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So my class has been given the task to find functions $f$ and $g$,both from R to R such that: $f+g$ is differentiable and either $f'(0)$ dne, $g'(0)$ dne or both.

I'm starting to believe, or at least convince myself that no such functions exist. That is, if we were to choose an $x$ in the intersection of the $dom(f)$ and $dom(g)$, then $(f+g)'(x)=f'(x) + g'(x)$

Thus, in order for $(f+g)$ to be differentiable, then both $f$ and $g$ must be differentiable on the intersection of their domain. Any thoughts/hints/explanations? Is my reasoning out of line here?? Thanks!

Ben Anderson
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Hint: Let $f:\mathbb R\to \mathbb R$ be any function. Then $f+(-f)$ is differentiable. So, you can take for $f$ any function that is not differentiable at $0$. There are well-known examples of functions that are nowhere continuous, and thus also nowhere differentiable, for instance the Dirichlet function. For a very elementary example though, you can take the function $f(x)=1/x$ for all $x\ne 0$, and $f(0)=17$. Then $f$ is not differentiable at $0$.

Ittay Weiss
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  • I've tried something along these lines, but got a little stuck. Say $f=1/x$ and g=$-1/x$ then $(f+g)(x)=0$, however the $dom(f intersect g)$ does not include 0, thus $f'(0)$ dne and $g'(0)$ dne but so does $(f+g)'(0)$ So I still haven't found functions. That is if zero is not in the intersection of the domains, then no matter what (f+g) is remains undefined...? – Ben Anderson Apr 01 '13 at 03:47
  • What you say is true. Your problem states though that the functions are from $\mathbb{R}$ to $\mathbb{R}$ so $f=1/x$ would not work as the same case is for $g$. –  Apr 01 '13 at 03:50
  • Have you perhaps seen an example of a function that is not continuous anywhere? – Ittay Weiss Apr 01 '13 at 03:52
  • Perhaps I missed something in the calculus sequence, I can't seem to think of a function that in not continuous anywhere and google show up with the Dirichlet Function which I've never seen before... – Ben Anderson Apr 01 '13 at 03:57
  • Then now you know about the Dirichlet function :) – Ittay Weiss Apr 01 '13 at 04:00
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    :) Works - if nothing else my prof will love it... – Ben Anderson Apr 01 '13 at 04:05
  • @BenAnderson, you can also look at the edit to my solution to see how to deal with tweaking your initial attempt to make it work as well. – Ittay Weiss Apr 01 '13 at 04:07