Countinous after derivative

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February 6th 2006, 06:45 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.
February 6th 2006, 09:44 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
February 7th 2006, 02:00 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?
February 7th 2006, 08:14 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
February 8th 2006, 06:22 PM
ThePerfectHacker

I agree with you,

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

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