exam1 (2) - Granillo, Yvette Exam 1 Due: Sep 29 2005, 11:00...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Granillo, Yvette – Exam 1 – Due: Sep 29 2005, 11:00 pm – Inst: Edward Odell 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points AFter t seconds the displacement, s ( t ), oF a particle moving rightwards along the x -axis is given (in Feet) by s ( t ) = 3 t 2 - 3 t + 7 . Determine the average velocity oF the particle over the time interval [1 , 2]. 1. average vel. = 6 Ft/sec correct 2. average vel. = 8 Ft/sec 3. average vel. = 4 Ft/sec 4. average vel. = 7 Ft/sec 5. average vel. = 5 Ft/sec Explanation: The average velocity over a time interval [ a, b ] is given by dist travelled time taken = s ( b ) - s ( a ) b - a . ±or the time interval [1 , 2], thereFore, average vel. = s (2) - s (1) 2 - 1 Ft/sec . Now s (2) = 3 × 4 - 3 × 2 + 7 = 13 Feet , while s (1) = 3 - 3 + 7 = 7 Feet . Consequently, average vel. = 13 - 7 = 6 Ft/sec . keywords: Stewart5e, 002 (part 1 oF 1) 10 points Below is the graph oF a Function f . 2 4 6 - 2 - 4 - 6 2 4 6 8 - 2 - 4 Use the graph to determine the leFt hand limit lim x 3 - f ( x ) . 1. lim x 3 - f ( x ) = 9 correct 2. lim x 3 - f ( x ) = 7 3. lim x 3 - f ( x ) = 8 4. the limit does not exist 5. lim x 3 - f ( x ) = - 4 Explanation: As the graph shows, lim x 3 - f ( x ) = 9. keywords: Stewart5e, 003 (part 1 oF 1) 10 points
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 When f is the function deFned by f ( x ) = 4 x - 5 , x 2 , 3 x - 4 , x > 2 , determine if the limit lim x 2+ f ( x ) exists, and if it does, Fnd its value. 1. limit = 2 correct 2. limit = 0 3. limit does not exist 4. limit = 4 5. limit = 3 6. limit = 1 Explanation: The right hand limit lim x 2+ f ( x ) depends only on the values of f to the right of 2. Thus lim x 2+ f ( x ) = lim x 2+ 3 x - 4 . Consequently, limit = 3 × 2 - 4 = 2 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Determine the limit lim x 1 3 ( x - 1) 2 . 1. limit = 3 2. limit = - 3 3. limit = correct 4. none of these 5. limit = -∞ Explanation: Since ( x - 1) 2 0 for all x , we see that lim x 1 3 ( x - 1) 2 = . keywords: Stewart5e, 005 (part 1 of 1) 10 points ±ind the value of lim x 3 1 x - 1 3 ·‡ 2 x - 3 · . 1. limit = - 1 3 2. limit = 1 3 3. limit does not exist 4. limit = - 2 9 correct 5. limit = 2 9 Explanation: After the Frst term in the product is brought to a common denominator, the given expression becomes 2(3 - x ) 3 x ( x - 3) = - 2 3 x so long as x 6 = 3. Thus lim x 3 1 x - 1 3 ·‡ 2 x - 3 · = - lim x 3 2 3 x . Consequently,
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 9

exam1 (2) - Granillo, Yvette Exam 1 Due: Sep 29 2005, 11:00...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online