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March 16th, 2018, 06:56 PM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Let $f(x) = 0$ ; $x \epsilon R${$0,1$}.
Let $f(x) = 0$ ; $x \epsilon R\{0,1\}$. Then a) $f$ is discontinuous at $0$ and $1$ b) $f$ is continuous but not differentiable at $0$ and $1$ c) $f$ is differentiable at $0$ and $1$ but $f'$ is not continuous at $0$ and $1$ d) none of the above If $f'(x) =0$, then is it continuous? Only option (c) I have doubt. If yes, then the answer would be option (d). Last edited by skipjack; August 8th, 2018 at 07:56 PM. 
March 16th, 2018, 07:22 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 531 Thanks: 304 Math Focus: Dynamical systems, analytic function theory, numerics 
I suppose you mean $f(x) = 0$ for $x \in \mathbb{R}  \{0,1\}$? If so, then the answer should be (d). You should find examples which violate each of (a)(c). Also, $f'(x) = 0$ is not necessary for any examples, nor does this assumption make any of the claims true. I'm not even sure how it's related. Last edited by skipjack; August 8th, 2018 at 07:57 PM. 
March 16th, 2018, 07:29 PM  #3  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
Last edited by skipjack; August 8th, 2018 at 07:59 PM.  
August 8th, 2018, 04:58 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
I assume that "x∈R−{0,1}" means that f is not defined at 0 and 1. In that case it is not continuous so not differentiable there.

August 8th, 2018, 02:15 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,660 Thanks: 647 
If f(x) is not defined at 0 or 1, then d) applies for lack of information.

August 8th, 2018, 08:09 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,089 Thanks: 1902 
What information did you have in mind for choice (a)?


Tags 
$fx, epsilon, r$$0, r$0 
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