The statement is false.
Two more examples are f ( x ) = sin ( x ) {\displaystyle f(x)=\sin(x)} , 0 < x < π {\displaystyle 0<x<\pi } , and f ( x ) = cos ( x ) {\displaystyle f(x)=\cos(x)} , − π / 2 < x < π / 2 {\displaystyle -\pi /2<x<\pi /2} . The argument is the same as in solution 1.
