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 Calculus Copyright © 2008 W. H. Freeman and Company

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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.

Rogawski CalculusCopyright © 2008 W. H. Freeman and Company

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

Rogawski CalculusCopyright © 2008 W. H. Freeman and Company

Homework

Homework Assignment #8 Review Section 2.8 Page 113, Exercises: 1 – 13 (EOO)

Rogawski CalculusCopyright © 2008 W. H. Freeman and Company