Exercise 4.14

(Matt “frito” Lundy)
Because both $h(x)=x$ and $f$ are continuous mappings on $I$, $g(x)=f(x)-x$ is also continuous on $I$. We are searching for an $x$ such that $g(x)=0$. $g(0)\in [0,1]$ and $g(1)\in [-1,0]$, and if either $g(0)=0$ or $g(1)=0$, we are done, so assume that both $g(0)\ne$ and $g(1)\ne 0$. Then $g(0)\in (0,1]$ and $g(1)\in [-1,0)$, and because $g$ is continuous on $I$, with $I$ connected, the intermediate value theorem applies, and there exists an $x\in (0,1)$ such that $f(x)=0$.