100% satisfaction guarantee Immediately available after payment Both online and in PDF No strings attached
logo-home
Previously searched by you
Previously searched by you
logo
Solution Manual for Thomas' Calculus, SI Units, 15th edition Joel R. Hass Christopher E. Heil Maurice D. Weir R331,47
Add to cart
Quickly navigate to
Preview
  2.  
  3. Solution Manual
  4.  
  5. Solution Manual

Exam (elaborations)

Solution Manual for Thomas' Calculus, SI Units, 15th edition Joel R. Hass Christopher E. Heil Maurice D. Weir

2 reviews
 3 purchases
  • Course
  • Solution Manual
  • Institution
  • Solution Manual
  • Book
  • Thomas\' Calculus, SI Units

Solution Manual for Thomas' Calculus, SI Units, 15th edition Joel R. Hass Christopher E. Heil Maurice D. Weir

Preview 4 out of 1335  pages

Document information

  • September 15, 2024
  • 1335
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers

Subjects

  • solution manual for thomas calculus si units 15

Connected book

book image

Book Title:

Author(s):

  • Edition:
  • ISBN:
  • Edition:

Written for

  • Solution Manual
  • Solution Manual

2  reviews

review-writer-avatar

By: atarendash15 • 1 month ago

review-writer-avatar

By: hohengchiwisdom • 5 months ago

It helps me to investigate in maths topic ! The book is too comprehensive

Seller

avatar-seller
SolutionsStuvia

Reviews received

Content preview

SOLUTION MANUAL
THOMAS' CALCULUS, SI UNITS, 15TH EDITION JOEL R. HASS
CHRISTOPHER E. HEIL MAURICE D. WEIR
CHAPTER 1-19



CHAPTER 1 FUNCTIONS

1.1 FUNCTIONS AND THEIR GRAPHS

1. domain  (, ); range  [1, ) 2. domain  [0, ); range  (, 1]

3. domain  [2, ); y in range and y  5 x  10  0  y can be any positive real number  range  [0,  ).


4. domain  (, 0]  [3, ); y in range and y  x 2  3x  0  y can be any positive real number 
range  [0, ).

5. domain  (, 3)  (3, ); y in range and y  3 4 t , now if t  3  3  t  0  3 4 t  0, or if t  3 
3  t  0  3 4 t  0  y can be any nonzero real number  range  (, 0)  (0, ).


6. domain  (,  4)  ( 4, 4)  (4, ); y in range and y  2 , now if t  4  t 2  16  0  2  0, or if
t 2  16 t 2  16
4  t  4  16  t 2  16  0   16
2  2 , or if t  4  t  16  0 
2 2  0  y can be any nonzero
t 2  16 t 2  16
real number  range  (,  18 ]  (0, ).

7. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Is the graph of a function of x since any vertical line intersects the graph at most once.

8. (a) Not the graph of a function of x since it fails the vertical line test.
(b) Not the graph of a function of x since it fails the vertical line test.



2
9. base  x; (height)2  2x  x 2  height  23 x; area is a( x)  12 (base)(height)  12 ( x) 23 x  43 x 2 ;  
perimeter is p( x)  x  x  x  3x.


10. s  side length  s 2  s 2  d 2  s  d ; and area is a  s 2  a  12 d 2 .
2


11. Let D  diagonal length of a face of the cube and  the length of an edge. Then 2  D2  d 2 and

 
2 2 3/2 3
D2  2 2  3 2  d 2   d . The surface area is 6 2
 6d3  2d 2 and the volume is 3
 d3  d .
3 3 3




© 2023 Pearson Education Ltd. All Rights Reserved.
1

,2 Chapter 1 Functions

 
12. The coordinates of P are x, x so the slope of the line joining P to the origin is m  x
x
 1 ( x  0).
x


Thus, x, x    1
m2
, 1
m .
13. 2 x  4 y  5  y   12 x  54 ; L  ( x  0)2  ( y  0)2  x 2  ( 12 x  54 )2  x 2  14 x 2  54 x  16
25

20 x 2  20 x  25 20 x 2  20 x  25
 54 x 2  54 x  16
25 
16
 4


14. y  x  3  y 2  3  x; L  ( x  4) 2  ( y  0) 2  ( y 2  3  4) 2  y 2  ( y 2  1) 2  y 2
 y4  2 y2  1  y2  y4  y2  1


15. The domain is ( ,  ). 16. The domain is ( ,  ).




17. The domain is ( ,  ). 18. The domain is ( , 0].




19. The domain is (, 0)  (0, ). 20. The domain is (, 0)  (0, ).




21. The domain is 22. The range is [5, ) .
(, 5)  (5, 3]  [3, 5)  (5, ).




© 2023 Pearson Education Ltd. All Rights Reserved.

, Section 1.1 Functions and Their Graphs 3


23. Neither graph passes the vertical line test.
(a) (b)




24. Neither graph passes the vertical line test.
(a) (b)




 x  y 1   y  1 x 
   
x y 1  or  or 
 x  y  1  y  1  x 
   

25. x 0 1 2 26. x 0 1 2
y 0 1 0 y 1 0 0




 4  x , x  1  , x  0
2 1
27. F ( x)   28. G ( x)   x
 x  2 x, x  1
2
 x, 0  x




29. (a) Line through (0, 0) and (1, 1): y  x; Line through (1, 1) and (2, 0): y   x  2
 x, 0  x  1
f ( x)  
 x  2, 1  x  2




© 2023 Pearson Education Ltd. All Rights Reserved.

, 4 Chapter 1 Functions

 2, 0  x 1
 0, 1 x  2

(b) f ( x)  
 2, 2 x3
 0, 3 x 4

30. (a) Line through (0, 2) and (2, 0): y   x  2
0 1
Line through (2, 1) and (5, 0): m  5  2  31   13 , so y   1 ( x  2)  1   1 x  5
3 3 3
  x  2, 0  x  2
f ( x)   1
 3 x  3 , 2  x  5
5


3  0
(b) Line through (1, 0) and (0, 3): m  0  (1)  3, so y  3x  3
1  3
Line through (0, 3) and (2, 1) : m  2  0  24  2, so y  2 x  3
3x  3,  1  x  0
f ( x)  
2 x  3, 0  x  2

31. (a) Line through (1, 1) and (0, 0): y   x
Line through (0, 1) and (1, 1): y  1
0 1
Line through (1, 1) and (3, 0): m  3  1  21   12 , so y   12 ( x  1)  1   12 x  23
 x 1  x  0

f ( x)   1 0  x 1
 1 1 x  3
 2 x  2
3  1x 2  x  0
 2
(b) Line through ( 2, 1) and (0, 0): y  12 x f ( x)  2 x  2 0  x  1
Line through (0, 2) and (1, 0): y  2 x  2  1 1 x  3
Line through (1, 1) and (3, 1): y  1 


  1 0
 
32. (a) Line through T2 , 0 and (T, 1): m  T  (T /2)  T2 , so y  T2 x  T2  0  T2 x  1
 0, 0  x  T2

f ( x)  
 T x  1, 2  x  T
2 T

 A, 0  x  T
 2
  A, T  x  T
 2
(b) f ( x)  
 A, T  x  32T

  A, 32T  x  2T

33. (a)  x   0 for x  [0, 1) (b)  x   0 for x  (1, 0]

34.  x    x  only when x is an integer.

35. For any real number x, n  x  n  1, where n is an integer. Now: n  x  n  1   (n  1)   x   n.
By definition:   x   n and  x   n    x   n. So   x     x  for all real x.




© 2023 Pearson Education Ltd. All Rights Reserved.

The benefits of buying summaries with Stuvia:

Guaranteed quality through customer reviews

Guaranteed quality through customer reviews

Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.

Quick and easy check-out

Quick and easy check-out

You can quickly pay through EFT, credit card or Stuvia-credit for the summaries. There is no membership needed.

Focus on what matters

Focus on what matters

Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!

Frequently asked questions

What do I get when I buy this document?

You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.

Satisfaction guarantee: how does it work?

Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.

Who am I buying this summary from?

Stuvia is a marketplace, so you are not buying this document from us, but from seller SolutionsStuvia. Stuvia facilitates payment to the seller.

Will I be stuck with a subscription?

No, you only buy this summary for R331,47. You're not tied to anything after your purchase.

Can Stuvia be trusted?

4.6 stars on Google & Trustpilot (+1000 reviews)

69411 documents were sold in the last 30 days

Founded in 2010, the go-to place to buy summaries for 15 years now

Start selling
R331,47  3x  sold
  • (2)
Add to cart
Added