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


$f(0)$ 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.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.