Show $\cosh{[0,\infty)}=[1,\infty)$ and $\sinh{\mathbb{R}}=\mathbb{R}$

Question

Show $\cosh{[0,\infty)}=[1,\infty)$ and $\sinh{\mathbb{R}}=\mathbb{R}$.

I have shown that both functions are strictly increasing (hyperbolic cosine strictly decreading on the negatives but that should’t matter here?). Also, it is shown that the functions are continuous. From thid, how can I show that they are indeed having the domain and thus range. I sm sure that I have to use the intermediate value theorem but do not know how to formally show this.

Answer 1

Take $y\in[0,\infty)$.

You have $\cosh(0)\leqslant y$ and, since $\lim_{x\to\infty}\cosh(x)=\infty$, there is some $x_0\in[0,\infty)$ such that $\cosh(x_0)>y$. So, by the intermediate value theorem, there is some $x\in[0,x_0]$ such that $\cosh(x_1)=y$. The case of $\sinh$ is similar.

Answer 2

First of all, both are continuous.

• $\cosh$ is strictly increasing and s.t. $\cosh(0)=1$ and $\lim_{x\to \infty }\cosh(x)=+\infty $. Therefore, Intermediate Value Theorem (IVT) allows you to conclude.

• Same with $\sinh$, it's strictly increasing and s.t. $\lim_{x\to -\infty }\sinh(x)=-\infty $ and $\lim_{x\to \infty }\sinh(x)=+\infty $. So, IVT allows you to conclude.