Its derivative fn' (x) = NX (n-1)+1>; 0, so the function increments on [0, 1]. And f (0) =- 1

Proof: sup(n & gt；; =2)xn= 1。

Counterevidence:

if sup(n & gt； = 2)xn = a & lt； 1, with n >；; 0 means any n & gta^n & lt； (1-a)/2, so

xn^n+xn- 1 & lt； a^n+a- 1 & lt； ( 1-a)/2+a- 1 =-( 1-a)/2 & lt； 0 this contradicts that xn is the root of x n+x-1= 0.

So the conclusion is established.