PDA

View Full Version : Is a most general antiderivative required to be continous?


General_Sax
Nov18-09, 07:57 PM
I've looked through my textbook and I've found no mention, but I've seen some posts on the Internet which explicitly state to check for continuity while taking the derivative certain functions.

Is it because: F'(x) = f(x)

I understand that a function is required to be continuous and differentiable, is that why we must check an antiderivative for continuity?

Dick
Nov18-09, 09:45 PM
The derivative of 'certain functions' aren't continuous, that's true. But why do you think you need to check antiderivatives for continuity because of 'some posts'? You are being really vague here.

HallsofIvy
Nov19-09, 04:25 AM
I've looked through my textbook and I've found no mention, but I've seen some posts on the Internet which explicitly state to check for continuity while taking the derivative certain functions.
While taking the "derivative"? Well, certainly, if a function is not continuous at a given point, it does not have a derivative there. But are you talking about dertivatives or antiderivagives?

Is it because: F'(x) = f(x)

I understand that a function is required to be continuous and differentiable, is that why we must check an antiderivative for continuity?
I don't understand what you are talking about. What do you mean by "F'(x)= f(x)"? And "a function is required to be continuous and differentiable" for what?

HallsofIvy
Nov19-09, 04:28 AM
Do you mean "Because, if F(x) is the anti-derivative of f(x) then F'(x)= f(x) and a function is differentiable at a given point only if it is continuous there, must the anti-derivative of a function be continuous at that point?" The answer to that is "yes". Although you can have situations where F(x) is the anti-derivative of f(x) except at a single point- and then F may not be continous at that point.

General_Sax
Nov19-09, 08:17 PM
Yes, that's exactly what I meant. Sorry for being so unclear.

Basically, I had to find the most general anti-derivative of an absolute value function, so I was searching on the Internet and found a similar question.

They advised taking the anti-derivative on the individual intervals and then checking F(x) for continuity be equating the applicable limits. This allows for one to solve for the constants (in a sense).

However, I could never find a reason for why this is so.

"Although you can have situations where F(x) is the anti-derivative of f(x) except at a single point- and then F may not be continous at that point."

If you had an definite integral and the anti-derivative was discontinuous would you sum the areas before and after the discontinuity?

lanedance
Nov19-09, 08:57 PM
without getting too rigourous... if you think of the anti-derivative as "the area under the curve", then if F(x) = \int_a^x dx f(x) and f is continuous, it should be reasonably obvious that F will also be continuous (and differentiable, with F'(x) = f(x))

if you allow some less well behaved functions (like the dirac delta fucntion \delta(x)) then the inetgral doesn't have to be conitinuous (\int_{-\infty}^x dx \delta(x) is the step function, though \delta(x)) is not acontinuous function