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I am trying to show that:

f(x) is a function and there exists real constants $a>b$ such that $$f(a+x)=f(a-x) $$ and $$f(b+x)=f(b-x)$$ for every real number ${x}$.
Show that $2(a-b)$ is a period of $f$.

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$$f(x) = f(a + x - a) = f(a - (x-a)) = f(2a - x) = f(b + (2a-b-x)) = f(b - (2a-b-x)) = f(2b-2a+x)$$

So $2b-2a$ is a period.

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Change $x$ to $x+a$ in the first equation to get $f(2a+x)=f(-x)$. Change $x$ to $x+b$ in the second equation to get $f(2b+x)=f(-x)$. We now have $f(2a+x)=f(2b+x)$. Change $x$ to $x-2b$ to finish the proof.

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