811Solutions3

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

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

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).

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern