Domain of f(x)= (1/(3+lnx)): (0,∞)

[Oliviaven]

Hi just a quick question
Differentiate f and find the domain of f. (Enter the domain in interval notation.)
f(x)=(1/(3+lnx))


Here is my Derive: ((-3+lnx)^-2)(1/x) which is correct

And for the domain (web assign wants interval notation)
When I was doing this part, I wasn't sure if it includes 0 or not since ln(0) cannot be defined.
I tried both (0,∞) and [0,∞) which both were incorrect.

Did I visualize the graph wrong or something? I even tried to use a graphics calculator

Please enlighten me ^^

 

[Dick]

1/(3+ln(x)) is also not defined if ln(x)=(-3), right? For what value of x does that happen?

 

[SammyS]

Hi just a quick question
Differentiate f and find the domain of f. (Enter the domain in interval notation.)
f(x)=(1/(3+lnx))Here is my Derive: ((-3+lnx)^-2)(1/x) which is correct Not quite correct. It should be (-(3+lnx)^-2)(1/x)

And for the domain (web assign wants interval notation)
When I was doing this part, I wasn't sure if it includes 0 or not since ln(0) cannot be defined.
I tried both (0,∞) and [0,∞) which both were incorrect.

Did I visualize the graph wrong or something? I even tried to use a graphics calculator

Please enlighten me ^^

\displaystyle f(x)=\frac{1}{3+\ln(x)}

What value of x would make the denominator zero ??

 

[Oliviaven]

so whenever x=e^-3 this equation will not exist? Since denominator will equal to 0 ^^

Im still kind of confused how i should input the interval notation...

will it be (0,e^-3)U(e^-3,∞) if I use interval notation? Should 0 be included since ln(0) is undefined

 

[SammyS]

so whenever x=e^-3 this equation will not exist? Since denominator will equal to 0 ^^

Im still kind of confused how i should input the interval notation...

will it be (0,e^-3)U(e^-3,∞) if I use interval notation? Should 0 be included since ln(0) is undefined

Yes, that's the way to write the domain in interval notation.

Of course 0 is not included. ln(0) is undefined just as division by zero is undefined.

 

[Oliviaven]

Thank you for helping me on solving the question ^^

From the number of posts, you can see that I'm a new user in this forum and I think this is one really awesome forum