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If $f$ and $g$ are continuous on $[a, b]$ and differentiable on $(a, b)$, then $\max(f, g)$ is continuous on $[a, b]$ and differentiable on $(a, b)$.

I'm asked to either prove or disprove this statement. So far I'm thinking that it's true because if a function is continuous and differentiable on an interval, it takes on it's max/min over the interval, but I don't know if this is correct, and if it is, I don't know how to prove it. Any help would be appreciated.

Trajan
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user181928
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    $f(x)=x$, $g(x)=-x$, $[a,b]=[-1,1]$ – Adam Hughes Dec 11 '14 at 01:25
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    The part of the question after "So far I'm thinking" looks to me like an attempt at "proof by pun". You're talking there about the max of a function $f$ on an interval, the value $f(c)$ that is $\geq$ all the other values $f(z)$ for the same $f$ and other $z$'s. The problem, on the other hand, concerns the max (or min) of two functions $f$ and $g$ which is a function whose value at any $x$ is the larger of $f(x)$ and $g(x)$, with no reference to any other $z$'s. Although "max" occurs in both, it's applied in entirely different ways. – Andreas Blass Dec 13 '14 at 22:19
  • @1234: please don't make so many trivial edits so quickly! – Carl Mummert Dec 13 '14 at 22:41

2 Answers2

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Idea

Consider $\lvert x \rvert = \max\{x, -x\}$.

davidlowryduda
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    So then I let f(x) = x and g(x) = -x. Then let |x| = max{x,-x} which is a contradiction because |x| is not differentiable on the interval with f and g? – user181928 Dec 11 '14 at 01:32
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Can use the fact that max (f,g) = {f + g + |f-g|}/2

user200793
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  • Could you add more detail into how you'd use that fact? (I think there's a valid answer here, and some insight in that this fails because $|f-g|$ might not be differentiable, but this needs to be fleshed out further to be useful) – Milo Brandt Dec 13 '14 at 00:32
  • @Meelo Yeah I overlook that part. it is true that |f-g| might not be differentiable. Does that mean we need to assume f>0 and g>0 here? – user200793 Dec 13 '14 at 01:33
  • @user Well, it means that we'd need that $f-g$ never changes signs - which would imply that $\max(f,g)$ equals $f$ or $g$. We could weaken that to requiring that if $f-g$ changes sign around a point, then at that point $f'-g'=0$. (In general, per mixedmath's answer, the statement is false) – Milo Brandt Dec 13 '14 at 01:36