Show that #x/2<##f(x)##<##xf'(x)# , #AAx>0# ?

#f:RR->RR# differentiable in #RR# with

  • #f(x)-e^(-f(x))=x-1# , #x##in##RR#

Show that

#x/2<##f(x)##<##xf'(x)# , #x>0#

#f:RR->RR# differentiable in #RR# with

  • #f(x)-e^(-f(x))=x-1# , #x##in##RR#

Show that

#x/2<##f(x)##<##xf'(x)# , #x>0#

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Jim S Share
Mar 8, 2018

Answer:

Check below for answer

Explanation:

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)#

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