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 ,

AAxinRR

Show that f is strictly increasing in RR

1 Answer
May 24, 2018

Sign/contradiction & Monotony

Explanation:

f is differentiable in RR and the property is true AAxinRR so by differentiating both parts in the given property we get

f'(f(x))f'(x)+f'(x)=2 (1)

If EEx_0inRR:f'(x_0)=0 then for x=x_0 in (1) we get

f'(f(x_0))cancel(f'(x_0))^0+cancel(f'(x_0))^0=2 <=>

0=2 -> Impossible

Hence, f'(x)!=0 AAxinRR

  • f' is continuous in RR
  • f'(x)!=0 AAxinRR

-> {(f'(x)>0", "),(f'(x)<0", "):} xinRR

If f'(x)<0 then f would be strictly decreasing

But we have 0<1 <=>^(fdarr) <=> f(0)>f(1) <=>

0>1 -> Impossible

Therefore, f'(x)>0, AAxinRR so f is strictly increasing in RR