Show that #f# is strictly increasing in #RR# ?

Question

#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#

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Answer 1 · 2018-05-24T18:37:34

Sign/contradiction & Monotony

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

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

If #EEx_0##in##RR: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# #AA##x##in##RR#

#-># #{(f'(x)>0", "),(f'(x)<0", "):}# #x##in##RR#

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#, #AA##x##in##RR# so #f# is strictly increasing in #RR#