EXAM 1-solutions - Version 028 EXAM 1 Helleloid(58250 This...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Version 028 – EXAM 1 – Helleloid – (58250) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A tank holds 1000 gallons of water, which drains from the bottom of the tank in half an hour. The values in the table t (min) 5 10 15 20 25 30 V (gal) 689 495 211 177 16 0 show the volume, V ( t ), of water remaining in the tank (in gallons) after t minutes. If P is the point (15 , V (15)) on the graph of V as a function of time t , find the slope of the secant line PQ when Q = (25 , V (25)). 1. slope = 39 2. slope = 28 . 4 3. slope = 3 . 4 4. slope = 47 . 8 5. slope = 19 . 5 correct Explanation: When P = (15 , V (15)) , Q = (25 , V (25)) the slope of the secant line PQ is given by rise run = V (25) V (15) 25 15 . From the table of values, therefore, we see that slope = 16 211 25 15 = 19 . 5 . 002 10.0 points Below is the graph of a function f . -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2 4 6 2 4 6 2 4 6 8 2 4 Use the graph to determine lim x 4 f ( x ) . 1. limit = 8 2. limit = 6 3. limit = 7 4. limit does not exist correct 5. limit = 2 Explanation: From the graph it is clear the f has a left hand limit at x = 4 which is equal to 8; and a right hand limit which is equal to 3. Since the two numbers do not coincide, the limit does not exist . 003 10.0 points Determine lim x 0 x 1 x 2 ( x + 2) . 1. limit = 0 2. limit = 3. limit = 1 2 4. none of the other answers 5. limit = −∞ correct
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Version 028 – EXAM 1 – Helleloid – (58250) 2 6. limit = 1 Explanation: Now lim x 0 x 1 = 1 . On the other hand, x 2 ( x + 2) > 0 for all small x , both positive and negative, while lim x 0 x 2 ( x + 2) = 0 . Consequently, limit = −∞ . keywords: evaluate limit, rational function 004 10.0 points Let F be the function defined by F ( x ) = x 2 64 | x 8 | . Determine if lim x 8 - F ( x ) exists, and if it does, find its value. 1. limit = 16 2. limit = 8 3. limit = 8 4. limit = 16 correct 5. limit does not exist Explanation: After factorization, x 2 64 | x 8 | = ( x + 8)( x 8) | x 8 | . But, for x < 8, | x 8 | = ( x 8) . Thus F ( x ) = ( x + 8) , x < 8 , By properties of limits, therefore, the limit exists and lim x 8 - F ( x ) = 16 . 005 10.0 points Determine lim x 6 x 6 x + 3 3 . 1. limit = 3 2. limit = 1 3 3. limit doesn’t exist 4. limit = 1 6 5. limit = 6 correct Explanation: After rationalizing the denominator we see that 1 x + 3 3 = x + 3 + 3 ( x + 3) 9 = x + 3 + 3 x 6 . Thus x 6 x + 3 3 = x + 3 + 3 for all x negationslash = 6. Consequently, limit = lim x 6 ( x + 3 + 3) = 6 . 006 10.0 points Find the value of lim x 3 parenleftBig 1 x 1 3 parenrightBigparenleftBig 4 x 3 parenrightBig .
Image of page 2
Version 028 – EXAM 1 – Helleloid – (58250) 3 1. limit = 4 9 correct 2. limit = 2 3 3. limit = 4 9 4. limit = 2 3 5. limit does not exist Explanation: After the first term in the product is
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern