Is $f(x) = e^x - e^{-x}$ invertible on $\Bbb R$?

Question

Consider the function $f : \Bbb R \to \Bbb R$ defined by

$$ f(x) = \exp(x)-\exp(-x)$$

Is $f$ an invertible function? Give reasons for your answer.

The answer given with this question states that since the function is continuous, it is surjective and since it is increasing, it is injective and so bijective and hence invertible. But is it true that all continuous functions are surjective and if this reason is correct?

Answer 1

This is an odd strictly increasing function (compute $f'$) that is bounded neither above nor below. It is therefore invertible.

Answer 2

The derivative $e^x+e^{-x}$ never cancels, hence the function is strictly increasing. And being the sum of two continuous functions, it is also continuous.

Answer 3

This amounts to solving the equation that $u=e^{x}-e^{-x}$ for a given $u\in{\bf{R}}$ for a unique solution. Let $v=e^{x}$, then $u=v-v^{-1}$, so $v^{2}-uv-1=0$, and the discriminant is $u^{2}+4>0$, we have two roots $v_{1}=\dfrac{u+\sqrt{u^{2}+4}}{2}$ and $v_{2}=\dfrac{u-\sqrt{u^{2}+4}}{2}$, pick the positive one $v_{1}$ because $e^{x}>0$. Because $e^{x}$ is bijective, then there is a unique $x$ that $e^{x}=v_{1}$.