Show that for $f:\mathbb{R}\to \mathbb{R}$ there is some value $x_0$ such that $f\left(x_0\right)=x_0$ Let's say I have the function $f:\mathbb{R}\to \mathbb{R}$ the function domain and range are $[0,1]$ and the function is continuous. Show that there is a value $x_0$ such that $f\left(x_0\right)=x_0$.
If I put the function $y=x$ on a graph I can see that this graph cut it into 2 halves and such any function that fulfil the conditions I gave has to pass through $y=x$ and that interaction point is $x_0$
How can I prove it without using graphs?

Consider the function $g(x)=f(x)-x$. Our goal is to prove that the function $g$ has a zero on $[0,1]$.
We know that $g(0)\ge 0$, but $g(1)\le 0$. Can you see why those two things must be true?
Now, if either value actually equals $0$, then we're done, so suppose $g(0)$ is strictly positive, and $g(1)$ strictly negative. Since $g$ is defined as the difference of two continuous functions, then it is a continuous function, and the Intermediate Value Theorem applies to it.
Can you take it from there?
