1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inequalities help

  1. Jul 1, 2007 #1
    stupid error in the "tex" thing, see first post.
     
    Last edited: Jul 1, 2007
  2. jcsd
  3. Jul 1, 2007 #2
    damn latex keeps screwing up see next post, soz
     
    Last edited: Jul 1, 2007
  4. Jul 1, 2007 #3
    just starting to learn some stuff on inequalities and need some help. Not sure where to put this topic, so move it if necessary. thnx

    Heres one problem I think I've solved but need clarification from you it's right or where i've gone wrong. I also may have missed an obvious solution but im going off a very short and "concise" introduction to inequalities so I havn't really got a full understanding of how to solve a given problem. Thnx

    Find the set of real numbers [tex] x \neq 0[/tex] such that [tex]2x + 1/x < 3[/tex]:

    I then manipulated 2x + 1/x < 3 to get to

    [tex]\frac{(2x-1)(x-1)}{x} < 0[/tex]

    Im pretty sure up till there is right, but it's the next stage that I think my working in dubious. I'm pretty sure I can't times both sides by x because it could be negative, could be positive.

    I know [tex]\frac{(2x-1)(x-1)}{x}[/tex] is negative. Because its less than O. This means that either 2x-1, x-1, or x is negative, or all 3. I kinda think all that makes sense.

    (this is where is gets real dubious lol).

    So, the following are the options.

    2x-1>0, x-1<0, x>0
    or
    2x-1>0, x-1>0, x<0
    or
    2x-1<0, x-1<0, x<0
    or
    2x-1<0, x-1>0, x>0

    which cancels too

    x>1/2, x<1, x>0
    or
    x>1/2, x>1, x<0
    or
    x<1/2, x<1, x<0
    or
    x<1/2, x>1, x>0

    the 2nd and 4th possiblilities are impossible so that leaves

    1/2>x<1
    or
    x<0

    the first doesn't work either because it says find real numbers and subbing in "1" give 3<3, which is false.

    so then i conclude the answer has to be [tex]x \leq -1[/tex]


    is this right? i know i probably did it wrong, but need to know where I went wrong, thnx.
     
  5. Jul 1, 2007 #4
    should be fine now, fixed the latex.
     
  6. Jul 1, 2007 #5

    uart

    User Avatar
    Science Advisor

    The easiest thing to do is multiply both sides by x^2, which you know will be non-negative. This gets you to,

    [tex] (2x-1)(x-1)(x) < 0[/tex]

    This a cubic with three real roots at x=0, x=1/2 and x=1. Make a quick sketch of it if you like, it's easy to see that it is negative for x < 0 and for 1/2 < x < 1.

    BTW: You don't really need to sketch it, that was just a suggestion in case you where not already familar with what a cubic looks like. If you are familar with a cubic you'll know that a positive cubic (x^3 term positive) will go from -ive to +ive through the first zero, from +ive to -ive through the second zero and so on. So it's not hard to figure out where it is negative.
     
    Last edited: Jul 1, 2007
  7. Jul 1, 2007 #6
    o rite i see

    was my answer correct?
     
  8. Jul 1, 2007 #7

    VietDao29

    User Avatar
    Homework Helper

    No, it's not 1/2 > x < 1, in fact, it should read:

    1/2 < x < 1

    No, you don't substitute "1" in. 1/2 < x < 1, x = 1 does not satisfy the requirement. The solution is 1/2 < x < 1, so if you want to check, well take any x that is greater than 1/2, and less than 1, you can choose .999, but definitely not 1. Can you see why? :)

    Btw, how did you get x <= -1? You should note that x is a real number, not an integer.

    There's a more common method, are you familiar with drawing tables, and testing the sign on each interval?
     
  9. Jul 1, 2007 #8
    o crap haha, misread the question lol. stupid me.

    yeah thnx for that, yeah i think the 1/2 > x < 1 was a typo.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Inequalities help
Loading...