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MATH 100 Fall 2012 midterm exam solutions
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MATH 100 Fall 2012 midterm exam solutions
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MATH 100—Midterm Examinationversion 1
Date: October 27, 2012 Time: 90 minutes
Surname: Given name(s):
(Please, print!)
ID#: Signature:
Please, check your section/instructor!
Section Instructor X
EA1 V. Yaskin
EB1 E. Leonard
EC1 D. Hrimiuc
ED1 H. Yahya
EE1 V. Yaskin
EF1 T. Liko
EG1 H. Freedman
EH1 E. Woolgar
EJ1 H. Yahya
Instructions
1. Place your U of A Student ID card on your table or desk.
2. The exam has 9 pages including this cover page and a back page for instructors’ use. Please
ensure that you have all pages, and write your name at the top of each page.
3. The exam will be marked out of 60 points. There are 5 questions. The points for each
question are indicated beside the question number, as well as on the last page.
4. Answers for questions 1 through 4 must be accompanied by adequate justification. If you run
out of space, use the back of any page for answers as needed. Clearly direct the marker to
answers that you provide on the back of a page. For multiple-choice questions you are not
required to show your work.
5. This is a closed book exam. No books, notes, calculators, cell phones or other
electronic aids are allowed!
, ID#: Name:
1. (a) [5 pts] Use the definition of the derivative to find the derivative of
1
f (x) = .
x2 + 1
No marks will be given if the definition is not used.
Solution.
1 1
0 (x+h)2 +1
− x2 +1 x2 + 1 − ((x + h)2 + 1)
f (x) = lim = lim
h→0 h h→0 (x2 + 1)((x + h)2 + 1)h
x2 + 1 − (x2 + 2xh + h2 + 1) −2xh − h2
= lim = lim
h→0 (x2 + 1)((x + h)2 + 1)h h→0 (x2 + 1)((x + h)2 + 1)h
−2x − h −2x
= lim = 2
h→0 (x2 2
+ 1)((x + h) + 1) (x + 1)2
(b) [5 pts] Find all horizontal asymptotes of the graph of the function
√ √
f (x) = x2 + 6x + 3 − x2 + 1
Solution.
√ √ √ √
√ √ ( x2 + 6x + 3 − x2 + 1)( x2 + 6x + 3 + x2 + 1)
lim ( x2 + 6x + 3− x2 + 1) = lim √ √
x→∞ x→∞ x2 + 6x + 3 + x2 + 1
x2 + 6x + 3 − (x2 + 1) 6x + 2
= lim √ √ = lim √ √
x→∞ x + 6x + 3 + x + 1 x→∞ x + 6x + 3 + x2 + 1
2 2 2
2
6+ x 6
= lim q q =√ √ =3
x→∞
1+ 6
+ 3
+ 1+ 1 1+ 1
x x2 x2
√ √ √ √
√ √ ( x2 + 6x + 3 − x2 + 1)( x2 + 6x + 3 + x2 + 1)
lim ( x2 + 6x + 3− x2 + 1) = lim √ √
x→−∞ x→−∞ x2 + 6x + 3 + x2 + 1
x2 + 6x + 3 − (x2 + 1) 6x + 2
= lim √ √ = lim √ √
x→−∞ x2 + 6x + 3 + x2 + 1 x→−∞ x2 + 6x + 3 + x2 + 1
2
6+ x 6
= lim q q = √ √ = −3
x→−∞
− 1+ 6
+ 3
− 1+ 1 − 1− 1
x x2 x2
Horizontal asymptotes are y = 3 and y = −3.
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