
The following theorem will be our basic tool for finding limits
algebraically
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 13/176
This theorem can be stated informally as follows:
a) The limit of a sum is the sum of the limits.
b) The limit of a difference is the difference of the limits.
c) The limits of a product is the product of the limits.
d) The limits of a quotient is the quotient of the limits,provided the limit of the denominator is not zero.
e) The limit of the nth root is the nth root of the limit.
A constant factor can be moved through a limit symbol.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 14/176
31
58
5 )4( 2
5lim xlim2
5lim x2lim 5 x2lim .1
4 x4 x
4 x4 x4 x
!
!
!!
!
pp
ppp
6
12-18
12 )3( 6
12lim x6 lim12 x6 lim .23 x3 x3 x
!
!!
!ppp
13
131
2 )3( 534
2lim xlim5 xlim4lim
2lim x5lim xlim4lim
2 x5lim x4lim )2 x5( x4lim .3
3 x3 x3 x3 x
3 x3 x3 x3 x
3 x3 x3 x
!
!
!!
!
!
pppp
pppp
ppp
EXAMPLE Evaluate the following limits.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 15/176
21
10
425
52
4lim xlim5
x lim2
4lim x5lim
x2 lim
4 x5
x2lim .4
5 x5 x
5 x
5 x5 x
5 x
5 x
!
!
!
!
pp
p
pp
p
p
3375
156 33
6 lim xlim3
6 lim x3lim
6 x3lim6 x3lim .5
33
3
3 x3 x
3
3 x3 x
3
3 x
3
3 x
!
!!
!
!
!
pp
pp
pp
2
3
4
9
3 x
1 x8lim
3 x
1 x8lim .6
1 x1 x
!!
!
pp
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 16/176
OR
When evaluating the limit of a function at a givenvalue, simply replace the variable by the indicated
limit then solve for the value of the function:
22
3lim 3 4 1 3 3 4 3 1
27 12 1
38
x x x
p !
!
!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 17/176
EXAMPLE Evaluate the following limits.
2 x
8 xlim .1
3
2 x
p
Solution:
0
0
0
88
22
82
2 x
8 xlim
33
2 x!
!
!
p
2
2
2
2
2
3
2
2 2 4lim
2
lim 2 4
2 2 2 4
4 4 4
12
8lim 122
x
x
x
x x x
x
x x
x
x
p
p
p
!
!
! !
!
@ !
Equivalent function:
(indeterminate)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 18/176
Note: In evaluating a limit of a quotient which
reduces to , simplify the fraction. Just removethe common factor in the numerator and
denominator which makes the quotient .
To do this use factoring or rationalizing thenumerator or denominator, wherever the radical is.
0
0
0
0
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 19/176
0
2 2 2 2 0lim
0 0 x
x
x p
! !
0 0
0
0
2 2 2 2 2 2lim lim
2 2 2 2
1 1 1 2
lim 42 2 2 2 2 2
2 2 2lim
4
x x
x
x
x x x
x x x x
x
x
x
p p
p
p
!
! ! !
@ !
x
22 xlim .2
0 x
p
Solution:
Rationalizing the numerator:
(indeterminate)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 20/176
9 x4
27 x8lim .3
2
3
0 x
p
Solution:
Rationalizing the numerator:
(indeterminate)3
2
3
3
22
38 27
8 27 27 27 02lim
4 9 9 9 034 9
2
x
x
x p
¨ ¸© ¹ ª º! ! !
¨ ¸ © ¹ª º
3 3
2 2
32
23
2
2
2 3 4 6 98 27lim lim
4 9 2 3 2 3
4 6 9 9 9 9lim
2 3 3 3
27 9 3 3 2
6 2 22
x x
x
x x x x
x x x
x x
x
p p
p
!
! !
! ! ! !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 21/176
5 x
3 x2 xlim .4
2
3
2 x
p
Solution:
33
222
2 2 2 32 3lim
5 2 5
8 4 34 5
15
9
15
3
x
x x
x p
!
!
!
!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 22/176
EXERCISES
5w4w
7 w7 wlim 10.
2 x
8 xlim .5
19 x9 x2lim 9. 4 y
y8 y4lim .4
1 y2 y
3 y2 y1 ylim 8.
1 x
4 x3 xlim .3
1 x
3 x2 x3 x2lim 7 .
4 x3 x
1 x2lim .2
1 x9
1 x3
lim 6 . 2 x5 x4lim .1
2
2
1w
3
2 x
2
134
5 x
3
13
2 y
2
2
1 y3
2
1 x
2
23
1 x21 x
2
3
1 x
2
3 x
¹¹ º
¸
©©ª
¨
pp
pp
pp
pp
pp
Evaluate the following limits.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 23/176
LIMITS
of FUNCTIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 24/176
ONE-SIDED LIMITS
OBJECTIVES:
define one-sided limits
illustrate one-sided limits
investigate the limit if it exist or not using theconcept of one-sided limits.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 25/176
DEFINITION: One-Sided Limits
T
he limit of a function is called two-sided limit if itrequires the values of f(x) to get closer and closer to
a number as the values of x are taken from either
side of x=a. However some functions exhibit
different behaviors on the two sides of an x-value ain which case it is necessary to distinguish whether
the values of x near a are on the left side or on the
right side of a for purposes of investigating limitingbehavior.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 26/176
C
onsider the function
°¯®
"!!
0 x ,1
0 x ,1
x
x ) x( f
Q
Q1
-1
As x approaches 0 from the right, the
values of f(x) approach a limit of 1, and
similarly , as x approaches 0 from the
left, the values of f(x) approach a
limit of -1.
1lim and 1lim
,
!!
pp x
x
x
x
symbols In
o xo x
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 27/176
1.1.2 (p. 72)
One-Sided Limits (An Informal View)
This leads to the general idea of a one-sided limit
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 28/176
1.1.3 (p. 73)
The Relationship Between One-Sided and Two-Sided Limits
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 29/176
EXAMPLE:
x
x ) x( f !1. Find if the two sided limits exist given .
Q
Q1
-1
exist.notdoes lim or
existnotdoes lim
limlim sin
1lim and 1lim
x
xit
sided t
wot
het
hen
x
x
x
xt hece
x
x
x
x
o x
o xo x
o xo x
p
pp
pp
{
!!SOLUTION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 30/176
EXAMPLE:
2. For the functions in Fig1.1.13, find the one-sided
limit and the two-sided limits at x=a if they exists.
The functions in all three figures have the same
one-sided limits as , since the functions are
Identical, except at x=a.
a x p
1 ) x( f lim and 3 ) x( f lim
are it slim These
a xa x!!
pp
In all three cases the two-sided limit does not exist as
because the one sided limits are not equal.a x p
SOLUTION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 31/176
Figure 1.1.13 (p. 73)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 32/176
3. Find if the two-sided limit exists and sketch the graph
of .2
6+x if x < -2( ) =
x if x -2g x
® ¾¯ ¿
u° À
4
26
x6 lim ) x( g lim.a2 x2 x
!
!
! pp
4
2-
xlim ) x( g lim.b
2
2
2 x2 x
!
!
! pp
4 ) x( g lim or 4t oequal isand exist it lim sided t wo t he t hen
) x( g lim ) x( g limt he ce sin
2 x
2 x2 x
!
!
p
pp
SOLUTION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 34/176
4. Find if the two-sided limit exists and sketch the graph
of and sketch the graph.
2
2
3 + x if x < -2
( ) = 0 if x = -2
11 - x if x > -2
f x
® ¾± ±¯ ¿
± ±° ÀSOLUTION
7
23
x3lim ) x( f lim.a
2
2
2 x2 x
!
!
! pp
7
2-11
x11lim ) x( f lim.b
2
2
2 x2 x
!
!
! pp
7 ) x( f lim or
7 t oequal isand exist it lim sided t wo t he t hen
) x( f lim ) x( f limt he ce sin
2 x
2 x2 x
!
!
p
pp
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 35/176
gr a ph.t he sket chand
,exist f(x)limif eminer det ,4 x23 ) x( f If .52 xn
!
3
4223
4 x23lim ) x( f lim .a2 x2 x
!
!
! pp
3
4223
4 x23lim ) x( f lim .b2 x2 x
!
!
! pp
3 ) x( f lim or
3t oequal isand exist it lim sided t wo t he t hen
) x( f lim ) x( f limt he ce sin
2 x
2 x2 x
!
!
p
pp
SOLUTION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 36/176
f(x)
x
(2,3)
2
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 37/176
EXERCISES Sketch the graph of the following functions and
the indicated limit if it exists. find
a.
.
) x( g lim.c g(x)lim.b g(x)lim.a
1 x if 2x-7
1 x if 2
1 x if 3 x2
) x( g .2
) x( f lim.c f(x)lim.b f(x)lim.a
4- x if 4 x
-4 x if x4 ) x( f .1
1 x1 x1 x
4 x4 x4 x
ppp
ppp
±°
±¯
®
"
!
!
°¯®
e
"!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 38/176
.
) x( f lim.c f(x)lim.b f(x)lim.a
1 x2 ) x( g .5
) x( f lim.c f(x)lim.b f(x)lim.a
x4 ) x( g .4
) x( f lim.c f(x)lim.b f(x)lim.a
0 x if 3
0 x if x ) x( f .3
2
1 x
2
1 x
2
1 x
4 x4 x4 x
0 x0 x0 x
ppp
ppp
ppp
!
!
°¯®
!
{!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 39/176
LIMITS
of FUNCTIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 40/176
INFINITE LIMITS
OBJECTIVES:
define infinite limits;
illustrate the infinite limits; and
use the graphs to evaluate the
limits of functions.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 41/176
DEFINITION: INFINITE LIMITS
Sometimes one-sided or two-sided limits fail to exist
because the value of the function increase or
decrease without bound.
For example, consider the behavior of for
values of x near 0. It is evident from the table and
graph in Fig 1.1.15 that as x values are taken closer
and closer to 0 from the right, the values of
are positive and increase without bound; and asx-values are taken closer and closer to 0 from the
left, the values of are negative and
decrease without bound.
x
1 ) x( f !
x
1 ) x( f !
x
1 ) x( f !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 42/176
In symbols, we write
g!g!pp
x
1lim and
x
1lim
0 x0 x
Note:
The symbols here are not real
numbers; they simply describe particular ways in
which the limits fail to exist. Thus it is incorrect to
write .
gg and
0!gg
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 43/176
Figure 1.1.15 (p. 74)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 44/176
1.1.4 (p. 75)
Infinite Limits (An Informal View)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 45/176
Figure 1.2.2 (p. 84)
Figure 1.1.2 illustrate graphically the limits for rational
functions of the form
22
a x
1 ,
a x
1 ,
a x
1
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 46/176
EXAMPLE: Evaluate the following limits:
40 x x
1 lim .1 p
40 x x
1 lim.2
p
50 x x
1 lim.4 p
g!!
p 01
x1 lim 5
0 x
g!!
p 0
1
x
1 lim
40 x
g!!p 0
1
x
1 lim
40 x
g!!
p 0
1
x
1 lim
50 x
50 x x
1 lim.5
p
g!p 40 x
x
1 lim .3 g!
p 50 x x
1 lim.6
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 47/176
2 x
x3 lim a.
.5
2 x p
2 x
x3 lim.b
2 x p
g
!
!!
p 0
6
0
23
2- x
3x lim-2 x
2.028.12 x
8.1 sa y ,lef t from2t oclose xof value
takewemeans2 x
!!
p
g!
!
!!
p 0
6
0
23
2- x
3x lim
2 x
1.021.22 x
1.2 sa y ,right from2t oclose xof value
takewemeans2 x
!!
p
g!p 2 x
x3 lim .c2 x
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 48/176
) x( f lima x p
) x( f lima x p
) x( f lima xp
g g g
g
g
g g ggg
gg
SUMMARY: ) x( Q
) x( R ) x( f If !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 49/176
EXAMPLE
g!¹
º
¸©
ª
¨
!g!
p
pp
3 x
2
3 x
2 limt hen
3
1
3 x
2
lim and 3 x
2
lim .1
3 x
3 x3 x
g!g g!g cc
nSubt r act io / Addit ion: Not e
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 50/176
g!gg!g
g!gg!g
c c
c c
: Not e
g!¹ º
¸©ª
¨
!
g!
p
pp
1 x
3 x
1 x
x2 limt hen
1
1 x
3 x lim and
1 x
x2 lim .2
1 x
1 x1 x
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 51/176
g!¹ º
¸©ª
¨
!
g!
p
pp
4 x
6 x2
2 x
x3 limt hen
3
1
4 x
6 x2 lim and
2 x
x3 lim .3
2 x
2 x2 x
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 52/176
32
2 lim.4
4
16 lim .3
4
2 lim .2
9
4 lim .1
2
2
3
2
4
22
2
2
3
p
p
p
p
x x
x x
x
x
t
t
x
x
x
x
x
x
EXERCISES: Evaluate the following limits:
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 53/176
LIMITS
of FUNCTIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 54/176
LIMITS AT INFINITY
OBJECTIVES:
define limits at infinity;
illustrate the limits at infinity; anduse the graphs to find the limits of
functions
DEFINITION LIMITS AT INFINITY
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 55/176
DEFINITION: LIMITS AT INFINITY
If the values of the variable x increase without
bound, then we write , and if the values of
x decrease without bound, then we write .
The behavior of a function as x increases or
decreases without bound is sometimes called the
end behavior of the function.
) x( f
gp xgp x
F
or example ,
0 x
1lim and 0
x
1lim
x x!!
gpgp
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 56/176
x
x
0 x
1
lim x !gp 0 x
1lim x !gp
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 57/176
1.3.1(p
.8
9)
Limits at Infinity (An Informal View)
In general, we will use the following notation.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 58/176
Figure 1.3.2 (p. 89)
Fig.1.3.2 illustrates the end behavior of the function f when
L ) x( f lim or L ) x( f lim x x
!!gpgp
EXAMPLE
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 59/176
Figure 1.3.4 (p. 90)
Fig.1.3.2 illustrates the graph of . As suggested by
this graph,
EXAMPLE x
x
11 y ¹
º
¸©ª
¨ !
e x
11lim
and e x
11lim
x
x
x
x
!¹ º
¸©ª
¨
!¹ º
¸©ª
¨
gp
gp
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 60/176
EXAMPLES:
x
x x x
x x
x
x
x
x
x
31
125lim.3
52
4lim .2
86
53lim .1
23
3
2
gp
gp
gp
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 61/176
LIMITS
of FUNCTIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 62/176
SQUEEZING THEOREM
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 65/176
LIMITS
of FUNCTIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 66/176
C ontinuity of Functions
OBJE C TIVES:
At the end of the lesson, the students
should be able to:
define continuity and discontinuity;
illustrate continuity and discontinuity; and
verify the continuity and discontinuity of a
function
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 72/176
THE DERIVATIVEOBJECTIVES:
to define the derivative of a function
to find the derivative of a function by
increment method(4-step rule)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 73/176
Derivative of a Function
The process of finding the derivative of a function
is called differentiation and the branch of calculus
that deals with this process is called differential
calculus. Differentiation is an important
mathematical tool in physics, mechanics,
economics and many other disciplines that involve
change and motion.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 74/176
Consider a point on the curve ))(,( 22 x f xQ ),( x f y !
that is distinct from and compute the
slope of the secant line through P and Q.
)),(,( 11 x f x P
PQm
x
x f x f m PQ
(
!
)()(12 where
12
x x x !(
x x x (! 12and
x
x f x x f
m PQ (
(
!
)()( 11
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 75/176
2 x ,
1 xIf we let approach then the point Q will
move along the curve and approach point P. Aspoint Q approaches P, the value of approaches
zero and the secant line through P and Q
approaches a limiting position, then we will
consider that position to be the position of the
tangent line at P.
x(
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 76/176
))(,(11
x f x P ))(,( 22 x f xQ
)( x f y !
x x x
x x x
(!
!(
12
12
y(
tangent line
secant line
x
y
Thus we make the following definition
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 77/176
Thus we make the following definition
DEFINITION:
Suppose that is in the domain of the function f the tangent line to the curve at the point
is the line with equation
1 x
)( x f y !
))(,(11 x f x P
)()( 11 x xm x f y !
x
x f x x f m
x (
(!
p(
)()(lim 1
0where provided the limit
exists, and is the point of tangency.))(,( 11 x f x P
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 78/176
DEFINITION
The derivative of at point P on the curve is
equal to the slope of the tangent line at P, thus thederivative of the function f given by with
respect to x at any x in its domain is defined as:
)( x f y !
)( x f y !
0 0
( ) ( )lim lim x x
dy y f x x f x
dx x x( p ( p
( ( ! !( (
provided the limit exists.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 79/176
Other notations for the derivative of a function are:
)(),(',','),(, x f dx
d and x f f y x f D y D x x
Note:
To find the slope of the tangent line to the curve at point P
means that we are to find the value of the derivative at that
point P.
There are two ways of finding the derivative of a function:
1. By using the increment method or the four-step rule
2. By using the differentiation formulas
THE INCREMENT METHOD OR THE FOUR STEP RULE
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 80/176
THE INCREMENT METHOD OR THE FOUR-STEP RULE
One method of determining the derivative of a
function is the increment method or more commonly
known as the four-step rule.
The procedure is as follows:
x x ( y y (
STEP 1: Substitute for x and
for y in )( x f y !
STEP 2: Subtract y=f(x) from the result of step 1 to
obtain in terms of x and y( . x(
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 81/176
STEP 3: Divide both sides of step 2 by
STEP 4: Find the limit of step 3 as approaches 0.
. x(
x(
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 83/176
DIFFERENTIATION OF
ALGEBRAIC FUNCTIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 84/176
OBJECTIVES:
to identify the different rules of differentiation
and distinguish one from the other;
prove the different rules of differentiation using
the increment method;find the derivative of an algebraic function using
the basic rules of differentiation; and
extend these basic rules to other complex
algebraic functions.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 85/176
DERIVATIVE USING FORMULAS
The increment-method (four-step rule) of finding
the derivative of a function gives us the basic procedures
of differentiation. However these rules are laborious and
tedious when the functions to be differentiated arecomplex, that is, functions with large exponents,
functions with fractional exponents and other rational
functions
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 86/176
Understanding of the theorems of differentiation is
very important. This is the heart of differential calculus. All
of the succeeding topics such as applications of derivatives,
differentiation of transcendental functions etc. will bedependent on these theorems. Understanding of these
theorems will enable us to calculate derivatives more
efficiently and will make calculus easy and enjoyable.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 87/176
IFFERENTIATION FORMULAS
Derivative of a Constant
Theorem T he derivative of a constant function
is 0; that is, if c is any real number, then
0][ !c
dx
d
IFFERENTIATION FORMULAS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 88/176
IFFERENTIATION FORMULAS
Derivatives of Power FunctionsTheorem ( Power Rule) I f n is a positive integer,
then1][ ! nn nx x
dx
d
In words, to differentiate a power function,
decrease the constant exponent by one and multiply
the resulting power function by the original exponent .
IFFERENTIATION FORMULAS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 89/176
IFFERENTIATION FORMULAS
Derivative of a Constant Times a Function
Theorem ( Constant Multiple Rule) I f is a
differentiable function at x and c is any real
number, then is also differentiable at x and cf
? A ? A)()( x f dx
d c xcf
dx
d !
In words, the derivative of a constant times a function
is the constant times the derivative of the function,if this derivative exists.
Proof:ffd ( )()(
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 90/176
? A x
xcf x xcf xcf
dx
d
x (
(!
p(
)()(lim)(
0
¼½
»¬-
«
(
(!p( x
x f x x f c
x
)()(lim
0
x
x f x x f c
x (
(!
p(
)()(lim
0
? A)( x f dx
d c!
IFFERENTIATION FORMULAS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 91/176
IFFERENTIATION FORMULAS
Derivatives of Sums or Differences
Theorem ( Sum or Difference Rule) I f and areboth differentiable functions at x, then so are
and , and
In words, the derivative of a sum or of a difference
equals the sum or difference of their derivatives, if these
derivatives exist.
f g
g f g f
g dxd f
dxd g f
dxd s!s
? A ? A¼
½
»¬
-
«s!s )()()()( x g
dx
d x f
dx
d x g x f
dx
d
or
Proof:
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 92/176
x
x g x f x x g x x f x g x f
dx
d
x (
s(s(!s
p(
)]()([)()([lim)]()([
0
x
x g x x g x f x x f
x (
(s(!
p(
)]()([)]()([lim
0
x
x g x x g
x
x f x x f
x x
(
(s
(
(!
p(p(
)()(lim
)()(lim
00
)]([)]([ x g dx
d x f
dx
d s!
IFFERENTIATION FORMULAS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 93/176
IFFERENTIATION FORMULAS
Derivative of a Product
Theorem (The Product Rule) I f and are bothdifferentiable functions at x, then so is the
product , and
In words the derivative of a product of two functionsis the first function times the derivative of the second
plus the second function times the derivative of the
first, if these derivatives exist.
f g
g f
dx
df g
dx
dg f g f
dx
d !
? A ? A)()()]([)()()( x f dx
d x g x g
dx
d x f x g x f
dx
d !
or
Proof:
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 94/176
x
x g x f x x g x x f x g x f
dx
d
x (
((!
p(
)()()()(lim)]()([
0
x
x g x f x g x x f x g x x f x x g x x f
x (
((((!
p(
)]()()()()()()()(lim
0
¼½»
¬-«
(
(
(
((!
p( x
x f x x f x g
x
x g x x g x x f
x
)()()(
)()()(lim
0
x
x f x x f x g
x
x g x x g x x f
x x x x (
(
(
((!
p(p(p(p(
)()(lim)(lim
)()(lim)(lim
0000
? A ? A ? A ? A)()(lim)((lim00
x f dx
d x g x g
dx
d x x f
x x p(p(
(!
IFFERENTIATION FORMULAS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 95/176
IFFERENTIATION FORMULAS
Derivative of a Quotient
Theorem (The Quotient Rule) I f and are bothdifferentiable functions at x, and if
then is differentiable at x and
f g ,0)( { x g
or
g
f
2 g
dx
dg f
dx
df g
g
f
dx
d
!¹¹ º
¸©©ª
¨
? A ? A
? A2
)(
)()()()(
)(
)(
x g
x g dx
d x f x f dx
d x g
x g
x f
dx
d
!¼½
»¬-
«
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 96/176
In words, the derivative of a quotient of two
functions is the fraction whose numerator is the
denominator times the derivative of the numerator minus
the numerator times the derivative of the denominatorand whose denominator is the square of the given
denominator
IFFERENTIATION FORMULAS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 97/176
IFFERENTIATION FORMULAS
Derivatives of Composition
Theorem (The Chain Rule) If is differentiable at x
and if f is differentiable at , then the
composition is differentiable at x. Moreover,
if and then and
g
)( x g
g f Q
))(( x g f y ! )( x g u ! )(u f y !
dx
du
du
dy
dx
dy!
IFFERENTIATION FORMULAS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 98/176
IFFERENTIATION FORMULAS
Derivative of a Radical with index equal to 2
If u is a differentiable function of x, then
u
dx
du
udx
d
2!
The derivative of a radical whose index is two, is afraction whose numerator is the derivative of the
radicand, and whose denominator is twice the given
radical, if the derivative exists.
IFFERENTIATION FORMULAS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 99/176
IFFERENTIATION FORMULAS
Derivative of a Radical with index other than 2
If n is any positive integer and u is a differentiable
function of x, then
The derivative of the nth root of a given function isthe exponent multiplied by the product of u whose
power is diminished by one and the derivative of u, if
this derivative exists.
dx
duu
nu
dx
d nn !¼
½
»¬-
« 111 1
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 101/176
HIGHER DERIVATIVES
AND
IMPLICIT DIFFERENTIATION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 102/176
OBJECTIVES:
to define higher der ivatives;to apply the knowledge of higher der ivatives
and implicit diff erentiation in pr oving relations;
to find the higher der ivative of algebr aicfunctions; and
to deter mine the der ivative of algebr aic
functions implicitly under the specified
conditions.
HIGHER DERIVATIVES
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 103/176
HIGHER DERIVATIVES
The derivative of a function f is itself a function
and hence may have a derivative of its own. If is
differentiable, then its derivative is denoted by
and is called the second derivative of f. As long as
we have differentiability, we can continue theprocess of differentiating to obtain the third,
fourth, fifth, and even higher derivatives of f .
' f
' f
" f
,......)'(,)''''(,)'''(''',)''('',' 454 f f f f f f f f f !!!!
These successive derivatives are denoted by
Oth t ti f hi h d i ti
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 104/176
Other common notations for higher derivatives are
the following:
fir st der ivative: )(),(),(',', x f D x f dx
d x f y
dx
dy x
second der ivative: )(),(),('','', 22
2
2
2
x f D x f dxd x f y
dx yd x
nth der ivative: )(),(),(,, x f D x f
dx
d x f y
dx
yd xn
n
nnn
n
n
The symbols ,dx
dy,
2
2
dx
yd
n
n
dx
y d are called Leibniz notations.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 105/176
IMPLICIT DIFFERENTIATION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 106/176
FUNCTIONS DEFINED EXPLICITLY AND IMPLICITLY
There are two ways to define functions, implicitly and explicitly . Most of the equations we have dealt
with have been explicit equations, such as y = 2 x -3,
so that we can write y = f ( x ) where f ( x ) = 2 x -3. But
the equation 2 x -y = 3 describes the same function.
This second equation is an implicit definition of y as a
function of x . As there is no real distinction between
the appearance of x or y in the second form, thisequation is also an implicit definition of x as a
function of y.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 107/176
An implicit function is a function in which the
dependent variable has not been given "explicitly" interms of the independent variable. To give a
function f explicitly is to provide a prescription for
determining the output value of the function y in
terms of the input value x : y = f ( x ). By contrast, the
function is implicit if the value of y is obtained
from x by solving an equation of the form: R( x ,y ) = 0.
An equation of the form is said to define x f y !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 108/176
q
explicitly as a function of x because the variable y
appears alone on one side of the equation and doesnot appear at all on the other side. However,
sometimes functions are defined by equations in
which is not alone on one side; for example the
equation is not of the form , but
still defines y as a function of x since it can be
rewritten as . Thus we say that
defines y implicitly as a function of x , the functionbeing .
x y yx ! 1 )( x f y !
1
1
! x
x y x y yx ! 1
1
1)(
! x
x x f
Suppose we have an equation f(x, y) = 0 where
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 109/176
pp q f( y)
neither variable could be expressed as a function of
the other.In other words, it wouldnt be possible,
by rearranging f(x, y) = 0, to separate out one of the
variables and express it as a function of the other.
Often we can solve an equation f(x, y) = 0 for one of
the variables obtaining multiple solutions
constituting multiple branches. Consider the
equation which defines y as an implicit
function of x . If we solve for y in terms of x , weobtain two solutions and
thus we have found two functions that are defined
implicitly by .
0122 ! y x
21 x y !
0122 ! y x
IMPLICIT DIFFERENTIATION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 110/176
In general, it is not necessary to solve an equationfor y in terms of x in order to differentiate the
functions defined implicitly by the equation.
To find the derivative of functions defined implicitly
we use implicit differentiation.
Steps in Implicit Differentiation
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 111/176
Steps in Implicit Differentiation
1. Differentiate both sides of the equation withrespect to x.
2. Collect all the terms with on one side of the
equation.
3. Factor out .
4. Solve for .
dx
dy
dx
dy
dx
dy
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 113/176
Slope of a C urve,
T angent,and Normal Lines
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 114/176
SPE C I F I C OBJE C TIVES:
At the end of this lesson, the students are expected to
accomplish the following:
determine the slope of a curve and the derivative of a
function at a specified point;
solve problems involving slope of a curve;
determine the equations of tangent and normal lines
using differentiation; and solve problems involving tangent and normal lines.
tangent line
y
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 115/176
))(,(11
x f x P ))(,(22
x f xQ
)( x f y !
x x x
x x x
(!
!(
12
12
y(
tangent line
secant line
x
Consider a point on the curve))(( xfxQ )(xfy !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 116/176
Consider a point on the curve ))(,( 22 x f xQ ),( x f y !
that is distinct from and compute the
slope of the secant line through P and Q.
)),(,( 11 x f x P
PQm
x
x f x f m PQ
(
!
)()(12 where 12 x x x !(
x x x (! 12and
x
x f x x f m PQ
(
(!
)()( 11
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 117/176
2 x ,
1 xIf we let approach then the point Q will
move along the curve and approach point P. Aspoint Q approaches P, the value of approaches
zero and the secant line through P and Q
approaches a limiting position, then we will
consider that position to be the position of the
tangent line at P.
x(
Thus we make the following definition
DEFINITION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 118/176
DEFINITION:
Suppose that is in the domain of the function f
the tangent line to the curve at the point
is the line with equation
1 x
)( x f y !))(,(
11 x f x P
)()(11
x xm x f y !
x
x f x x f m
x (
(!
p(
)()(lim 1
0where provided the limit
exists, and is the point of tangency.))(,( 11 x f x P
DEFINITION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 119/176
The derivative of at point P on the curve is
equal to the slope of the tangent line at P, thus thederivative of the function f given by with
respect to x at any x in its domain is defined as:
)( x f y !
)( x f y !
0 0
( ) ( )lim lim x x
dy y f x x f x
dx x x( p ( p
( ( ! !( (
provided the limit exists.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 121/176
THE DERIVATIVE IN
GRAPHING ANDAPPLICATIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 122/176
ANALYSIS OF FUNCTIONS I:
INCREASE, DECREASE and CONCAVITY
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 123/176
OBJECTIVES:
define increasing and decreasing functions;
define concavity and direction of bending that isconcave upward or concave downward; and
determine the point of inflection.
.
Th t i i d i d t t d t
INCREASING and DECREASING FUNCTIONS
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 124/176
02 4
increasing decreasing increasing constant
The term increasing, decreasing, and constant are used to
describe the behavior of a function as we travel left to right
along its graph. An example is shown below.
The following definition, which is illustrated in
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 125/176
Definition 4.1.1 (p. 233)
Figure 4.1.2, expresses these intuitive ideas precisely.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 126/176
Figure 4.1.2 (p. 233)
y y
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 127/176
x
y
y
Each tangent line
has po
sitive
slo
pe;function is increasing
x
y
y
Each tangent line
hasne
gative
slo
pe;function is decreasingy
x
y y
Each tangent line
Has zer o slope,
function is constant
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 128/176
Theorem 4.1.2 (p. 233)
. g sindecreaand g sinincreais3 x4 x f(x)whichon int ervalst he Find .1 2 !
EXAMPLE
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 129/176
y
A
? °¯®
g""
g
!!
2,onincrea sing is f 2 xwhen0 x' f
,2-ondecrea sing is f 2 xwhen0 x' f t hus
2 x24 x2 x' f
-1
2
3
x
3 x4 x ) x( f 2
!
increasing
decreasing
. g sindecreaand g sinincreais x f(x)whichon int ervalst he Find .2 3!
EXAMPLE
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 130/176
-4
3
4
x
y 3 x ) x( f !
A
? °¯®
g""
g"
!
0, onincrea sing is f 0 xwhen0 x' f
,0- onincrea sing is f 0 xwhen0 x' f t hus
x x' f 2
in
cre
asin
g
increasing
-3
CONCAVITY
Although the sign of the derivative of f reveals where
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 131/176
Although the sign of the derivative of f reveals where
the graph of f is increasing or decreasing , it does not
reveal the direction of the curvature.
Figure 4.1.8 suggests two ways to characterize the concavity
of a differentiable f on an open interval:
f is concave up on an open interval if its tangent lines have
increasing slopes on that interval and is concave down if
they have decreasing slopes.
f is concave up on an open interval if its graph lies above the
its tangent line and concave down if it lies below its tangent
lines.
y
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 132/176
y
y
y
y
y
concave up
yy
y
y y
concave down
x x
y
increasing slopesdecreasing slopes
Figure 4.1.8
CONCAVITY
Formal definition of the concave up and
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 133/176
Formal definition of the concave up and
concave down .
Definition 4.1.3 (p. 235)
Since the slopes of the tangent lines to the gr aph of a
differentiable function f are the values of its derivative f¶
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 134/176
Theorem 4.1.4 (p. 235)
diff erentiable function f are the values of its der ivative f ,
it f ollows f r om Theorem 4.1.2 (applied to f¶ r ather than f )
that f¶ will be increasing on intervals where f¶¶ is positive and that f¶ will be decreasing on intervals where f¶¶ is
negative. Thus we have the f ollowing theorem.
EXAMPLE:
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 135/176
-1
2
3
x
3 x4 x ) x( f 2
!
increasing
decreasing
0 x' ' f 2 x' ' f and 4 x2 x' f
. ,-int erval t heonup concaveis3 x4 x y funct iont het hat suggest sabove figureThe
2
"!!
gg!
EXAMPLE
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 136/176
-4
3
4
x
y 3 x ) x( f !
increasing
increasing
-3
°¯®
""
!!
gg
!
0 x if 0 x' ' f
and 0 x if 0 x' ' f x6 x' ' f and x3 x' f
0,int erval t heonupconcaveand .0-int erval t heondown concaveis
x y funct iont het hat suggest sabove figureThe
2
3
INFLECTION POINTS:
Points where the curve changes from concave up
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 137/176
Points where the curve changes from concave up
to concave down or vice-versa are called points of
inflection.
Definition 4.1.5 (p. 236)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 138/176
Figure 4.1.9 (p. 236)
The figure shows the graph of the function .
Use the 1st and 2nd derivatives of f to determine the intervals
1 x3 x x f 23
!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 139/176
2-1 3
1
2
-3
y
y
x
on which f is increasing, decreasing, concave up and concave
down. Locate all inflection points and confirm that yourconclusions are consistent with the graph.
1 x6 6 x6 x' ' f
2 x x3 x6 x3 x' f
:SO LU TION
2
!!
!!
INTERVAL (3x)(x-2) f(x) CONCLUSION
x<0 ( )( ) + f is increasing on A0g
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 140/176
x<0 (-)(-) + f is increasing on
0<x<2 (+)(-) - f is decreasing on
x>2 (+)(+) + f is increasing on
A0 ,g
? A2 ,0
? g ,2
INTERVAL (6)(x-1) f(x) CONCLUSION
x<1 (-) - f is concave down on
x>1 (+) + f is concave up on g ,1
1 ,g
The 2nd table shows that there is a point of inf lection at x=1,
since f changes f r om concave up to concave down at that point.
The point of inf lection is (1,-1).
. x x f of any,if point s,inflect iont he Find 4!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 141/176
upconcave;0 x when 0 x' ' f
upconcave0; x when0 x' ' f x12 x' ' f
x4 x' f
:SO LU TION
2
3
°¯®
""
"!
!
.00' f' t hougheven0, xat point inflec
t
ionnohencea
nd conca
vit
yincha
ngenoist
hereThus!!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 142/176
ANALYSIS OF FUNCTIONS II:
RELATIVE EXTREMA;GRAPHING
POLYNOMIALS
OBJECTIVES:
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 143/176
define maximum, minimum, inflection, stationaryand critical points, relative maximum and relativeminimum;
determine the critical, maximum and minimum
points of any given curve using the first and secondderivative tests;
draw the curve using the first and second derivativetests; and
describe the behavior of any given graph in terms of concavity and relative extrema
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 144/176
Definition 4.2.1 (p. 244)
Figure 4.2.1
y1x4xx
4x
1y 234 !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 145/176
2
-1
3
1
2
-3
x
-3
3
1
1 xat ma ximumrel at iveaand
2 xand 1 xat minimarel at iveaha s
1 x4 x x3
x2
y
!
!!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 146/176
Figure 4.2.3 (p. 245)
The points x 1, x 2, x 3, x 4, and x 5 are critical points. Of these, x 1, x 2, and x 5 are
stationary points.
Figure 4.2.3 illustrates that a relative extremum
can also occur at a point where a function is not
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 147/176
differentiable.
In general, we define a critical point for a function f
to be a point in the domain of f at which either the
graph of f has a horizontal tangent line or f is not
differentiable (line is vertical).to distinguish between the two types of critical
points we call x a stationary point of f if f(x)=0.
Thus we have the following theorem:
1 x3 x x f of point scrit ical t heall Find 3 !
EXAMPLE
tangent line
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 148/176
°¯®
!p!
!p!
!!!
1 x01 x
1 x01 x ,t hus
01 x1 x33 x3 x' f
:SO LU TION
2
1 xand 1 x at occur
point scrit ical t heTherefore
1 x whenand
-1 x when0 x f' t hat meansThis
!!
!
!!
tangent line
tangent line
EXAMPLE
3
2
3
5
x15 x3 x f of point scrit ical t heall Find !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 149/176
±°
±¯
®
{p{
!p!
!!!!
0 x0 x
2 x02 x
,t hus
0
x
2 x52 x x5 x10 x5 x' f
:SO LU TION
3
1
3
1
31
31
32
point. y stat ionar ais2 x t hat
and 2 xand 0 x at occur
point scrit ical t heTherefore
0 x when x f' and
2 x when0 x f' t hat meansThis
!
!!
!g!
!!
5
2-1 3
-4
1
-3
y
x
-2
2
1
y
tangent line
FIRST DERIVATIVE TEST
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 150/176
Theorem 4.2.2 asserts that the relative extrema must occur
at critical points, but it does not say that a relative
extremum occurs at every critical point.
sign.changes f' where point scrit ical
t hoseat ext remum rel at ive a ha s funct ion A
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 151/176
Figure 4.2.6 (p. 246)
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 152/176
Theorem 4.2.3 (p. 247) First Derivative Test
The above theorem simply say that for a continuous
function, relative extrema occur at critical points where the
derivative changes from (+) to () and relative minima
where it changes from () to (+).
EXAMPLE
3
2
3
5
x15 x3 x f of point scrit ical t heall Find !
:SO LU TION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 153/176
point. y stat ionar ais2 x t hat
and 2 xand 0 x at occur
point scrit ical The
!!!
y
5
2-1 3
-4
1
-3
y
x
-2
2
1
y
tangent line
3
1
3
1
3
1
3
2
x
2- x5
2 x x5 x10 x5 x' f
t hat shownhavewe
!
!!
y
:below shownisderivat ivet hisof analysis sign A
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 154/176
INTERVAL f(x)
x<0 (-)/(-) +
0<x<2 (-)/(+) _
x>2 (+)/(+) +
3
1
x / 2 x5
imumminrel at ive2 x at t ochanges f' of signThe
imumma xrel at ive0 x at -t ochanges f' of signThe
p!
p!
SECOND DERIVATIVE TEST
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 155/176
There is another test for relative extrema that is based on
the following geometric observation:
a function f has a relative maximum at stationary point if
the graph of f is concave down on an open interval
containing that point
a function f has a relative minimum at stationary point if
the graph of f is concave up on an open interval
containing that point
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 156/176
Figure 4.2.7
Note: The second der ivative test is applicable only to stationar y points
where the 2nd der ivative exists.
EXAMPLE 35 x5 x3 x f of ext remarel at ivet he Find !
:SO LU TION
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 157/176
12x30x30x-60x x' ' f
1 x -1; x 0; x
01 x ;01 x ;015x
01 x1 x15x0 x' f when
1 x1 x15x1 x15x15x-15x x' f
23
2
2
22224
!!
!!!
!!!
!p!!!!
STATIONARY
POINTS f
2nd DERIVATIVE TEST
x 1 30 f has a relative maximum
1 x2 x30 2
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 158/176
x=-1 -30 - f has a relative maximum
x=0 0 0 inconclusive
x=1 30 + f has a relative minimum
INTERVAL f(x) Conclusion
x<-1 (+)(-)(-) +
x=-1 f has a relative maximum
-1<x<0 (+)(+)(-) -
x=0 f has neither a relative max nor min
0<x<1 (+)(+)(-) -
x=1 f has a relative minimum
x>1 (+)(+)(+) +
1 x1 x x152
y
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 159/176
2-1
1
x
-2
2
1
3 x x3 y of curve t he t r ace and Analyze .1 !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 160/176
1 x and -1 x
0 x1 and 0 x1
0 x1 x13 x-13
x33' y
x x3 y
2
2
3
!!
!!
!!
!
!
0 x
0 x6
x6 ' ' y
!
!
!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 161/176
INTERVAL Conclusion
x<1 (+)(-)(+)= - + f is decreasing; concave upward
x=-1 -2 0 + f has a relative minimum
-1<x<0 (+)(+)(+)= + + f is increasing; concave upward
x=0 0 3 0 f has a point of inflection
0<x<1 (+)(+)(+)= + - f is increasing; concave downward
x=1 2 0 - f is has a relative maximum
x>1 (+)(+)(-)= - - f is decreasing; concave downward
2 x3 x
x f
x1 x13
x' f
x6 x' ' f
y
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 162/176
-2
y
y
2-1
1
x
-2
2
1
3 x3 x3 y !
2 x4
x4 y of curve t he t r ace and Analyze .2
!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 163/176
2 x and -2 x
0 x2 and 0 x2
0 x4
x2 x24
x4
x44
' y
x4
x2 x44' y
x4
x2 x44 x4' y
x4
x4
y
2222
2
22
22
22
2
2
!!
!!!
!
!
!
!
!
x2 x4216 x4 x8 x4''y
222 !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 164/176
? A
32 x and 0 x
012 x x4 x8
024 x2 x44x-
016 x4 x28 x44x-016 x4 x42 x44x-
016 x4 x4 x4 x8 x4
x4
y
22
22
222
222
2222
42
s!!
!
!
! !
!
!
INTERVAL Conclusion x f x' f x' ' f
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 165/176
x= -2 -1
-2<x<0
x=0 0
0<x<2
x=2 1
32 x
32 x !
2 x32
32 x2
32 x "
32 x !
2
3
23
y2
x4 y !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 166/176
-2
y
y
2-1
1
x
-2
2
1-3-4 3 4
`
2 x4y
7x12x3x2y1 23 234x6x8x3y6
More examples:
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 167/176
7 x12 x3 x2 y .1 3 !
4 x
2 y .2
2 !
234 x12 x4 x3 y .3 !
2
2
x
1 x y .4 !
35 x
3
2 x
5
1 y .5 !
x6 x8 x3 y .6 !
) x2( x y .7 22 !
2
x1
x y .8
!
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 168/176
ROLLES THEOREMAND
THE MEAN-VALUETHEOREM
ROLLES THEOREM
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 169/176
This theorem states the geometrically obvious factthat if the graph of a differentiable function
intersects the x-axis at two places, a and b there
must be at least one place where the tangent line ishorizontal.
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 170/176
Theorem 4.8.1 (p. 302)Rolle's Theorem
Figure 4.8.1
EXAMPLE
Find the two x-intercepts of the function
and confirm that f(c) = 0 at some point between those
4 x5 x x f 2 !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 171/176
and confirm that f (c) = 0 at some point between those
Intercepts.
Solution:
? A
.0c f' which at
1,4 int erval t heon point ais25 cso ,
25 x ;05 x2 x f'
0' c f' t hat such 1,4
int erval t hein c point onelea st at of exist encet he guar ant eed areweThus
.1,4 inet rval t heon sat isfied areTheorem s Rolle' of hypot hesest he
,everywhere iabledifferent and cont inuousis f l plolynomiat he since
4 xand 1 xareint ercept s- xt he so ,4 x1 x 4 x5 x x f 2
!
!!!!
!
!!!!
y
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 172/176
1 2 3 4
1
2
-1
-2
x
02
5' f !¹
º
¸©ª
¨
THE MEAN-VALUE THEOREM
Rolles Theorem is a special case of a more general
lt ll d th M l Th
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 173/176
result, called the Mean-value T heorem.
Geometrically, this theorem states that between anytwo points A (a,f(a)) and B(b,f(b)) on the graph of a
differentiable function f, there is at least one place
where the tangent line to the graph is parallel to thesecant line joining A and B.(Fig 4.8.5)
Figure 4.8.5 (p. 304)
Note that the slope of the secant line joining A(a,f(a)) and B(b,f(b)) is
afbf
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 174/176
ab
a f b f m
!
and that the slope of the tangent line at c in Figure 4.5.8a is f(c).
Similarly, in Figure 4.5.8b the slopes of the tangent lines at joining
A(a,f(a)) and B(b,f(b)) is
Since nonvertical parallel lines have the same slope, the Mean-Value
Theorem can be stated precisely as follows
.lyrespect ive ,c f' and c f' arecand c 2121
Theorem 4.8.2 (p. 304)Mean-Value Theorem
EXAMPLEShow that the function satisfies the hypotheses
of the mean-value theorem over the inteval [0,2], and find 1 x
4
1 x f 3 !
8/3/2019 MATH21 Coursewares
http://slidepdf.com/reader/full/math21-coursewares 175/176
all values of c in the interval (0,2) at which the tangent line
to the graph of f is parallel to the secant line joining the
points (0, f (0)) and (2, f (2)).
Solution:
? A
0,2int erval t heinlies1.15 only ,15.13
32
3
4c Therefore
4c3 0213
4c3
ab
a f b f c f' Thus
4
c3c f' and ,
4
x3 x f' 32 f b f ,10 f a f But
2.band 0awit h sat isfied areTheoremV alue- Meant heof
hypot hesest he so ,0,2onabledifferent iand 0,2oncont inuousis f par t icul ar In
. polynomial aisit becauseeverywhere iabledifferent and cont inuousis f
22
22
s}s!s!
!!
!
!!!!!!
!!
y1 x
4
1 y 3 !
4