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I have found that $f(-2)=\frac{-1}{3}$ and that $f(-2)=0$. But that means that $\frac{-1}{3}=0$. But this is clearly false. Does this means that there is no solution to the function? Is it enough to conclude?

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  • $\begingroup$ It might help if you give the original problem. $\endgroup$
    – Brian Tung
    Sep 22 '16 at 16:49
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I'm not sure if "no solution" is the term you're looking for. If you get $f(-2) = \frac{-1}{3} = 0$ then this contradicts the very definition of what a function is. So, you can conclude that your function is not defined.

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  • $\begingroup$ Yea, so there is no solution for the function $f(x)$, no? $\endgroup$ Sep 22 '16 at 16:52
  • $\begingroup$ what are you trying to solve? you usually only talk about something having "no solution" when you're trying to solve a system, etc. If you found that $f(x)$ equals two different values, for the same $x$, then your function is not defined. This could imply that the system you're trying to solve has no solution, but we'd have to know more. $\endgroup$ Sep 22 '16 at 16:56
  • $\begingroup$ OK, that was what I were searching. Thanks +1 $\endgroup$ Sep 22 '16 at 17:33
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Depends on the conditions imposed on $f(x)$. If $f(x)$ is a multi-valued or many-valued function, then there is no problem with the above statement.

In the strict sense , however, a well-defined function associates one, and only one, output to any particular input. The term multi-valued function is, therefore, a misnomer because functions are single-valued. Multi-valued functions often arise as inverses of functions that are not injective.

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