## continuous and differentiable function

1. The problem statement, all variables and given/known data
function f:R->R can be written as a sum f=f1+f2 where f1 is even and f2 is odd。show that if f is continuous then f1 and f2 may be chosen continuous, and if f is differentiable then f1 and f2 can be chosen differentiable

2. The attempt at a solution
i have try some examples, but i still cannot get the idea from that
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 Quote by frankpupu 1. The problem statement, all variables and given/known data function f:R->R can be written as a sum f=f1+f2 where f1 is even and f2 is odd。show that if f is continuous then f1 and f2 may be chosen continuous, and if f is differentiable then f1 and f2 can be chosen differentiable 2. The attempt at a solution i have try some examples, but i still cannot get the idea from that
You should be able to find an expression for f1 and f2 given f. Think about f(x)+f(-x) and f(x)-f(-x).
 Recognitions: Gold Member Science Advisor Staff Emeritus Given function f, let $f_e(x)= (f(x)+ f(-x))/2$ and $f_o(x)= (f(x)- f(-x))/2$.

## continuous and differentiable function

 Quote by HallsofIvy Given function f, let $f_e(x)= (f(x)+ f(-x))/2$ and $f_o(x)= (f(x)- f(-x))/2$.
yes f(x)=((f(x)+f(−x))/2 )+(f(x)−f(−x))/2) then (f(x)+f(−x))/2 is even and (f(x)−f(−x))/2 is odd,then i don't know how to argue their continuity and differentiability?

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 Quote by frankpupu yes f(x)=((f(x)+f(−x))/2 )+(f(x)−f(−x))/2) then (f(x)+f(−x))/2 is even and (f(x)−f(−x))/2 is odd,then i don't know how to argue their continuity and differentiability?
I don't think you have to prove it from scratch. If f(x) is continuous then f(-x) is continuous. What theorem about continuous functions might you use to prove that?

 Quote by Dick I don't think you have to prove it from scratch. If f(x) is continuous then f(-x) is continuous. What theorem about continuous functions might you use to prove that?
if f(x)is continuous then f(-x)is continuous ,(f(x)+f(-x))/2 is continuous by sum rule (f(x)-f(-x))/2 is continuous as well .but they should be chosen continuous

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