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Homework Help: Finding the set of values for x

  1. Apr 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the set of values for x which

    [tex]\frac{(x-3)^{2}}{x+1} < 2[/tex]

    I turned this into

    [tex]\frac{(x-7)(x-1)}{x+1} < 0 [/tex]

    Now this implies that the critical angles are x = 7, 1 and -1 but how do I create the sets. I think I have to do somthing with sign change?

    Thanks :)

  2. jcsd
  3. Apr 25, 2009 #2
    There is nothing wrong with your latex; the forum is having problems at the moment.

    It is a product of 3 terms, so you have the conditions abc < 0 iff (ab < 0 and c > 0) or (bc < 0 and a > 0) or (ac < 0 and b > 0). Inequalities usually generate logical sets like this.
  4. Apr 25, 2009 #3


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    Hi thomas49th! :smile:
    Yes, that's fine …

    (and of course it's the same as the product (x-7)(x-1)(x+1) < 0 … do you see why? :wink:)

    and you know that the sign can only change at 7 1 or -1 (but why are you calling them angles? :confused:), and will change each time (since there are no squared factors) :wink:
  5. Apr 25, 2009 #4
    Sorry I meant values NOT angles. My bad :S

    Never done the sets like that. Pretty sure i've never been taught that method, though I cant remeber the method so that's why im asking :)

    I don't know why you can multiply them like that. (x+1) is the denominator !!! you could say (x+1)-1 maybe?

    The way I say it outloud is:

    y values of (x-7)(x-1) are divide by the y values x+1 graph and I want to find where the new composite graph is below the x-axis. I'm not sure what the most sensible way of doing it is... Im sure there was a method of drawing out a table???

    Thanks :)
  6. Apr 25, 2009 #5


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    Because (x+1)2 is always positive, so A/(x+1) has the same sign as (A/(x+1))(x+1)2, = A(x+1).
    Why are you making this so complicated?

    Use slider142's :smile: method …

    the product of three things will be negative if exactly an odd number of them is negative … ie if either one is negative, or all three are negative …

    so in this case … ? :smile:
  7. Apr 26, 2009 #6
    sorted. Thanks :)
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