sketch intervals described by given inequalities: lx- 1/2 l < 1
I'm assuming that it's \[\large \left| x-\frac{ 1 }{ 2 } \right|<1\] So remember that absolute value is DISTANCE FROM 0. So if some {stuff} has to be LESS THAN a units from 0 (a is a distance here so has to be positive), what that really means is that the stuff satisfies: \[\large -a<stuff<a\]
yes it is haha sorry about that, I dont know how to type it onto the computer
So for example\[\large \left| t \right|<7 \]really means \rightarrow -7<t<7\[ -7<t<7\] since in order for t to be within 7 units of 0, it must be between -7 and +7. right?
(that's ok - that's exactly what you typed but I just wanted to confirm, because sometimes people DON'T mean exactly what they type! lol)
oh haha, okay. okay so the 0 is about one unit, or point, away ?
@DebbieG Okay, so .... I get this then \[1/2 < 0 > 1 \frac{ 1 }{ 2 } ? Right?\]
\[\left| x-\frac{ 1 }{ 2 } \right|<1\] \[-1 < x-\frac{ 1 }{ 2 } <1\] then you add 1/2 to all sides. what do you get?
^^ right.... what @surdawi said. Your inequality can NEVER have the inequality signs going in opposite directions, that's a big no-no!
i get \[-\frac{ 1 }{ 2 } <x >1\frac{ 1 }{ 2 }\]
oh haha I didnt notice I did opposite signs
why do you flip the inequality sign ?
I pressed the wrong button, sorry @surdawi
No.... look at my example above. if |stuff|<a then -a<stuff<a 1. the inequalities go in the same direction 2. the STUFF is exactly what was inside the absolute value 3. use -a on the left and a on the right.
THEN solve that compound inequality by isolating the variable in the middle.
ya, so its - 1/2 < x < 1 1/2; right?
Yes, there you go. :)
so x> -1/2 and x< 1 1/2 ?
so by showing these calculations, is this is what it means to sketch or would I have to sketach a number line?
Write. If you are to graph it on a number line, you draw in the line between those two numbers. Since it is a strict inequality, you should have a "()" at each endpoing (or an open circle) to indicate that the endpoint is not included.
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