Suppose $f:[a,b]\rightarrow \mathbb R $ is differentiable on (a,b) and continuous on [a,b]. Does it follow that $f$ is right-differentiable at $a$ and left-differentiable at $b$?

I guess it does not follow since $f:[a,b]\rightarrow \mathbb R $ can be differentiable at a point, call it $ c \in (a,b)$ but not at the end points since the interval is open.

Is it right? If so, how do i formally prove?


  • 1
    $\begingroup$ Think of square roots... $\endgroup$ – Amitai Yuval Dec 4 '14 at 23:29
  • $\begingroup$ am I assuming right? $\endgroup$ – needhelp Dec 4 '14 at 23:31
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    $\begingroup$ Yes yes you are. $\endgroup$ – Amitai Yuval Dec 4 '14 at 23:31

Consider the function given by $$f(x)=\cases{x\cos(1/x)&if $0<x\le1$\cr 0&if $x=0$.\cr}$$ It is clear that $f$ is continuous on $(0,1]$ and differentiable on $(0,1)$. Moreover, $$\lim_{x\to0^+}f(x)=0$$ by the pinching theorem (sandwich theorem, squeeze theorem), so $f$ is also continuous at $0$ and the assumptions in your question are satisfied. However, $f$ is not right-differentiable at $0$ since $$\frac{f(x)-f(0)}{x-0}=\cos\Bigl(\frac1x\Bigr)$$ has no limit as $x\to0^+$.

  • $\begingroup$ Would it be still valid if $f$ is continuous on [0,1]? $\endgroup$ – needhelp Dec 4 '14 at 23:42
  • $\begingroup$ In my example, $f$ is continuous on $[0,1]$. $\endgroup$ – David Dec 4 '14 at 23:43
  • $\begingroup$ Ok, i just saw that "...f is continuous on (0,1] ..." thanks! $\endgroup$ – needhelp Dec 4 '14 at 23:46

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