Let $f, g$ be two bijective continuous functions $f\left(g^{-1}(x)\right)+g\left(f^{-1}(x)\right)=2 x$ question - 
Let $f, g$ be two bijective continuous functions on $\mathbb{R}$ such that
$$
f\left(g^{-1}(x)\right)+g\left(f^{-1}(x)\right)=2 x
$$
for all $x \in \mathbb{R} .$ Suppose there exists $x_{0} \in \mathbb{R}$ such that $f\left(x_{0}\right)=$ $g\left(x_{0}\right) .$ Prove that $f(x)=g(x)$ for all real numbers $x$
my try - 
by Putting $h(x)=f\left(g^{-1}(x)\right) .$ i get  $h(x)+h^{-1}(x)=2 x .$ ....(1)
where H is also bijective and continuous .....
after this the hint says that Conclude that $h(x)=x+c .$ Show that $c=0$ and hence $f(x)=g(x)$ for all $x$ 
i am not getting how to prove that $h(x)=x+c$ ...i tried replacing x by h(x) in 1)...but did not able to conclude ....
any help will be helpful 
thankyou
 A: 1st APMO 1989, Problem 5

Find all strictly increasing bijective functions $h:\mathbb{R}\to\mathbb{R}$ such that $$h(x) + h^{-1}(x) = 2x$$ for all $x\in\Bbb R$.  Here $h^{-1}:\mathbb{R}\to\mathbb{R}$ denotes the inverse function of $h$.

Answer: $h(x) = x + b$ for some fixed real $b$.
Solution
Suppose for some $a$, we have $h(a) ≠ a$. Then for some $b ≠ 0$, we have $h(a) = a + b$. Hence $h(a + b) = a + 2b$ (because $h\big( h(a) \big) + h^{-1}\big( h(a) \big) = 2 h(a)$, so $h(a + b) + a = 2a + 2b$) and by two easy inductions, $$h(a + nb) = a + (n+1)b$$ for all integers $n$ (positive or negative).
Now take any $x$ between $a$ and $a + b$. Suppose $h(x) = x + c$. The same argument shows that $$h(x + nc) = x + (n+1)c.$$ Since $h$ is strictly increasing $x + c$ must lie between $h(a) = a + b$ and $h(a+b) = a + 2b$. So by a simple induction $x + nc$ must lie between $a + nb$ and $a + (n+1)b$. So $c$ lies between $b + \frac{a-x}n$ and $b + \frac{a+b-x}n$ for all $n$. Hence $c = b$. Hence $h(x) = x + b$ for all $x$.
If there is no $a$ for which $h(a) ≠ a$, then we have $h(x) = x$ for all $x$.
Remark:  This is a copy of the solution given in this link.  This website moved locations a few times in the past, so I decided to make a copy of the solution in case it is moved again. The condition that $h$ is a strictly increasing bijection can be replaced by the condition that $h$ is a continuous bijection.
