On what cases i get 0 denomiator

[transgalactic]

$$f(x)=\frac{1}{1+ln|x|}$$

the ln|x| part could be -1 when x=e^-1
correct??

 

[arildno]

Certainly!

Do you have any other solutions to the equation 1+ln|x|=0?

 

[transgalactic]

no
??

 

[arildno]

Well, if ln|x|=-1, what must |x| equal?

 

[transgalactic]

|x|=e^-1
x=-e^-1
x=e^-1

 

[tiny-tim]

$$f(x)=\frac{1}{1+ln|x|}$$

the ln|x| part could be -1 when x=e^-1
correct??

Yes.

 

[transgalactic]

thanks

 

[HallsofIvy]

But Arildno's point is that that is not the only value. As you said, the denominator is 0 for $e^{-1}$ or $-e^{-1}$.