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Sorry for the unclear title. Hopefully i'll clear it up here.

If for one $x$ value a part of my function becomes undefined, does that mean that the whole function is undefined at that part?

For example look at this function

$$f(x)=\ln(x^2+1)-\ln(x-1)$$

$f(0)$ makes the $\ln(1) = 0$, but the second one $\ln(-1)$ which we know is not solvable. Plotting the function on a graph makes the domain go from $1$ to $\infty$. Im kind of vexed here, because i've done functions like $f(x)=\frac{x-2}{x+2} +4x$ and this is function would still be defined for $f(-2)$.

Any help would be greatly appreciated.

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  • $\begingroup$ Yes it will be undefined, you cannot perform operations on undefined numbers, including addition, multiplication, etc. $\endgroup$ – Rishi Dec 4 '16 at 14:21
  • $\begingroup$ Alright, what about the second function i included in the second paragraph. $\endgroup$ – Jane Doe Dec 4 '16 at 14:26
  • $\begingroup$ I think i kinda get the jist of it, it just does not make intuitive sense. I feel like i've also done basic arithmetic with a function whose value has sometimes been undefined, maybe i factored out the asymptote. I dont really know, thanks anyhow. $\endgroup$ – Jane Doe Dec 4 '16 at 14:29
  • $\begingroup$ Regarding the second function, that is also undefined at $x=-2$, despite the fact that $4x$ is still defined when $x=-2$ $\endgroup$ – Rishi Dec 4 '16 at 14:49
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it must be $$x\geq 1$$ since the definition of the logarithm function. In your second function we have $$x\ne -2$$ else we have a quotient with zero in the denominator

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