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Study23

  • 2 years ago

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?

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  1. Study23
    • 2 years ago
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    \(\ \Huge \left| \frac{1}{x} - \frac{1}{2} \right| < 0.2 \)

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

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

  4. Study23
    • 2 years ago
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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 \(\ \delta \).

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

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

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

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

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

  10. satellite73
    • 2 years ago
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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 \[|\frac{2-x}{x}|<3|x-2|\]

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

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