Differentiability implying continuity

[ahorn]

Does differentiability on (a,b) imply continuity on [a,b]?
If yes, why does it not imply continuity only on (a,b)?
If a function is continuous only on (a,b), is it possible for it to be differentiable on (a,b)?
Put differently, I can prove that a function f is continuous at a if f'(a) exists, but does that say anything about f near a?

 

[pka]

Does differentiability on (a,b) imply continuity on [a,b]?
If yes, why does it not imply continuity only on (a,b)?
If a function is continuous only on (a,b), is it possible for it to be differentiable on (a,b)?
Put differently, I can prove that a function f is continuous at a if f'(a) exists, but does that say anything about f near a?

\(\displaystyle f(x) = \left\{ {\begin{array}{*{20}{rl}}
{\sqrt {1 - {x^2}} ,}&{0 \le x \le 1}\\
{b,}&{\text{else}}
\end{array}} \right.\)

The function \(\displaystyle f\) is clearly differentiable on \(\displaystyle (0,1)\) but not continuous at \(\displaystyle 1\) unless \(\displaystyle b=0\).
But in that case is it differentiable at \(\displaystyle x=1~?\)

 

[ahorn]

\(\displaystyle f(x) = \left\{ {\begin{array}{*{20}{rl}}
{\sqrt {1 - {x^2}} ,}&{0 \le x \le 1}\\
{b,}&{\text{else}}
\end{array}} \right.\)

The function \(\displaystyle f\) is clearly differentiable on \(\displaystyle (0,1)\) but not continuous at \(\displaystyle 1\) unless \(\displaystyle b=0\).
But in that case is it differentiable at \(\displaystyle x=1~?\)

I'd say this function is not differentiable at x=1 because \(\displaystyle \lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} \neq \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h}=0\) iff \(\displaystyle b \neq 0\)

Are you saying that this is a case where differentiability on (a,b) does not imply continuity on [a,b]?

 

[pka]

I'd say this function is not differentiable at x=1 because \(\displaystyle \lim_{h \to 0^+} \frac{f(1+h)-f(1)}{h} \neq \lim_{h \to 0^-} \frac{f(1+h)-f(1)}{h}=0\) iff \(\displaystyle b \neq 0\)
Are you saying that this is a case where differentiability on (a,b) does not imply continuity on [a,b]?

If \(\displaystyle b=1\) then \(\displaystyle f\) is differentiable on \(\displaystyle \mathbb{R}\setminus\{0,1\}\)

 

[ahorn]

If f'(x) = 0 on an interval (a,b), then f is constant on [a,b]

I created this thread when working on the proof for "If f'(x)=0 on an interval (a,b), then f is constant on [a,b]". Now I see from your example that if the endpoints are not continuous, f can still be differentiable on (a,b). So, doesn't that mean there should be an extra proviso in the statement above: either that f is constant on (a,b), or f is constant on [a,b] provided that \(\displaystyle \lim_{x \to a^+}f(x)=f(a)\) and \(\displaystyle \lim_{x \to b^-}f(x)=f(b)\)?

 

[HallsofIvy]

What I suspect was intended was "if f'= 0 on (a, b) then f is constant on (a, b)" (not [a, b]). Knowing about f' on (a, b) tells you nothing about what happens at a and b. The function f(x)= 1 on (0, 1), defined any way you like otherwise, satisfies the conditions of this theorem.