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Midterm 1

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University of Connecticut Department of Mathematics

Math 1131

Sample Exam 1

Spring 2014

Name:
This sample exam is just a guide to prepare for the actual exam. Questions on the actual exam may or may not be of the same type, nature, or even points. Don’t prepare only by taking this sample exam. You also need to review your class notes, homework and quizzes on WebAssign, quizzes in discussion section, and worksheets. The exam will cover up through section 3.2 (product and quotient rule). Read This First! • Please read each question carefully. Other than the question of true/false items, show all work clearly in the space provided. In order to receive full credit on a problem, solution methods must be complete, logical and understandable. • Answers must be clearly labeled in the spaces provided after each question. Please cross out or fully erase any work that you do not want graded. The point value of each question is indicated after its statement. No books or other references are permitted. • Give any numerical answers in exact form, not as approximations. For example, one-third 1 is 3 , not .33 or .33333. And one-half of π is 1 π, not 1.57 or 1.57079. 2 • Turn smart phones, cell phones, and other electronic devices off (not just in sleep mode) and store them away. • Calculators are allowed but you must show all your work in order to receive credit on the problem. • If you finish early then you can hand in your exam early.

Grading - For Administrative Use Only
Question: Points: Score: 1 12 2 6 3 9 4 12 5 9 6 6 7 24 8 10 9 6 10 6 Total 100

Math 1131

Sample Exam 1

1. If the statement is always true, circle the printed capital T. If the statement is sometimes false, circle the printed capital F. In each case, write a careful and clear justification or a counterexample. (a) The line x = 1 is a vertical asymptote of the graph of y = Justification: x2 − 1 . x2 − 2x + 1 T F [3]

(b) The function y(t) = 3 + 6e−kt , with k a positive constant, has horizontal asymptote y = 6. Justification:

T

F

[3]

(c) A function that is not continuous at a point can not be defined at that point. Justification:

T

F

[3]

(d) If f (x) is differentiable then f (x) and f (x) − 2 have the same derivative. Justification:

T

F

[3]

Page 1 of 10

Math 1131 2. (a) If f (x) =

Sample Exam 1 [3]

x for x = −3, then find its inverse function f −1 (x) and determine the x+3 domain of the inverse function.

(b) If log2 a = .83 and log2 b = .56, compute log2 (a3 /b2 ) and give your numerical answer exactly.

[3]

Page 2 of 10

Math 1131 3. Let f (x) = ln (4x + 1) + 1, g(x) = (a) Find the domain of f (x). √ 3x + 6, and h(x) = x3 .

Sample Exam 1

[3]

(b) Find the domain of (g ◦ h)(x)

[3]

(c) Find f −1 (x) and its domain.

[3]

Page 3 of 10

Math 1131

Sample Exam 1

4. Determine the following limits, using algebraic methods to simplify the expression before finding the limit. Evaluate each of the following limits. If the limit does not exist but goes to ∞ or −∞, indicate so. If the limit does not exist for any other reason, write DNE with a justification. (a) lim 3−x x→2 1/3 − 1/x [4]

(b) lim

x2 − 1 x→0 x2 − x

[4]

x−4 (c) lim √ x→4 x−2

[4]

Page 4 of 10

Math 1131 5. Let f (x) = √ 2x3 + x2 − 7 . 49x6 + 9x2 + x (a) Compute lim f (x). x→∞ Sample Exam 1

[3]

(b) Compute lim f (x). x→−∞ [3]

(c) What are all the horizontal asymptotes of the graph of y = f (x)?

[3]

Page 5 of 10

Math 1131 3 − kx, k − x, if x > 1, if x ≤ 1.

Sample Exam 1

6. Let f (x) =

(a) lim f (x) = x→1− [2]

(b) lim f (x) = x→1+ [2]

(c) Find the constant k so that f (x) is continuous at every point.

k=

[2]

Page 6 of 10

Math 1131

Sample Exam 1

7. Find the derivatives of the following functions. Simplify your answers (e.g., combine like powers of x in a polynomial). √ (a) f (x) = 9x − 7 x.

[6]

(b) f (x) = x4 ex .

[6]

(c) f (x) =

x2 + x . x3 − 3

[6]

(d) f (x) =

ex . x2 + 1

[6]

Page 7 of 10

Math 1131

Sample Exam 1 [6]

8. (a) Find the slope of the tangent line to the graph of y = (x2 − 1)2 at (2, 27).

(b) Find the equation of the tangent line in part a, writing the answer as y = mx + b.

[4]

Page 8 of 10

Math 1131 9. The picture below is the graph of f (x) = x3 − x, −x2 + ax + b, if x ≤ 1, if x > 1,

Sample Exam 1 [6]

where a and b are chosen to make the function differentiable at x = 1. Determine a and b. (Hint: the function has to be continuous at x = 1 also, since it is differentiable.) y

x 1

Page 9 of 10

Math 1131

Sample Exam 1 [6]

10. In the graph of y = x2 /3, which is illustrated below, find the number c such that the tangent lines to the graph at the points where x = 1 and x = c are perpendicular. y (c, c2 /3) (1, 1/3) x c 1

Page 10 of 10

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...អាណាចក្រភ្នំ អាណាចក្រភ្នំ គស​ 50-630 ទីតាំង * ៣០០លី លិចលីនីយ(ជនជាតិចាម) * ៧០០០លី Jenan(តុងកឹង) * ឈូងសមុទ្រធំមួយ * ទន្លេរធំមួយ លិចនឹងពាយព្យគឺសមុទ្រ * ១លី=៥៧៦ម=១៧២៨គម=> 1. កម្ពុជា 2. កម្ពុជាក្រោម 3. ថៃ(ភាគកណ្តាល) រាជវង្សមាន៖ 1. លីវយី(៥០-៦៨) 2. ហ៊ុនទៀន(៦៨) 3. ហ៊ុនប៉ានហួង៖ដែលជាមេទ័ពបានប្រើល្បិចវាយក្រុងទាំង៧នឹងបានដណ្តើមអំណាចពីព្រះ បាទហ៊ុនទៀន 4. ហ៊ុនប៉ានប៉ាង៖ជាកូនហ៊ុនប៉ានហួង 5. ហ្វាន់ជេម៉ាន់៖ជាអ្នកសំលាប់សោយរាជ្យបន្តរឺក៍ហ៊ុនប៉ានប៉ាងផ្ទេរអំណាចអោយ 6. គិនចេង(២២៥)៖ត្រូវជាកូនរបស់របស់ហ្វាន់ជេម៉ាន់ពីព្រោះគាត់បានស្លាប់ពេលវាយ នៅ គិនស៊ីន 7. ហ្វានឆាន(២២៥-២៤៥)៖បានសំលាប់គិនចេងដើម្បីសោយរាជ្យបន្តដែលត្រូវជាក្មួយហ្វាន់ជេម៉ាន់នឹងត្រូវជាបងប្អូនគិនចេង 8. ហ្វានឆាង(២៤៥-២៥០)៖ជាកូនពៅរបស់ហ្វានជេម៉ាន់បានមកសងសឹកនឹងសោយរាជ្យបន្ត 9. ហ្វានស៊ីយ៉ុន(២៥០-២៨៩)៖បានសំលាប់ហ្វានឆាងសោយរាជ្យបន្ត 10. ធៀនឈូឆានតាន(៣៥៧) 11. កៅណ្ឌិន្យ(៣៥៧)៖គាត់មានកូនពីរគឺស្រីឥន្រ្ទវរ្ម័ននឹងស្រេស្ធវរ្ម័ន 12. កៅណ្ឌិន្យជ័យវរ្ម័ន(៤៤២-៥១៤)៖មានបុត្រាពីរគឺគុណវរ្ម័នជាប្អូននឹងរុទ្រវរ្ម័នជាបងក៍ប៉ុន្តែគុណវរ្ម័នជាអ្នកសោយរាជ្យដែលត្រូវជាកូនកុលប្រភាវតីជាមហេសីរីឯរុទ្រវរ្ម័នជាកូនស្នំ។ដោយមិនសុខចិត្តព្រោះខ្លួនជាបងមិនបានសោយរាជ្យក៍ប្រើល្បិចសំលាប់ប្អូនដើម្បី សោយរាជ្យម្តង។ 13. គុណវរ្ម័ន 14. ចេនឡា ចេនឡា រុទ្រវរ្ម័ន(៥១៤-៥៥៦) គស ៥៥០-៨០២ 15. ឥសីកម្ពុស្វយម្ហូវ៖ 16. ស្រុតវរ្ម័ន៖ 17. ស្រស្ធវរ្ម័ន៖ 18. វីរវរ្ម័ន៖ 19. ភវរ្ម័ន(៩០០-៩២២)៖ * ទីតាំងរបស់ចេនឡានៅត្រង់តំបន់បាសាក់តាមដងទន្លេរមេគង្គដែលច្ចុប្បន្ននៅភាគ អាគ្នេយ៍ប្រទេសឡាវ...

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General

...– – – – • – – • – • • • • – – – – • • • – – – – – – Wood Stick Holder Premium Wood Stick Holder Glow in the Dark Stick Holder Brass Burner Premium Brass Burner Aroma Ring Votive Holder NIPPON KODO INCENSE HERB & EARTH STICKS STICKS Classic STICKS CONES AFRICAN AMERICAN Family Unity – – Rhythm Sensuality Spirituality • • • • Bergamot Cedar • • • Fashion & Style • Number 4 - 100-st Number 6 - 100-st Chamomile Frankincense Jasmine Lavender Orange Patchouli Peppermint Rose Sandalwood Vanilla GONESH DIFFUSER SETS 3 fl.oz. Coconut Lime Mango Peach Sweet Apple Vanilla Cream REFILLS - 6 oz Coconut Lime Mango Peach Sweet Apple Vanilla Cream HOLIDAY TRADITIONS Number 8 - 100-st Number 10 Number 12 Number 14 Variety 1 (6,8,12) - 30 st Variety 2 (2,4,10) - 30 st MORNING STAR STICKS GONESH® EXTRA RICH Amber Apple Cider Jasmine Lavender Sandalwood Christmas Dream (Winter) Nutcracker Dance (Winter) Snowy Sensations (Winter) Holiday Memories (Winter) SCENTED REEDS & OILS REEDS OILS Black Cherry Cedarwood Cherry Blossom Cinnamon Coconut Dragon’s Blood...

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Robotics Collision Lab

...Mr. Weidenboerner Period 7 Purpose: To explore sensors and use them to knock down a box filled with bean bags without going over the edge of a precipice. Hypothesis: I think that designs with a high point of impact and and sensor placed out in front of the robot will have the best results. Group 2 | Trial | Distance from the Egde | 1 | 28 mm | 2 | 32 mm | 3 | 35 mm | 4 | 22 mm | 5 | fail | Average | 32 mm | Competion | Group | Average | 1 | 23 mm | 2 | 32 mm | 3 | fail | 4 | 7 mm | Program Flow: 1. #Include “Main.h” 2. 3. void main (void) 4. { 5. int limitswitch; 6. 7. // 0 is pressed 8. // 1 is not pressed 9. Wait (5000) 10. while (1==1) 11. { 12. limitswitch = Get DigitalInput (1); 13. if (limitswitch==1) 14. { 15. Set Motor (1.0); 16. Set Motor (10.0); 17. Wait (200) 18. } 19. else 20. } 21. Set Motor (1.-40); 22. Set Motor (10.40); 23. } 24. } 25. } Results: Group 1 cam in second place with an average of 23 mm from 5 trials. Group 2 (my group), came in third place with an average of 32 mm from the edge of the table. Group 3 came in last place with one fail and not having completed the rest of the trials yet. Group 4 came in first place with an average of 7 mm from the edge of the table. Conclusion: I think that...

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