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March 16th, 2018, 05:56 PM   #1
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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.
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March 16th, 2018, 06:22 PM   #2
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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.
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Last edited by skipjack; August 8th, 2018 at 06:57 PM.
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March 16th, 2018, 06:29 PM   #3
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Quote:
Originally Posted by SDK View Post
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.
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August 8th, 2018, 03:58 AM   #4
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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.
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August 8th, 2018, 01:15 PM   #5
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If f(x) is not defined at 0 or 1, then d) applies for lack of information.
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August 8th, 2018, 07:09 PM   #6
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What information did you have in mind for choice (a)?
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