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For $f(x) = 1/(2-x)$.

Apologies for this overtly simplistic question; am I missing something deeper/more meaningful in this homework question or is it as simple as $m=1$?

In terms of context, the previous questions asked to determine the domain of function and it's compositions with other functions. I'm not sure how this question (as part of the same set) has to do with domains if any.

If I'm overlooking something, I would greatly appreciate a nudge/hint in the right direction as to what should consider before applying Occam's razor.

Thank you!

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    m has to be 1 only,2012-03-18
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    @5T0M: There is another solution.2012-03-18
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    @xlm Is $m=1$ the only possibility? Why?2012-03-18
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    for m=1, any value of x satisfies the equation, for m=0, only x=0 satisfies it.2012-03-18

2 Answers 2

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The question can be rephrased as follows: for which $m$ does the equation $$\frac1{2-mx}=\frac1{2-x}$$ have a solution? If you try to solve this equation, you find that you must have $$2-x=2-mx\;,$$ and hence $x=mx$. This certainly has a solution when $m=1$, but in fact it has a solution for lots of other values of $m$ as well. What are they? (Note: for these values it has only one particular solution.)

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    is it when m tends to -$\infty$2012-03-18
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    @5T0M: No, it’s an honest-go-goodness number. But whereas **every** value of $x$ works for $m=1$, for this value of $m$ there’s only one value of $x$ that’s a solution.2012-03-18
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    Is it $m=-1$, and $x = 0$2012-03-18
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    for x=0, any value of m satisfies the equation.2012-03-18
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    @5T0M: Sorry: I had a mental hiccup, and my answer wasn’t quite right. Yes, you’ve found one solution. Are there more along the same lines?2012-03-18
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    (m,x) can be (1.R), or (-1,0) or (R,0), R denotes the set of real numbers. I think this looks fine @BrianM.Scott, which is partially derived from quartz's comments2012-03-18
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    @5T0M: Right, except that your notation is incorrect. What you mean is that $(m,x)$ can be $(1,r)$ or $(r,0)$ for any $r\in\Bbb R$.2012-03-18
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    Yes thats what I meant, @BrianM.Scott, by the way how did you generated the symbol to represent the set of real numbers?2012-03-18
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    $x = m x \implies x - m x = 0 \implies x(1-m) = 0.$ So two cases (1) $m = 1$ (2) $x = 0 \implies m =\ \color{red}{??}$ @xlm can you find $\color{red}{??}$2012-03-18
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    @BrianM.Scott, Thanks for the great answer! I'm so glad I asked this, your hint really opened my eyes!2012-03-18
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    @J.D., thanks for setting it out. To check if I got it: if m=1 then x is any real except 2? (divide by zero); if x=0 then m is any real.2012-03-18
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    @BrianM.Scott I just cooked an answer compiling all the ingredients from the comments.2012-03-18
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$(m,x)$ can have the following values,

when $m=1$ , $x$ can be any real number (except 2)

when $x=0$ , $m$ can be any real number,

when $m=-1$, $x$ has to be $0$.

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    The answer has been derived from comments by quartz and Brian M Scott.2012-03-18
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    Thanks for summarizing. Should we consider also if m=1, x can be any real except 2?2012-03-18
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    Yes very true, I just edited the answer, thanks for pointing that out @xlm2012-03-18