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Construct a function which is continuous in $[1,5]$ but not differentiable at $2, 3, 4$.

This question is just after the definition of differentiation and the theorem that if $f$ is finitely derivable at $c$, then $f$ is also continuous at $c$. Please help, my textbook does not have the answer.

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Intuitively, a function is differentiable at a point if the graph of the function at that point is "smooth". –  JavaMan Oct 20 '11 at 17:54
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Just make some kind of saw-tooth with peeks in 2, 3, 4. –  Jonas Teuwen Oct 20 '11 at 17:55
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Intuitively, a function is continuous if you can "walk" on the graph and it is differentiable if you can see where you came from and where you are going. –  AD. Oct 25 '11 at 18:05
    
@AD. Intuitively, a real-valued function of one real variable is differentiable if you can "walk" on its graph without stopping. –  Amitesh Datta Oct 26 '11 at 8:59

4 Answers 4

up vote 111 down vote accepted

$|x|$ is continuous, and differentiable everywhere except at 0. Can you see why?

From this we can build up the functions you need: $|x-2| + |x-3| + |x-4|$ is continuous (why?) and differentiable everywhere except at 2, 3, and 4.

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$\uparrow$ Change first + sign to a - sign for the infamous $\mathsf{W}$ solution...:) –  Qmechanic Oct 22 at 13:36

Remember that if a function isn't continuous at a point, then its antiderivative is not differentiable there. Knowing this, you could construct the desired function by taking the integral of a function that's discontinuous at $x=2,3,4$: $$ f(x) = \int_1^x F(y) \,\mathrm{d}y, $$ where $$ F(x) = \begin{cases} -1 & 1\le x\le 2 \\ +1 & 2<x\le 3 \\ -1 & 3<x\le 4 \\ +1 & 4<x\le 5 \\ \end{cases} $$

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$$\ \ \ \ \mathsf{W}\ \ \ \ $$

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OK, I confess: I always dreamt of reading/posting an answer with only one character... so I could not resist the occasion. –  Did Oct 20 '11 at 19:48
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+1 for succinctness. –  aardvarkk Oct 20 '11 at 19:50
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Best answer ever! –  J. M. Oct 20 '11 at 22:36
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This answer finally made me sign up to math.se (so that I could upvote). –  Heinzi Oct 21 '11 at 8:37
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-1 "This answer is not useful" to any one who would actually ask this question. –  Graphth Oct 24 '11 at 3:12

How about $f(x) = \max(\sin(n\pi x),0)$ or perhaps $g(x) = |\sin(n\pi x)|$?

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You're right, I think, because we're considering only one-sided derivatives at $1$ and $5$. –  Saaqib Mahmuud Aug 16 '13 at 9:45

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