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

## Given that for $x \ge 1 , {e}^{f} \left(x\right) + x f \left(x\right) = x \ln x + 2 x - 1$

Nov 16, 2017

#### Explanation:

For $x \ge 1 , {e}^{f} \frac{x}{x} + f \left(x\right) = \ln x + 2 - \frac{1}{x}$ ,

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

$g ' \left(x\right) = \frac{1}{x} + \frac{1}{x} ^ 2 > 0 , x \ge 1$

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

I will work with proof of contradiction.
Supposed there is $a , b$ with $a$<$b$ and $f \left(a\right) \ge f \left(b\right)$
${e}^{f} \left(a\right) \ge$${e}^{f} \left(b\right)$$\implies$${e}^{f} \frac{a}{a}$$>$${e}^{f} \frac{b}{b}$$\implies$${e}^{f} \frac{a}{a} + f \left(a\right)$$>$${e}^{f} \frac{b}{b} + f \left(b\right)$ , because $\frac{1}{a}$$> \frac{1}{b} > 0$
so, $g \left(a\right) > g \left(b\right)$ $\implies$$a > b$ , $g$ strictly increasing
This is a contradiction, because we supposed that $a <$$b$
So that makes $f$ strictly increasing too. And as a result $1 - 1$