Consider f:R -> R, f = x/ sqrt(x^2+1). Prove that for any a between -1 and 1, f(x)=a has only one solution.

f'(x)=( sqrt (x^2+1) - x * ( x / sqrt (x^2 +1) ) ) / (x^2+1) = (x^2 + 1 + x^2) / ( (x^2 + 1) * sqrt ( x^2 + 1) ) =  1 / ( (x^2 + 1) * sqrt (x^2 + 1) ). 

f'(x) > 0 for any x => f is increasing. When x-> -infinite, lim f(x) = -1. When x -> infinite, lim f(x) = 1. f is a composition of continuous functions, so f is continuous. Therefore, for any a between -1 and 1, f(x) = a has one solution.