M408K Hm. 13 Solutions

M408K Hm. 13 Solutions - Mahon Kevin Homework 13 Due 3:00...

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

View Full Document Right Arrow Icon
Mahon, Kevin – Homework 13 – Due: Nov 24 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Determine which, if any, of the following f ( x ) = 4 x + 8 , g ( x ) = 2 2 x +4 , h ( x ) = 16 (4 x ) , define the same function. 1. only g, f 2. only f, h 3. f, g, and h 4. only g, h correct 5. none of f, g, or h Explanation: By the Laws of Exponents, f ( x ) = 4 x + 8 = (2 2 ) x + 8 = 2 2 x + 8 , while g ( x ) = 2 2 x +4 and h ( x ) = 16 (4 x ) = (2 4 )(2 2 ) x = (2 4 )(2 2 x ) = 2 2 x +4 . Thus g and h define the same function. On the other hand, f (0) = 9 , g (0) = 16 = h (0) , so neither g nor h can define the same function as f . Consequently only g, h define the same function. keywords: Stewart5e, 002 (part 1 of 1) 10 points If x 1 , x 2 are the solutions of the equation 6 3 x 2 = 1 36 5 x +2 , compute the value of | x 1 + x 2 | . 1. | x 1 + x 2 | = - 10 3 2. | x 1 + x 2 | = 2 13 3 3. | x 1 + x 2 | = 5 3 4. | x 1 + x 2 | = 10 3 correct 5. | x 1 + x 2 | = - 2 13 3 Explanation: Since 1 36 5 x +2 = 6 - 10 x - 4 , the equation can be written as 6 3 x 2 = 6 - 10 x - 4 , which in turn can be rewritten as 3 x 2 = - 10 x - 4 by taking logs to the base 6 of both sides. By the quadratic formula, therefore, x 1 , x 2 = - 5 ± 13 3 . Thus | x 1 + x 2 | = 10 3 . keywords: Stewart5e, 003 (part 1 of 1) 10 points
Image of page 1

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

View Full Document Right Arrow Icon
Mahon, Kevin – Homework 13 – Due: Nov 24 2005, 3:00 am – Inst: Edward Odell 2 Which function has 2 4 - 2 - 4 2 4 - 2 - 4 as its graph? 1. f ( x ) = 2 x - 1 - 3 2. f ( x ) = 2 - x - 1 - 2 3. f ( x ) = 3 x - 3 correct 4. f ( x ) = 3 - x - 2 5. f ( x ) = 2 - 2 - x - 1 6. f ( x ) = 2 - 3 - x Explanation: The given graph has the property that lim x → -∞ f ( x ) = - 3 . But lim x → ∞ 2 - x = 0 = lim x → ∞ 3 - x , while lim x → -∞ 2 x = 0 = lim x → -∞ 3 x , so f ( x ) must be one of 3 x - 3 , 2 x - 1 - 3 . On the other hand, the y -intercept of the given graph is at y = - 2. Consequently, the graph is that of f ( x ) = 3 x - 3 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Find the value of lim x →∞ e 2 x + 2 e - 2 x 2 e 2 x - 5 e - 2 x . 1. limit = - 2 2. limit = - 1 2 3. limit = 2 4. limit = - 1 7 5. limit = 1 2 correct 6. limit = 1 7 Explanation: From the graph of e x it follows that lim x →∞ e - kx = 0 , lim x →-∞ e kx = 0 for each k > 0. Now e 2 x + 2 e - 2 x 2 e 2 x - 5 e - 2 x = 1 + 2 e - 4 x 2 - 5 e - 4 x . But by properties of limits lim x →∞ 1 + 2 e - 4 x 2 - 5 e - 4 x = 1 2 Consequently, limit = 1 2 . keywords: Stewart5e,
Image of page 2
Mahon, Kevin – Homework 13 – Due: Nov 24 2005, 3:00 am – Inst: Edward Odell 3 005 (part 1 of 1) 10 points Find the value of f 0 (0) when f ( x ) = 1 3 e 2 x + 1 4 e - x . 1. f 0 (0) = 7 12 2. f 0 (0) = 2 3 3. f 0 (0) = 1 2 4. f 0 (0) = 1 6 5. f 0 (0) = 5 12 correct Explanation: By the Chain rule, f 0 ( x ) = 2 3 e 2 x - 1 4 e - x Consequently, f 0 (0) = 5 12 . keywords: Stewart5e, 006 (part 1 of 1) 10 points Determine f 0 ( x ) when f ( x ) = e 3 x +5 .
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