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LIMITS

of FUNCTIONS



LIMITS OF FUNCTIONS

OBJECTIVES:

define limits;

illustrate limits and its theorems; and

evaluate limits applying the given theorems.



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



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



1.1.1 (p. 70)Limits (An Informal View)

This leads us to the following general idea.



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



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



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.



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



Figure 1.2.1 (p. 80)



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.



Theorem 1.2.2 (p. 81)

The following theorem will be our basic tool for finding limits

algebraically



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.



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.



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



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 !

!

!



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)



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



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)



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

!

! !

! ! ! !



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

!

!

!

!



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.



LIMITS

of FUNCTIONS



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.



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.



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



1.1.2 (p. 72)

One-Sided Limits (An Informal View)

This leads to the general idea of a one-sided limit



1.1.3 (p. 73)

The Relationship Between One-Sided and Two-Sided Limits



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



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



Figure 1.1.13 (p. 73)



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



x

-2-6 4



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



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



f(x)

x

(2,3)

2



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

"!



.


1 x2 ) x( g .5


x4 ) x( g .4


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

!

!

°¯®

!

{!



LIMITS

of FUNCTIONS



INFINITE LIMITS

OBJECTIVES:

define infinite limits;

illustrate the infinite limits; and

use the graphs to evaluate the

limits of functions.



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 !



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



Figure 1.1.15 (p. 74)



1.1.4 (p. 75)

Infinite Limits (An Informal View)



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



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



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



) 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 !



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



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



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



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:



LIMITS

of FUNCTIONS



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



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



x

x

0 x

1

lim x !gp 0 x

1lim x !gp



1.3.1(p

.8

9)

Limits at Infinity (An Informal View)

In general, we will use the following notation.



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



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



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



LIMITS

of FUNCTIONS



SQUEEZING THEOREM



THEOREM

a.

b.



LIMITS

of FUNCTIONS



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



THE DERIVATIVEOBJECTIVES:

to define the derivative of a function

to find the derivative of a function by

increment method(4-step rule)



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.



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



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(



))(,(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




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

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.



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



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(



STEP 3: Divide both sides of step 2 by

STEP 4: Find the limit of step 3 as approaches 0.

. x(

x(



DIFFERENTIATION OF

ALGEBRAIC FUNCTIONS



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.



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



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.



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





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 .





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 ( )()(



? 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!





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:



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!





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:



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(

(!





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

»¬-

«



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





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!





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.





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



HIGHER DERIVATIVES

AND

IMPLICIT DIFFERENTIATION



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



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



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.



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.



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 !



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



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




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



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



Slope of a C urve,

T angent,and Normal Lines



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



))(,(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 !



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



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(


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

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.



THE DERIVATIVE IN

GRAPHING ANDAPPLICATIONS



ANALYSIS OF FUNCTIONS I:

INCREASE, DECREASE and CONCAVITY



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



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



Definition 4.1.1 (p. 233)

Figure 4.1.2, expresses these intuitive ideas precisely.



Figure 4.1.2 (p. 233)

y y



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



Theorem 4.1.2 (p. 233)

. g sindecreaand g sinincreais3 x4 x f(x)whichon int ervalst he Find .1 2 !

EXAMPLE



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



-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



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



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



Formal definition of the concave up and

concave down .


Since the slopes of the tangent lines to the gr aph of a

differentiable function f are the values of its derivative f¶



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:



-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



-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



Points where the curve changes from concave up

to concave down or vice-versa are called points of

inflection.




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

!



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



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!



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!!



ANALYSIS OF FUNCTIONS II:

RELATIVE EXTREMA;GRAPHING

POLYNOMIALS

OBJECTIVES:



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




Figure 4.2.1

y1x4xx

4x

1y 234 !



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

!

!!



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



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



°¯®

!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 !



±°

±¯

®

{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



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



Figure 4.2.6 (p. 246)



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



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



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



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



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



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



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



2-1

1

x

-2

2

1

3 x x3 y of curve t he t r ace and Analyze .1 !



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

!

!

!



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



-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

!



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 !



? 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



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 !



-2

y

y

2-1

1

x

-2

2

1-3-4 3 4

`

2 x4y

7x12x3x2y1 23 234x6x8x3y6

More examples:



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

!



ROLLES THEOREMAND

THE MEAN-VALUETHEOREM

ROLLES THEOREM



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.



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 !



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



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



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



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 !



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



1 2 3 4

1

2

-1

-2

x

3

4

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