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 September 7th, 2011, 08:15 AM #1 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 Write inequality in the form a < x < b A question asks me to write the inequality in the form a < x < b. How do I do this using the inequality |x| < 3 as an example?
 September 7th, 2011, 09:44 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,660 Thanks: 965 Math Focus: Elementary mathematics and beyond Re: Write inequality in the form a < x < b -3 < x < 3.
 September 7th, 2011, 11:42 AM #3 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 Re: Write inequality in the form a < x < b Thank you for the reply, but you only solved the equation. Can you explain it?
 September 7th, 2011, 01:57 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,660 Thanks: 965 Math Focus: Elementary mathematics and beyond Re: Write inequality in the form a < x < b |x| < 3 means x is within three units from zero, i.e. -3 < x < 3.
 September 8th, 2011, 07:23 PM #5 Senior Member   Joined: Apr 2011 From: USA Posts: 782 Thanks: 1 Re: Write inequality in the form a < x < b Remember that you have an absolute value of x. The "x < 3" part of it probably makes sense. The absolute value of a positive is just itself, so you can keep that "as is." But the absolute value of a negative number is the positive of that same number. So if you had -2, the absolute value is +2, which is less than 3, so that "fits." But the absolute value of -5 is +5, which is great than 3 so that doesn't work. Anything that is -3 or less, is going to be +3 or greater when you use the absolute value. So it also has to be greater than -3. Think of it like a mirror.
 September 8th, 2011, 09:06 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs Re: Write inequality in the form a < x < b You can also use $|x|\equiv\sqrt{x^2}$ to write: $\sqrt{x^2}<3$ Square both sides, since both are positive: $x^2<9$ $x^2-9<0$ $f(x)=(x+3)(x-3)<0=$ We have two critical values: -3, 3 which creates 3 intervals: $$$-\infty,-3$$$ f(-4) : (-)(-) = + not part of solution. $$$-3,3$$$ f(0) : (+)(-) = - part of solution $$$3,\infty$$$ : f(4) : (+)(+) = + not part of solution Thus, we find x is in the interval (-3,3) which written as an inequality is: -3 < x < 3

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