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Study23 Group Title

How do I isolate the x in this expression? (ie. so x is by itself). In other words, how do I factor that fraction out?

  • 2 years ago
  • 2 years ago

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  1. Study23 Group Title
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    \(\ \Huge \left| \frac{1}{x} - \frac{1}{2} \right| < 0.2 \)

    • 2 years ago
  2. satellite73 Group Title
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    what are you trying to solve?

    • 2 years ago
  3. satellite73 Group Title
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    i think you must start with \[|\frac{2-x}{2x}|<0.2\]

    • 2 years ago
  4. Study23 Group Title
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    It's in the format of \(\ \Huge \left| f(x) - L \right| < \epsilon \). Im trying to factor that expression so I can use the end result to find \(\ \delta \).

    • 2 years ago
  5. satellite73 Group Title
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    \[-0.2<\frac{2-x}{2x}<0.2\] is the next step

    • 2 years ago
  6. satellite73 Group Title
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    you probably are taking the limit as \(x\to 2\) so you have control over the size of the numerator

    • 2 years ago
  7. Study23 Group Title
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    Yes, I am taking the lim as x approaches 2

    • 2 years ago
  8. satellite73 Group Title
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    or you can write \[|\frac{2-x}{2x}|=\frac{1}{2}|\frac{x-2}{x}|<0.2\] so that \[|\frac{x-2}{x}|<0.4\]

    • 2 years ago
  9. Study23 Group Title
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    If I do that, then where would I go from there?

    • 2 years ago
  10. satellite73 Group Title
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    you have control over \(|x-2|\) that is you can make it as small as you like as for the \(x\) in the denominator, you can simply say that since you are taking the limit as \(x\to \frac{1}{2}\) you can assert that it is say between \(\frac{1}{3}\) and \(\frac{2}{3}\) so that the whole thing will be largest when the denominator is smallest, namely when it is \(\frac{1}{3}\) giving the inequality \[|\frac{2-x}{x}|<3|x-2|\]

    • 2 years ago
  11. Study23 Group Title
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    How would I find \(\ \delta ?\)

    • 2 years ago
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