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Study23
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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
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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Study23 Group TitleBest ResponseYou've already chosen the best response.0
\(\ \Huge \left \frac{1}{x}  \frac{1}{2} \right < 0.2 \)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
what are you trying to solve?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i think you must start with \[\frac{2x}{2x}<0.2\]
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
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

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[0.2<\frac{2x}{2x}<0.2\] is the next step
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
you probably are taking the limit as \(x\to 2\) so you have control over the size of the numerator
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Yes, I am taking the lim as x approaches 2
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
or you can write \[\frac{2x}{2x}=\frac{1}{2}\frac{x2}{x}<0.2\] so that \[\frac{x2}{x}<0.4\]
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
If I do that, then where would I go from there?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
you have control over \(x2\) 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 \[\frac{2x}{x}<3x2\]
 2 years ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
How would I find \(\ \delta ?\)
 2 years ago
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