811Solutions3

811Solutions3 - Homework 3 Solutions Kyle Chapman 31.4...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework 3 Solutions Kyle Chapman October 20, 2011 31.4 Suppose 2 x + 1 /x < 3. Then there are two cases, either x is positive, or negative. If x is positive, then multiplying both sides of the equation by x leaves the orientation of the inequality, giving us 2 x 2 + 1 < 3 x so 2 x 2 - 3 x + 1 < 0 meaning (2 x - 1)( x - 1) < 0. This means exactly one of the two must be negative. Since in this case x > 0 we know 2 x > x so 2 x - 1 > x - 1 so if the first term is negative, then the second is also. Thus, the first case is exactly (2 x - 1 > 0 > x - 1 so 2 x > 1 > x so 1 / 2 < x < 1. This fully solves the first case. For the second case, x being negative means multiplying through by x reverses the inequality. This gives us the equation (2 x - 1)( x - 1) > 0. Now, both terms must have the same sign, and since x < 0, x - 1 < - 1 < 0 so one term must be negative making both negative. Thus, 2 x - 1 < 0 forms the solution space for the second case, which is x < 1 / 2. We have to restrict this to negative entries, and so we get that all negative x values are solutions. Thus, the full solution is ( -∞ , 0) (1 / 2 , 1).
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
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