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Let $f,g : (a;b) \to \mathbb{R}$ differentiable, and $\max(f,g) : (a;b) \to \mathbb{R} $ defined by $$\max(f,g)(x) := \max(f(x),g(x)).$$

Show that $\max(f,g)$ is differentiable at $ c \in (a;b)$, for all $c \in (a;b)$ such that $f(c) \neq g(c)$.

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    $\begingroup$ In a neighbourhood of such $c$, $\max(f,g)$ is either equal to $f$ or equal to $g$. In either case, it's differentiable. $\endgroup$ – user3482749 Nov 29 '18 at 15:38
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    $\begingroup$ What have you tried yourself so far, and what jumps out as the roadblock in the proof to you? Also, how does your course define “differentiable”? (There are a few equivalent definitions.) $\endgroup$ – Lynn Nov 29 '18 at 15:40
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Since $f(c)\neq g(c)$, so assume that $f(c)>g(c)$.

This implies, $max(f,g)(x) = f(x)$, in a neighbourhood of $c$.

So, $$\lim\limits_{x\rightarrow c^-}\frac{max(f,g)(x)-max(f,g)(c)}{x-c}=\lim\limits_{x\rightarrow c^-}\frac{f(x)-f(c)}{x-c}$$ and $$\lim\limits_{x\rightarrow c^+}\frac{max(f,g)(x)-max(f,g)(c)}{x-c}=\lim\limits_{x\rightarrow c^+}\frac{f(x)-f(c)}{x-c}.$$

Since $f$ is differentiable at $c$, $\lim\limits_{x\rightarrow c^-}\frac{f(x)-f(c)}{x-c}=\lim\limits_{x\rightarrow c^+}\frac{f(x)-f(c)}{x-c}$.

Hence,$$\lim\limits_{x\rightarrow c^-}\frac{max(f,g)(x)-max(f,g)(c)}{x-c}=\lim\limits_{x\rightarrow c^+}\frac{max(f,g)(x)-max(f,g)(c)}{x-c}=\{max(f,g)(c)\}'.$$

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$max(f,g)=\frac{1}{2}(f+g+|f-g|)$. So to prove differentiability of $max(f,g)$ at $x=c$, where $f(c)\neq g(c)$, it is enough to prove differentiability of $|f-g|$. Note that the modulus function $|\;|:\mathbb{R}\setminus\{0\}\rightarrow \mathbb{R}$ defined by $|\;|(x):=|x|$ is differentiable, and so is $f-g$. Now $|f-g|=|\;|\circ(f-g)$, being a composition of two differentiable functions, is differentiable.

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