How to show that #f# is "1-1" ? (one-to one)

Given that for #x>=1 , e^f(x) + xf(x) = xlnx + 2x - 1#

1 Answer
Nov 16, 2017

Proof of contradiction & Monotony

Explanation:

For # x>=1 , e^f(x)/x + f(x) = lnx + 2 - 1/x# ,

#g(x) = lnx + 2 - 1/x , x>=1 #

#g'(x) = 1/x + 1/x^2 > 0 , x>=1#

so #g# is strictly increasing
I need to prove that for each #a,b# with
#1<=a<β# it's #f(a)<##f(b)#

I will work with proof of contradiction.
Supposed there is #a,b# with #a#<#b# and #f(a)>=f(b)#
#e^f(a)>=##e^f(b)##=>##e^f(a)/a##>##e^f(b)/b##=>##e^f(a)/a +f(a)##>##e^f(b)/b+f(b)# , because #1/a##>1/b>0#
so, #g(a)>g(b)# #=>##a>b# , #g# strictly increasing
This is a contradiction, because we supposed that #a<##b#
So that makes #f# strictly increasing too. And as a result #1-1#