# If I is an open interval, f is differentiable on I ....

• Apr 13th 2010, 01:06 PM
sw2010
If I is an open interval, f is differentiable on I ....
If I is an open interval, f is differentiable on I and a is in I, then there is a sequence a_n in I\{a} such that a_n-> a and f '(a_n)->f '(a)
• Apr 15th 2010, 02:21 PM
sw2010
• Apr 15th 2010, 02:51 PM
Drexel28
Quote:

Originally Posted by sw2010

Work. Let's see some.
• Apr 15th 2010, 03:14 PM
sw2010
if f is differentiable, then f is continuous.
• Apr 15th 2010, 03:55 PM
Plato
Quote:

Originally Posted by sw2010
if f is differentiable, then f is continuous.

Good grief. Come on. Where is any real effort on your part?
If so you do, give us something more than that!
Think about the mean value theorem.
• Apr 15th 2010, 05:10 PM
sw2010
we need to assume that f is a continuous, real-valued function, defined on an interval. If the derivative of f at every interior point of the interval I exists and is zero, then f is constant.
Proof: Assume the derivative of f at every interior point of the interval I exists and is zero. Let (a, b) be an arbitrary open interval in I. By the mean value theorem, there exists a point c in (a,b) such that 0=f '(c)=f(b)-f(a)/(b-a)

So f(a) = f(b). Thus, f is constant on the interior of I and thus is constant on I by continuity
• Apr 15th 2010, 05:28 PM
mabruka
so how does that help? ??
• Apr 15th 2010, 05:31 PM
sw2010
If I is an open interval, f is differentiable on I and a is in I, then there is a sequence a_n in I\{a} such that a_n-> a and f '(a_n)->f '(a)

if f is differentiable on an interval then it is continuous on that interval. the mean value theorem says that there is a c in the the interval.