Conversely, suppose a continuous $f:I\to\{-1,1\}$ satisfies $f(-1)=-1$ and $f(1)=1$. The function $g=-f:I\to I$ is continuous, so part (d) gives $x=g(x)=-f(x)$. If $f(x)=1$, this forces $x=-1$, contradicting $f(-1)=-1$; if $f(x)=-1$, it forces $x=1$, contradicting $f(1)=1$. Hence no such $f$ exists.


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 11H / e / Solution

 Solution (source code)