# Countinous after derivative

• Feb 6th 2006, 07:45 PM
ThePerfectHacker
Countinous after derivative
If a function $f$is countinous on the closed interval $(a,b)$ and diffrenciable on the open interval $(a,b)$. Then $f'$ is countinous on the closed interval $[a,b]$. It seems true, I was not able to find any counterexamples.
• Feb 6th 2006, 10:44 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
If a function $f$is countinous on the closed interval $(a,b)$ and diffrenciable on the open interval $(a,b)$. Then $f'$ is countinous on the closed interval $[a,b]$. It seems true, I was not able to find any counterexamples.

I may have the wrong end of the stick here but what about:

$f(x)=\sqrt x,\ x \in [0,1]\ ?$

RonL
• Feb 7th 2006, 03:00 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
I may have the wrong end of the stick here but what about:

$f(x)=\sqrt x,\ x \in [0,1]\ ?$

RonL

"Wrong end of the stick" is that supposed to be a pun :D
Well, it is true that the endpoint is not countinous (end of stick) okay thus, what about the open interval?
• Feb 7th 2006, 09:14 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
"Wrong end of the stick" is that supposed to be a pun :D
Well, it is true that the endpoint is not countinous (end of stick) okay thus, what about the open interval?

Not sure what you mean. It's a counter example; its continuous on a closed
interval $[0,1]$, differentiable on $(0,1)$, but its derivative is not continuous
on $[0,1]$.

RonL
• Feb 8th 2006, 07:22 PM
ThePerfectHacker
I agree with you,

but what about if discountinous on an open interval? Is that possible?