3 - 1 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 3 The Derivative.
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Transcript of 3 - 1 © 2012 Pearson Education, Inc.. All rights reserved. Chapter 3 The Derivative.
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3 - 1
© 2012 Pearson Education, Inc.. All rights reserved.
Chapter 3
The Derivative
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© 2012 Pearson Education, Inc.. All rights reserved.
Section 3.1
Limits
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Figure 2
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Notation *from Spivak’s Calculus
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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.
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Left and Right
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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 ?
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infinity
lim 1/x as x infinity ?
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Two Tools with Limits
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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
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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
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Your Turn
2 1
60
31
lim nx x
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Section 3.2
Continuity
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Figure 14
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Figure 15 - 16
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Figure 19
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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
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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
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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
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Figure 20
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Section 3.3
Rates of Change
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Section 3.4
Definition of the Derivative
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Section 3.5
Graphical Differentiation
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Chapter 3
Extended Application: A Model for
Drugs Administered Intravenously
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