Prove that $\cosh(x):[0,\infty)\rightarrow[1,\infty)$ is bijective

Question

Let $\cosh(x) = \frac{e^x + e^{-x}}{2}$. Prove that $\cosh(x):[0,\infty)\rightarrow[1,\infty)$ is bijective

How do I do this?

So to prove that it is surjective:

$\cosh(x)=y,\quad \frac{e^x + e^{-x}}{2} = y,\quad x =\ln(y\pm\sqrt{y^2 + 1})$

and then I plug this into the function:

$$\cosh(\ln(y\pm\sqrt{y^2 + 1})) = \frac{e^{\ln(y\pm\sqrt{y^2 + 1})} + e^{-\ln(y\pm\sqrt{y^2 + 1})}}{2}$$

If I'm not mistaken, this can be written as $$\frac{y\pm\sqrt{y^2 + 1} -(y\pm\sqrt{y^2 + 1})}{2}$$ and isn't this just $0$?

Shouldn't I be showing that it is equal to $y$?

I think I got a hold of how to show that it is injective, but surjective - no.

Answer 1

You are on the right track. Just consider as inverse map of $\cosh:[0,\infty)\rightarrow[1,\infty)$, the map $$g(y)=\ln(y+\sqrt{y^2 + 1}).$$ The one with the minus sign is not defined (the argument of $\ln$ is not positive).

Answer 2

Since $f(x)=\cosh x$ has $$ f(0)=1, \qquad \lim_{x\to\infty}f(x)=\infty, \qquad f'(x)=\sinh x $$ and $\sinh x>0$ for $x>0$, the function $f$ is strictly increasing, so injective and the image is $[1,\infty)$ by the intermediate value theorem.

An “algebraic” proof can be obtained by solving the equation $$ y=\frac{e^x+e^{-x}}{2} $$ with respect to $x$, given that $y\ge1$. Setting $t=e^x$ and noticing that $t>0$, we can rewrite it as $$ y=\frac{t^2+1}{2t} $$ or $$ t^2-2yt+1=0 $$ that has a single root in $[1,\infty)$ for every $y\ge1$, namely $$ t=y+\sqrt{y^2-1} $$ so $$ x=\log(y+\sqrt{y^2-1}) $$ This proves that $\cosh\colon[0,\infty)\to[1,\infty)$ has an inverse, so it is bijective.