If $ f$ is continuous on $ [a,b] $ and differentiable on $ (a,b) $ does this imply $f'_{+}(a)$ exist?

Asked Jun 17 '21 at 14:30

Active Jun 17 '21 at 15:06

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Question: If $f$ is continuous on $ [a,b] $ and differentiable on $ (a,b) $ does this imply $f'_{+}(a)\in \mathbb{R} $ exist?

I'm not sure, here's a counter-example when $ f $ is not continuous on $ [a,b] $:
Let $ f(x) = 1/x $ for all $ x \in (0,1) $. Then $ f $ is differentiable on $ (0,1) $ but the right-hand sided limit at $ 0 $ ( i.e. the limit of $ f'(x) = \frac{1}{x} $ as $ x \to 0^+ $ ) does not exist.

edited Jun 17 '21 at 15:06

asked Jun 17 '21 at 14:30

hazelnut_116

1,699

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$\sqrt{x}$ on $[0,1]$? – eyeballfrog Jun 17 '21 at 14:32
The derivative of $ \sqrt{x} $ does not have limit at $ 0^+ $ – hazelnut_116 Jun 17 '21 at 14:34
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Also $f(x)=x\sin\frac1x$ on $[0,1],$ if you want q case where $f’(0)=\infty$ doesn’t make sense. – Thomas Andrews Jun 17 '21 at 14:35
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Ok, I haven't thought of that counter-example. You are right, the function is continuous on $ [0,1] $ and differentiable on $ (0,1] $ but the limit does not exist at $ 0^+ $. – hazelnut_116 Jun 17 '21 at 14:38
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What exactly do you mean by $f'_{a^+}(x)$? That's some very strange notation... – Hans Lundmark Jun 17 '21 at 14:59
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@HansLundmark You are right, I meant to write $ f'{+}(a)= lim{x \to a^+} \frac{f(x) - f(a)}{x-a} $ , sorry... – hazelnut_116 Jun 17 '21 at 15:06
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OK. But note that $f'+(a)$ is not necessarily equal to $\lim{x \to a^+} f'(x)$. The former may exist even if the latter doesn't (but not the other way around if $f$ is continuous on $[a,b]$, see here for example). – Hans Lundmark Jun 17 '21 at 15:15

If $ f$ is continuous on $ [a,b] $ and differentiable on $ (a,b) $ does this imply $f'_{+}(a)$ exist?

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