Show that ff is strictly increasing in RR ?
f:RR->RR differentiable with f' continuous in RR , f(0)=0 , f(1)=1
f(f(x))+f(x)=2x ,
AA x in RR
Show that f is strictly increasing in RR
f(f(x))+f(x)=2x ,
Show that
1 Answer
May 24, 2018
Sign/contradiction & Monotony
Explanation:
If
Hence,
f' is continuous inRR f'(x)!=0 AA x in RR
If
But we have
Therefore,