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Thread: Related range of values of f(x) necessary/sufficient/ with absolute values

  1. #1
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    Thumbs up Related range of values of f(x) necessary/sufficient/ with absolute values

    Hi Forum.

    I've asked a lot of times about absolute values.
    But here I go again:

    Consider two conditions
    $\displaystyle x^2-3x-10$ and $\displaystyle |x-2|<a$

    Where x is a real number, and a is a positive real number.

    The range of a, so that $\displaystyle |x-2|<a$ is a necessary condition for $\displaystyle x^2-3x-10<0$

    The range of a, so that $\displaystyle |x-2|<a$ is a sufficient condition for $\displaystyle x^2-3x-10<0$

    This problem is mainly about logic, right?
    Here is something that I thought, after calculating the range of $\displaystyle x^2-3x-10<0$, that is $\displaystyle -2<x<5$

    If we have $\displaystyle -2<x<5$
    -2________________________________5
    |________________________________|
    ___________________________a=0 ______________________________ a=3
    |_________________________________________________ _| <-|x-2|<a


    Edit: This didn't work so well but, the a=0 is right below -2 and the a=3 is right below 5


    This is our sufficient condition: 0<a<3
    I'm not sure what the absolute value does here.
    What does the absolute value mean?
    Absolute values are only the distance to x=0, right?

    After calculating -correctly- you get a nicer solution that is:
    $\displaystyle 0<a\leqslant 3$

    But how do we get such solution?

    $\displaystyle |x-2|<a $ is
    $\displaystyle x-2$ if $\displaystyle x-2\geqslant a$ and
    $\displaystyle -x+2$ if $\displaystyle x-2<a$

    This relates to two range of values so this got a little out of control.
    Somehow confusing, I wish I knew somewhere to find exercises of this kind!

    I'm confused about that necessary/sufficient part

    Sufficient
    If you mow the lawn, you receive 10 dollars.
    ____________________Necessary

    The $\displaystyle 0<a\leqslant 3$ range is sufficient for f(x) to work, or something like that?
    But where does necessity comes in this? I know that if something is necessary it is not sufficient.

    A--->B
    A implies B
    A sufficient, B necessary

    I appreciate any help!
    Thanks!
    Last edited by Zellator; May 31st 2011 at 07:50 PM. Reason: Correction
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  2. #2
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    Interesting problem. Your problem is dealing with implication. Let's let $\displaystyle P$ be the proposition that

    $\displaystyle x^{2}-3x-10<0,$

    and let's let $\displaystyle Q(a)$ be the proposition that

    $\displaystyle |x-2|<a.$

    Now, if you want to find the range of $\displaystyle a$ such that $\displaystyle Q(a)$ is sufficient for $\displaystyle P,$ then that's the same thing as finding the range of $\displaystyle a$ such that $\displaystyle Q(a)$ implies $\displaystyle P.$

    Conversely (literally!), if you want to find the range of $\displaystyle a$ such that $\displaystyle Q(a)$ is necessary for $\displaystyle P,$ then that's the same thing as finding the range of $\displaystyle a$ such that $\displaystyle P$ implies $\displaystyle Q(a).$ So here, you just reverse the direction of the implication.

    I agree with your 'sufficient' answer. $\displaystyle 0<a\le 3$ is exactly correct.

    But now, with the 'necessary' answer, you want $\displaystyle P$ to imply $\displaystyle Q(a).$ How could that happen?
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