Hi,

There's a theorem in my book that states the following;

Let$\displaystyle f$be a real-valued function, defined and continuous on a bounded closed interval$\displaystyle [a,b]$of the real line. Assume, further, that$\displaystyle f(a)f(b)\leq 0$; then, there exists$\displaystyle \xi$in$\displaystyle [a,b]$such that$\displaystyle f(\xi)=0$.

I feel quite comfortable with that theorem. Later on in the book the authors write:

"An alternative sufficient condition for the existence of a solution to the equation $\displaystyle f(x)=0$ is arrived at by rewriting it in the equivalent form $\displaystyle x-g(x)=0$ where $\displaystyle g$ is a certain real-valued function, defined and continuous on $\displaystyle [a,b]$. The problem of solving the equation $\displaystyle f(x)=0$ is converted into one of finding $\displaystyle \xi$ such that $\displaystyle \xi-g(\xi)=0$."

Now that I do not understand. I don't really see how x-g(x)=0 is equivalent to f(x)=0. Could someone please explain this to me, or point me to some resources?

Thanks.