My Math Forum Let $f(x) = 0$ ; $x \epsilon R-${$0,1$}.

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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)
 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. Thanks from Lalitha183
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 its 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 $f'(x)$.

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