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Find all functions $f(x)$ for $x\in \mathbb{R},$ such that $f(1+x) = f(1-x)$ and $f(2+x) = f(2-x)$.

A little bit of arrangement in the first equality will give $f(x) = f(2-x)$. $\implies f(x) = f(x+2)$. This is a function with periodicity $2$. So all functions with periodicity $2$ are $f(x)$.

Another obvious solution is $f(x)=c$ where $c$ is a constant. Are there any other possible functions?

Shaurya Gupta
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  • Just as a thought, the first equation says your graph has an axis of symmetry of the line x=1. The second says your graph has an axis of symmetry of x=2 – Alan Oct 04 '14 at 20:37

2 Answers2

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All functions $f$ satisfying your requirements can be constructed by choosing any function $f:[0,1]\to \mathbb{R}$, then extending this to $[0,2]$ using the identity $f(1-x)=f(1+x)$ (i.e. reflecting over line $x=1$), then extending this periodically to a period 2 function on the whole real line.

There are many such functions, for example, $f(x)=\sin^2(\pi x/2)$.

Mike Earnest
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Yes. In fact every function that satisfies in those equations also satisfies in $$ f(x) = f(x+2) \\ f(x) = f(-x) $$ and vise versa.to see this

$$ f(1 +(1+y)) = f(1-(1+y))\Rightarrow f(2+y) = f(-y) = f(2-y)$$ and of course $$ f(x) = f(x+2)=f(2-x)=f(-x) $$ conversely if we have $f(x) = f(-x)$ and $f(x) = f(x+2)$ we have $$f(1+x) = f(-x-1) = f(1-x) $$ and $$ f(2+x) = f(-2-x) = f(-x) = f(2-x) $$

But for examples just consider an arbitrary function on $[0,1]$. It's uniquely defines by above equations on $\mathbb{R}$.

Abdollah
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