
Real Analysis Real Analysis Math Forum 
 LinkBack  Thread Tools  Display Modes 
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) 
March 16th, 2018, 06:22 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 377 Thanks: 205 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 its related. 
March 16th, 2018, 06:29 PM  #3  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
 

Tags 
$fx, epsilon, r$$0, r$0 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Epsilon Delta?  daigo  Calculus  2  June 23rd, 2013 02:52 PM 
N(epsilon) for lim n^(1/n) = 1  yms  Calculus  3  November 17th, 2012 12:37 PM 
epsilon delta  mymathgeo  Real Analysis  17  January 26th, 2012 10:00 PM 
epsilon proof  guroten  Calculus  1  September 14th, 2010 01:19 PM 
less than or equal to a given epsilon  ElMarsh  Real Analysis  8  October 4th, 2009 01:11 PM 