Question

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

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

Answer 1

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`