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Reciprocal property inequality

  1. May 10, 2009 #1
    1. How to solve -1<1/(x+1)<2

    2. Now if I switched both sign and took the reciprocal such as 1/-1>(x+1)/1>1/2 I get the right answer: x<-2 and x>-1/2.

    3. Without using the reciprocal property would be lengthy and confusing, but I have very little and contradictory information found on the reciprocal property when solving inequalities. If I could get an explanation on what I'm doing that would be great
  2. jcsd
  3. May 10, 2009 #2
    If I break it down, however, into 4 equations not using the reciprocal but using:

    -1 < 1/(x+1) with x+1<0
    -1 < 1/(x+1) with x+1>0


    1/(x+1)<2 with x+1<0
    1/(x+1)<2 with x+1>0

    Would I take the intersection of the 2 different sets?
  4. May 10, 2009 #3
    Ok I drew a number line....
  5. May 11, 2009 #4


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    There are really only two cases.

    If x+1> 0 then -1< 0< 1/(x+1)< 2 so 1< 2(x+1)

    If x+1< 0 then -1< 1/(x+1)< 0< 2 so -(x+1)> 1

    You can also argue that a rational function such as 1/(x+1) is continuous everywhere except where the denominator is 0, which, for this function, is x= -1. Further a continuous function can change from ">" to "<" only at "= ". That is, the intervals on which -1< 1/(x+1) and on which -1> 1/(x+1) can be separated only where x+1= 0 or 1/(x+1)= -1. The first is at x= -1 and the second where -(x+1)= 1. Similarly, intervals on which 1/(x+1)< 2 and on which 1/(x+1)> 2 can be separated only where x+1= 0 or 1/(x+1)= 2. Use those equations to determine points separating the intervals and try one point in each interval to see which is true.
  6. May 11, 2009 #5
    Thanks for the reply. A number line always serves but it is the steps that can confuse me.

    Another question I've tried working on but I don't know if I made a critical error is

    (|x+1|/x) -2 > 3

    From here I did |x+1| -2x > 3x


    |x+1| -2x < 3x

    I am wondering about the above if I made any critical error.

    For if I further break down the two equation with respect to the absolute value I'd just be moving around the inequality sign again so I didn't.

    from the first eqn i got x< 1/4
    and x> 1/4 for the second

    I put 0 and 1/4 on a number line and tested numbers within the regions n<0, 0<n<1/4, and 1/4<n

    and came out with the final solution 0<x<1/4

    Is this correct? Did I use the correct procedure?

    Any help would be greatly appreciated.
  7. May 12, 2009 #6
    A "yes" or "no" would have a least been nice :(
  8. May 12, 2009 #7


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    This is correct so far, but I would have written it as |x+1|>5x if x>0 and |x+1|<5x if x<0

    This sentence makes no sense to me. You have two cases to examine:

    (1)Your first case is x>0 and hence |x+1|>5x.. clearly |x+1|=x+1 for this case and so it is simple.

    (2) Your second case is not so simple, for if -1<x<0, |x+1|=x+1 but when x<-1, |x+1|=-(x+1)


    you also need to examine the case where x<-1

    Your final answer is correct, but your procedure was not.
  9. May 12, 2009 #8


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    (|x+1|/x) -2 > 3. You might as well have expressed this (|x+1|/x) > 5 but that is a minor point.

    I don't know what you mean by the real number line, but if you mean sketch the functions, have an idea what they look like, that will be a check and can show you up errors. E.g. the above one has infinities and asymptotes which are pretty indicative.

    But for doing it formally as you are asked, for all of these inequalities needing manipulation I suggest you just have in mind anything like 4 < 7. If you multiply that by -1 or - anything it is not true, you have to change also the < into >. The trap is the when thing you are multiplying by is not obviously positive or negative like 5 or -2 for instance, but contains x. We say is a function of x, f(x) say. So you have two separate conditions, the original condition... oh words are too complicated, let's do formulae.

    If g(x)/f(x) > h(x).

    then g(x) > f(x)h(x) if f(x) > 0 , two conditions that you can combine on go on to find x > or < something.

    whereas g(x) < f(x)h(x) if f(x) < 0, a different two conditions which you can also combine to find a condition for the original inequality to hold.
    Last edited: May 13, 2009
  10. May 12, 2009 #9
    Thank you for the replies. Even simple corrections can make a great deal. Leaving one to wonder how on earth they got so mixed up in the first place.

    Thank you to everyone.
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