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Absolute function problem.

  1. Sep 13, 2014 #1

    adjacent

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    1. The problem statement, all variables and given/known data
    ##|x^2-2| \leq 1##
    Find ##x##

    3. The attempt at a solution
    ##x^2-2\leq 1##
    ##x \leq \sqrt{3}##

    and

    ##-(x^2-2)\leq 1##
    ##-x^2+2 \leq 1##
    ##x \geq 1##

    Therefore the solution is ##1 \leq x \leq \sqrt{3}##
    However the book gives another solution which is ##-1 \leq x \leq -\sqrt{3}##
    I don't know how the latter solution is found out. How can a range be multiplied by ##-1##?
     
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  3. Sep 13, 2014 #2

    BruceW

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    be careful when taking the square root.
    [tex]x^2 \geq 1[/tex]
    does not imply x is positive.
     
  4. Sep 13, 2014 #3

    Ray Vickson

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    Draw the graph of ##y = x^2-2## over some reasonable x-range. On the same plot, draw horizontal lines ##y = 1## and ##y =-1##. What does your plot show?
     
  5. Sep 13, 2014 #4

    adjacent

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    If that's the case, can't we write ##-1 \leq x \leq \sqrt{3}## or ##1 \leq x \leq -\sqrt{3}##
    Why should we make the two square roots have the same sign?
     
  6. Sep 13, 2014 #5

    adjacent

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    Here it is.

    I don't see anything special in it :confused:
     

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  7. Sep 13, 2014 #6

    Ray Vickson

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    Oh? You can't see what points on the graph ##y = x^2-2## lie between y = -1 and y = +1?
     
  8. Sep 13, 2014 #7

    adjacent

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    ##1## to ##\sqrt{3}## when ##x>0## and ##-\sqrt{3}## to ##-1## when ##x<0##?
    But I am not supposed to draw a graph. :confused:
     
  9. Sep 13, 2014 #8

    Ray Vickson

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    So, don't tell anybody. You just use the graph to get a "feel" for what is going on, and to help you keep your thinking straight. After you have used the graph like that you can throw it away. That is what we all do!
     
  10. Sep 13, 2014 #9

    statdad

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    You should see something wrong immediately with one of these: it is really true that [itex] 1 \le -\sqrt 3 [/itex]?
     
  11. Sep 13, 2014 #10

    vela

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    It's easier if you look at it this way: You got to the point
    $$1 \le x^2 \le 3.$$ Now take the square root, remembering that ##\sqrt {x^2} = \lvert x \rvert##.
     
  12. Sep 13, 2014 #11

    BruceW

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    Well, your choice at the start of your solution is an if statement.
    IF ##x^2 \geq 2## then ##x^2-2 \leq 1## and a bunch of stuff follows
    ELSE IF ##x^2 \leq 2## then ##-x^2+2 \leq 1## and a bunch of stuff follows
    So you have to treat the two IF statements as completely separate things. You cannot carry results over from one IF statement into the other IF statement. I hope that answers your question. I wasn't totally sure what you meant.

    p.s. I think the other solution given by the book should be ##-\sqrt{3} \leq x \leq -1## not ##-1 \leq x \leq -\sqrt{3}##
     
  13. Sep 14, 2014 #12

    adjacent

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    Ok :smile:
    :shy: No. What about [itex] -1 \le \sqrt 3 [/itex]
    Oh. This seems to be a good way. Thanks
    Where did ##x^2 \geq 2## come from?
    Yes. It was a typo. :shy:
     
  14. Sep 14, 2014 #13

    BruceW

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    This was your implied "choice" when you said ##|x^2-2|=x^2-2##. The other possible choice was ##|x^2-2|=-x^2+2## which occurs iff ##x^2 \leq 2##. So anyway, the point I was hoping to get across is that the two cases ##x^2 \geq 2## and ##x^2 \leq 2## should be considered as separate solutions to the problem. It is not OK to carry over results from one case to the other. Also, as I said before, I am not totally sure if this was where you were having trouble anyway, but I was hoping it might be useful.
     
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