$f(x)= \frac{\sqrt{x^2-1}}{x+\log x}$ in the set $E=[1,+ \infty)$ I have the function $f(x)= \frac{\sqrt{x^2-1}}{x+\log x}$ in the set $E=[1,+ \infty)$
(i) I have to prove that $\forall y_0 \in [0,1) $exists only one $x_0 \in E $ such that $f(x_0)=y_0$
(ii) to discuss the uniform continuity of f in E.
Supposing that there are two different values  $x_1$ and $x_2 \in E $ such that 
$f(x_1)=y_0=f(x_2)$, I have to prove that $   f(x_1)=y_0=f(x_2)>1$
Can someone help me to understand what to do?

1) first show that $f$ is a strictly increasing function; hence, (i) follows.
2) note that $f$ is also bounded and continuous. Now show that any bounded continuous increasing function is uniformly continuous. Hence $f$ is also uniformly continuous.
