Sum of all the $x$- and $y$-intercepts of the graph of $f(x)$ is equal to the sum of all the $x$- and $y$-intercepts of the graph of $f^{-1}(x)$ The function $f$ is  one-to-one. Prove that the sum of all the $x$- and $y$-intercepts of the graph of $f(x)$ is equal to the sum of all the $x$- and $y$-intercepts of the graph of $f^{-1}(x)$.

The given statement seems obvious and true but I dont know how to prove it theoretically.
I tried:  
Let $x_1,f(0)$ be the $x$-intercept and $y$-intercept of the graph of the function $f(x)$.
The sum of all the $x$- and $y$-intercepts of the graph of $f(x)$ is: $$x_1+f(0)$$
I dont know how to solve it further. I am stuck here.

You said $f(x)$ is one-to-one, so $f^{-1}(x)$ is well defined. (I will also asume that $f(x)$ actually intersepts the $x$- and $y$-axis, for obvious reasons.) We have the sum $$ x_1+f(0),$$ where $f(x_1)=0$ and let's say $f(0)=y_1$.  
This means that $f^{-1}(0)=x_1$ and $f^{-1}(y_1)=0$. This means that $x_1$ is the $y$-intersept of $f^{-1}(x)$ and that $y_1$ is its $x$-intersept.
If we count the $x$- and $y$-intersept of $f^{-1}(x)$ in the same way we did for $f(x)$, we get
$$y_1+f^{-1}(0)=f(0)+x_1,$$ which is the same sum indeed.
