3 - 1 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 3 The Derivative.

3 - 1

© 2012 Pearson Education, Inc.. All rights reserved.

Chapter 3

The Derivative

3 - 2


Section 3.1

Limits

3 - 3 © 2012 Pearson Education, Inc.. All rights reserved.


Figure 2


Notation *from Spivak’s Calculus


Your Turn

Solution: since

The numerator also approaches 0 as x approaches −3, and 0/0

is meaningless. For x ≠ − 3 we can, however, simplify the

function by rewriting the fraction as

Now

2

3

12Find lim .

3x

x xx

3lim 3 0.x

x

2 12 ( 4)( 3)4.

3 ( 3)x x x x

xx x

2

3 3

12lim lim 4

3x x

x xx

x

3 4 7.


Left and Right


Left and Right

What can you say about lim f(x) as x 10 if lim f(x) as x 10- (from the left) is 5 lim f(x) as x 10+ (from the right) is 5 ?


infinity

lim 1/x as x infinity ?


Two Tools with Limits


Your Turn 5

Suppose

and

find

Solution:

2lim ( ) 3x

f x

2lim ( ) 4.x

g x

2

2lim[ ( ) ( )] .x

f x g x

27 49



Your Turn 8

Solution: Here, the highest power of x is x2, which is used to divide

each term in the numerator and denominator.

2

2

2 3 4Find lim .

6 5 7x

x xx x

2 2 2

2

2

2

2

2

2 3 4

lim6 5 7x

x x x

x x x

x x

x x

2

2

3 42

lim 5 7

6x

x x

x x

2 1

60

31

lim nx x


Your Turn

2 1

60

31

lim nx x

3 - 15


Section 3.2

Continuity


Figure 14



Figure 15 - 16


Figure 17


Figure 18


Figure 19





Your Turn 1

Find all values x = a where the function

is discontinuous.

Solution: This root function is discontinuous wherever the

radicand is negative.

There is a discontinuity when 5x + 3 < 0

( ) 5 3f x x

3.

5x


Your Turn 2

Find all values of x where the piecewise function is discontinuous.

Solution: Since each piece of this function is a polynomial, the

only x-values where f might be discontinuous here are 0 and 3.

We investigate at x = 0 first. From the left, where x-values are less

than 0,

From the right, where x-values are greater than 0

Continued

2

5 4 if 0

( ) if 0 3

6 if 3

x x

f x x x

x x

0 0lim ( ) lim 5 4 4.x x

f x x

2

0 0lim ( ) lim 0.x x

f x x


Your Turn 2 Continued

Because

the limit does not exist, so f is discontinuous at x = 0 regardless

of the value of f(0).

Now let us investigate at x = 3.

Thus, f is continuous at x = 3.

0 0lim ( ) lim ( )x x

f x f x

2

3 3lim ( ) lim 9.x x

f x x

3 3lim ( ) lim 6 9.x x

f x x

2Furthermore, (3) 3 9.f


Figure 20


Figure 21


Figure 22

3 - 31


Section 3.3

Rates of Change



Figure 23


Figure 24



Figure 25



Figure 26

3 - 39


Section 3.4

Definition of the Derivative


Figure 27


Figure 28


Figure 29


Figure 30



Figure 31


Figure 32


Figure 33 - 34


Figure 35


Figure 36 - 37





Figure 38




Figure 39



Figure 40


Figure 41 - 42


Figure 43



Figure 44

3 - 63


Section 3.5

Graphical Differentiation


Figure 45


Figure 46


Figure 47 - 48


Figure 49


Figure 50


Figure 51


Figure 52


Figure 53


Figure 54

3 - 73


Chapter 3

Extended Application: A Model for

Drugs Administered Intravenously


Figure 55


Figure 56


Figure 57

3 - 1 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 3 The Derivative.

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