homework homework assignment #7 read section 2.8 page 106, exercises: 1 – 25 (eoo) rogawski...
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Homework
Homework Assignment #7 Read Section 2.8 Page 106, Exercises: 1 – 25 (EOO)
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
Homework, Page 1061. Use the IVT to show that f (x) = x3 + x takes on the value of 9 for some x in [1, 2].
.
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
3
3
1 1 1 4
2 2 2 11 4 9 11
by the IVT, 9 for some on [1, 2].
f
f
f x x
Homework, Page 1065. Show that cos x = x has a solution in the interval [0, 1].
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
Let cos
0 0 cos0 1
1 1 cos1 0
By the IVT, 0 for some on [0, 1]
and cos on the same interval.
f x x x
f
f
f x x
x x
Homework, Page 106Use the IVT to prove each of the following statements.
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
9. 2 exists.
2
2 2
Let on the interval [0, 2]
0 0 0 2 2 4
Polynomials by definition are continuous and 0 2 4
by IVT 2 must exist.
f x x
f f
Homework, Page 106Use the IVT to prove each of the following statements.
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
13. 2 has a solution for all 0. x b b
2 is a continuous, exponential function with domain .
lim 0 lim by the IVT, 2
for all 0.
x
x
x x
f x
f x f x b
b
Homework, Page 10617. Carry out three steps of the Bisection Method for f (x) =2x – x3 as follows:
(a) Show that f (x) has a zero in [1, 1.5].
(b) Show that f (x) has a zero in [1.25, 1.5].
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
3 31 1.51 2 1 1 1.5 2 1.5 0.546
0.546 0 1 by the IVT, there must be a zero in 1,1.5
f f
31.251.25 2 1.25 0.425 1.5 0.546
0.546 0 0.425
by the IVT, there must be a zero in 1.25,1.5
f f
Homework, Page 10617. (c) Determine whether [1.25, 1.375] or [1.375, 1.5] contains a zero.
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
31.375
1.25 0.425 1.5 0.546
1.375 2 1.375 0.006
0.006 0 0.425
by the IVT, there must be a zero in 1.25,1.375
f f
f
Homework, Page 106Draw the graph of a function f (x) on [0, 4] with the given properties.
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
21. Jump discontinuity at 2 and does not satisfy the
conclusion of the IVT.
x
Homework, Page 10625. Corollary 2 is not foolproof. Let f (x) = x2 – 1 and explain why the corollary fails to detect the roots at x = ± 1 if [a, b] contains [1, –1].
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
2 1 is positive for all on , 1 or 1, . If the interval
contains both of these numbers, there will be no sign change from
to and thus the corollary will fail to detect the roots
f x x x
f a
f b
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
Jon Rogawski
Calculus, ET
First Edition
Chapter 2: LimitsSection 2.8: The Formal Definition of a
Limit
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
In Figure 1(A), we see that the value of y approaches 1 as x approaches 0.
In Figure 1(B), we see that the value of |f (x) – 1| < 0.2, if – 1 < x < 1.In Figure 1(C), we see that the value of |f (x) – 1| < 0.004, if – 0.15 < x < 0.15. As we continue to narrow the gap around 0, welessen the value of |f (x) – 1| , leading to the formal definitionof a limit.
The formal definition of the limit of a function
lim ( ) 0 ,
0
( 0
:
.)
x a
f x L means for any given to me
I can find such that
whene er xv af x L
Courtesy of Tom Reardon: [email protected]
a
( )y f x
( )y f x L y L
y L
a a
} }
x a( ) f x L
( )means f x L
( )OR L f x L
means x a
OR a x aCourtesy of Tom Reardon: [email protected]
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
For y = 8x + 3, the formal definition gives us the limit as x approaches 3 as follows:
and as illustrated in Figure 2 below.
Example, Page 113
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
22. Consider lim , where 4 1.
(a) Show that 7 4 if 2 .
b Find a such that: 7 0.01 if 2
xf x f x x
f x x
f x x
Example, Page 113
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
2
2. (c) Prove that lim 7x
f x
Example, Page 113
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
0
18. The number has the following property: lim 1.
1Use a plot of to find a value of 0 such that
1 0.01 if 1 .
x
x
x
ee
x
ef x
x
f x x
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
For the parabola y = x2, we have a similar result.
Example, Page 11312. Based on the information conveyed in the graph, find values of c, L, ε, and δ > 0 such that the following statement holds:
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
if f x L x
4.7
4
3.3
y = f (x)
-0.1 0.1 0.2 0.3
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
Figure 4 illustrates how a limit does not exist at a jump discontinuity.
Example, Page 113
Rogawski CalculusCopyright © 2008 W. H. Freeman and Company
0
sin 227. Use sin sin cos 22
sin sin to prove that: lim cos
h
hha h a h a
h
a h aa
h