jayc said:Sure, there is. Just take the triangle function on the interval [-1, 1]:
\[\Lambda(x) = \left\{ \begin{array}{l l} x & \quad \mbox{if }x \leq 0\\ 1-x & \quad \mbox{if }x >0\\ \end{array} \right.\]
The derivative to this function is not continuous:
\[\frac{d\Lambda}{dx}(x) = \left\{ \begin{array}{l l} 1 & \quad \mbox{if }x \leq 0\\ -1 & \quad \mbox{if }x >0\\ \end{array} \right.\]
The derivative exists everywhere in the interval but is discontinuous at x=0.
main_road said:jayc,
What's your interval? Have you drawn that function? It's discontinuous in any open interval that includes 0.
jayc said:The derivative to this function is not continuous:
\[\frac{d\Lambda}{dx}(x) = \left\{ \begin{array}{l l} 1 & \quad \mbox{if }x \leq 0\\ -1 & \quad \mbox{if }x >0\\ \end{array} \right.\]
The derivative exists everywhere in the interval but is discontinuous at x=0.
eecs4ever said:...
but wait! , the statement also implies that f'(x) is not differentiable at A.
why?
because the definition for differentiability of f(x) imples that
lim f'(x) is equal when you approach from both sides for all x in the interval.
...
eecs4ever said:...
=> lim f'(x) as x --> A- from left side
is NOT EQUAL TO lim f'(x) as x---> A+ from right side
...
zox11 said:I may be wrong. Did not read the question carefully.Sorry.
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