Thread: 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)

please help me give me a hint

Originally Posted by sw2010

please help me give me a hint

Work. Let's see some.

if f is differentiable, then f is continuous.

Originally Posted by sw2010

if f is differentiable, then f is continuous.

Good grief. Come on. Where is any real effort on your part?
Do you understand anything about this question?
If so you do, give us something more than that!
Think about the mean value theorem.

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

so how does that help? ??

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.