# How can I show $\max(f,g)$ is differentiable at $c$?

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 $$f(c) \neq g(c)$$.

• In a neighbourhood of such $c$, $\max(f,g)$ is either equal to $f$ or equal to $g$. In either case, it's differentiable. – user3482749 Nov 29 '18 at 15:38
• 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.) – Lynn Nov 29 '18 at 15:40

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,$$\lim\limits_{x\rightarrow c^-}\frac{max(f,g)(x)-max(f,g)(c)}{x-c}=\lim\limits_{x\rightarrow c^+}\frac{max(f,g)(x)-max(f,g)(c)}{x-c}=\{max(f,g)(c)\}'.$$
$$max(f,g)=\frac{1}{2}(f+g+|f-g|)$$. 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.