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

Given that for x>=1 , e^f(x) + xf(x) = xlnx + 2x - 1x1,ef(x)+xf(x)=xlnx+2x1

1 Answer
Nov 16, 2017

Proof of contradiction & Monotony

Explanation:

For x>=1 , e^f(x)/x + f(x) = lnx + 2 - 1/xx1,ef(x)x+f(x)=lnx+21x ,

g(x) = lnx + 2 - 1/x , x>=1 g(x)=lnx+21x,x1

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