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 March 16th, 2018, 05:56 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 06:56 PM. March 16th, 2018, 06:22 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 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. Thanks from Lalitha183 Last edited by skipjack; August 8th, 2018 at 06:57 PM. March 16th, 2018, 06:29 PM   #3
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 Originally Posted by SDK 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.
Thanks for the help. I have checked that the function is differentiable at $0$. In the option (c), it says $f'(x)$ is not continuous at $0$ and $1$. I have $f(x)$ as $0$. So I thought if it is continuous or not in case of $f'(x)$.

Last edited by skipjack; August 8th, 2018 at 06:59 PM. August 8th, 2018, 03:58 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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, 01:15 PM #5 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 669 If f(x) is not defined at 0 or 1, then d) applies for lack of information. August 8th, 2018, 07:09 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 What information did you have in mind for choice (a)? Tags $fx, epsilon, r$$0, r$0 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post daigo Calculus 2 June 23rd, 2013 02:52 PM yms Calculus 3 November 17th, 2012 12:37 PM mymathgeo Real Analysis 17 January 26th, 2012 10:00 PM guroten Calculus 1 September 14th, 2010 01:19 PM ElMarsh Real Analysis 8 October 4th, 2009 01:11 PM

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