# Continuity problem

• Jan 26th 2010, 01:36 PM
nautica17
Continuity problem
I'm not sure how to find these intervals of continuity:

ln(x-2)

sin(x^2 - 2)

I don't understand how I am supposed to find the continuity. I look at the graphs of each and I'm not sure what to look for. I understand functions like tan(x), etc. but with these I'm a little confused. Can anyone point me in the right direction?
• Jan 26th 2010, 01:39 PM
Jhevon
Quote:

Originally Posted by nautica17
I'm not sure how to find these intervals of continuity:

ln(x-2)

sin(x^2 - 2)

I don't understand how I am supposed to find the continuity. I look at the graphs of each and I'm not sure what to look for. I understand functions like tan(x), etc. but with these I'm a little confused. Can anyone point me in the right direction?

these functions are continuous where they are defined. the log function is continuous where its argument is positive: for example, $\ln x$ is continuous where $x > 0$

the sine function is continuous everywhere, and so it is continuous wherever its argument is: $\sin (f(x))$ is continuous where $f(x)$ is continuous.

so what can you do with that information?
• Jan 26th 2010, 01:55 PM
nautica17
Quote:

Originally Posted by Jhevon
these functions are continuous where they are defined. the log function is continuous where its argument is positive: for example, $\ln x$ is continuous where $x > 0$

the sine function is continuous everywhere, and so it is continuous wherever its argument is: $\sin (f(x))$ is continuous where $f(x)$ is continuous.

so what can you do with that information?

Okay, so from what you have said I have come up with this:

ln(x-2) is continuous when x is greater than 2?

sin(x^2 - 2) is continuous no matter what and on whatever interval you sort of choose to put it on?
• Jan 26th 2010, 01:59 PM
Plato
Quote:

Originally Posted by nautica17
Okay, so from what you have said I have come up with this:

ln(x-2) is continuous when x is greater than 2?

sin(x^2 - 2) is continuous no matter what and on whatever interval you sort of choose to put it on?

Correct.
• Jan 26th 2010, 02:02 PM
nautica17
Alright thanks guys. :)

Sorry if it's a simple question, but it looked confusing at first.
• Jan 26th 2010, 08:19 PM
Pulock2009
Quote:

Originally Posted by Jhevon
these functions are continuous where they are defined. the log function is continuous where its argument is positive: for example, $\ln x$ is continuous where $x > 0$

the sine function is continuous everywhere, and so it is continuous wherever its argument is: $\sin (f(x))$ is continuous where $f(x)$ is continuous.

so what can you do with that information?

hey isn't -1<=sin(x^2-2)<=1
=>-pi/2<=x^2-2<=pi/2
=>sqrt(2-pi/2)<=x<=sqrt(pi/2+2)????
• Jan 26th 2010, 08:27 PM
Drexel28
Quote:

Originally Posted by Pulock2009
hey isn't -1<=sin(x^2-2)<=1
=>-pi/2<=x^2-2<=pi/2
=>sqrt(2-pi/2)<=x<=sqrt(pi/2+2)????

What?!
• Jan 27th 2010, 04:20 AM
HallsofIvy
Quote:

Originally Posted by Pulock2009
hey isn't -1<=sin(x^2-2)<=1
=>-pi/2<=x^2-2<=pi/2
=>sqrt(2-pi/2)<=x<=sqrt(pi/2+2)????

No, that is an interval on which the function is one-to-one and has nothing to do with where it is continous. $x^2- 2$ is continuous for all x, sin(x) is continous for all x, and the composition of two continuous functions is continuous.

In a certain, very specific, sense, "almost all" functions are never continuous. But continuous functions are so nice that our ways of writing functions have developed so that functions written as a single "formula" are continuous where ever they are defined.