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












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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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  • 5




    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






  • 1




    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
















1














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






real-analysis derivatives







share|cite|improve this question




















  • 5




    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






  • 1




    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














1












1








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






real-analysis derivatives







share|cite|improve this question















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






real-analysis derivatives




real-analysis derivatives






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edited Nov 29 '18 at 15:42









Lynn

2,5801526




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asked Nov 29 '18 at 15:36









DadaDada

7010




7010








  • 5




    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






  • 1




    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














  • 5




    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






  • 1




    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








5




5




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




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




1




1




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




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










2 Answers
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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, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$






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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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      2 Answers
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      2 Answers
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      active

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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, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



      Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



      Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$






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        2














        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, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



        Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



        Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$






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          2












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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, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



          Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



          Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$






          share|cite|improve this answer












          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, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



          Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



          Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$







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          answered Nov 29 '18 at 15:58









          GoldyGoldy

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              3














              $max(f,g)=frac{1}{2}(f+g+|f-g|)$. 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.






              share|cite|improve this answer




























                3














                $max(f,g)=frac{1}{2}(f+g+|f-g|)$. 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.






                share|cite|improve this answer


























                  3












                  3








                  3






                  $max(f,g)=frac{1}{2}(f+g+|f-g|)$. 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.






                  share|cite|improve this answer














                  $max(f,g)=frac{1}{2}(f+g+|f-g|)$. 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.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 '18 at 16:26

























                  answered Nov 29 '18 at 16:04









                  SurajitSurajit

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