Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function 
Given $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$, where $a,b$ are positive constants $(a>b)$, prove that $f(x)$ is a periodic function

I have done the following: 
$f(a-x)=f(a+x)$....(1)  
or, $f(-(x-a))=f(x+a). $
Putting $x=x+a$, we get
$f(-x)=f(x+2a) ....(i)$
Similarly, 
$f(b-x)=f(b+x).....(2) $
Putting $x=x+b,$ we get
$f(-x)=f(x+2b)......(ii)$ 
Equating $(i)$ & $(ii)$, we get:
$f(x+2a)=f(x+2b).$ 
Given, $a>b$ 
Therefore, $2a=2b+h.$ 
Putting $x=x-2a$, we get
$f(x+h)=f(x).$
Is this mathematically correct? Also, what other ways are there to solve it?

I guess that $f(a-x)=f(a+x)$ and $f(b-x)=f(b+x)$ hold for every $x$.
In particular, evaluating at $x+a$ instead of $x$,
$$
f(-x)=f(x+2a)
$$
Similarly $f(-x)=f(x+2b)$. Therefore
$$
f(x+2a)=f(x+2b)
$$
for every $x$. Evaluating at $x-2b$ instead of $x$, we obtain
$$
f(x+2a-2b)=f(x)
$$
and, setting $T=2a-2b$,
$$
f(x)=f(x+T)
$$
for every $x$.
Good job!
