For #x=0# we have
#f(0)-e^(-f(0))=-1#
We consider a new function #g(x)=x-e^(-x)+1# , #x##in##RR#
#g(0)=0# ,
#g'(x)=1+e^(-x)>0# , #x##in##RR#
As a result #g# is increasing in #RR#. Thus because it's strictly increasing #g# is "#1-1#" (one to one)
So, #f(0)-e^(-f(0))+1=0# #<=># #g(f(0))=g(0)# #<=># #f(0)=0#
We need to show that #x/2<##f(x)<##xf'(x)# #<=>^(x>0)#
#1/2<##f(x)/x<##f'(x)# #<=>#
#1/2<##(f(x)-f(0))/(x-0)<##f'(x)#
- #f# is continuous at #[0,x]#
- #f# is differentiable in #(0,x)#
According to the mean value theorem there is #x_0##in##(0,x)#
for which #f'(x_0)=(f(x)-f(0))/(x-0)#
#f(x)-e^(-f(x))=x-1# , #x##in##RR# so
by differentiating both parts we get
#f'(x)-e^(-f(x))(-f(x))'=1# #<=># #f'(x)+f'(x)e^(-f(x))=1# #<=>#
#f'(x)(1+e^(-f(x)))=1# #<=>^(1+e^(-f(x))>0)#
#f'(x)=1/(1+e^(-f(x)))#
The function #1/(1+e^(-f(x)))# is differentiable. As a result #f'# is differentiable and #f# is 2 times differentiable with
#f''(x)=-((1+e^(-f(x)))')/(1+e^(-f(x)))^2# #=#
#(f'(x)e^(-f(x)))/((1+e^(-f(x)))^2# #>0# , #x##in##RR#
-> #f'# is strictly increasing in #RR# which means
#x_0##in##(0,x)# #<=># #0<##x_0<##x# #<=>#
#f'(0)<##f'(x_0)<##f'(x)# #<=>#
#1/(1+e^(-f(0)))##<##f(x)/x<##f'(x)# #<=>#
#1/2<##f(x)/x<##f'(x)# #<=>^(x>0)#
#x/2<##f(x)<##xf'(x)#