Is it right to ignore the negative result in the derivation of the inverse hyperbolic cosine function?

Question

Please can someone explain to me why the inverse function is defined as the positive value to "avoid ambiguity"? Is it wrong to define the inverse function using both the positive and negative results?

Answer 1

There is a certain "sloppyness" in how we usually write down a lot of math, assuming that "you know what I mean". In your case, it makes sense to be a bit more strict.

So, when we say function, we usually mean a regular function. Every regular function $f$ has a domain $D$ and a codomain $C$, and we write $f:D\rightarrow C$. A regular function then associates to each element $x\in D$ exactly one element $y\in C$, and we write $f(x)=y$.

Now, for every set $S$, there is a special function, the identity $id_S:S\rightarrow S$, $id(x)=x$.

Now, a left-inverse of a given function $f:D\rightarrow C$ is a function $l:C\rightarrow D$ (i.e. a function with domain=codomain of f and codomain=domain of f), such that $l\circ f:D\rightarrow D = id_D$. In general, for every given function, there can exist zero, one or many left inverses. There exists at least one left inverse if $f$ is one to one (injective).

Similarly, a right-inverse of $f$ is a function $r:C\rightarrow D$ such that $f\circ r:C\rightarrow C=id_C$. Again, a $f$ might possess zero one or many right inverses. There exists at least one right inverse if $f$ is onto (surjective).

Finally, if a function $g:C\rightarrow D$ is both the left and the right inverse of $f$, then we say that $g$ is the inverse of $f$, and usually denote it by $f^{-1}=g$. Not every function posseses an inverse, but if it does, the inverse is unique. Trivially, the inverse exists if and only if $f$ is one to one onto (a bijection).

Now, let's look at the function $cosh$. The first question you should ask is: what is the domain and codomain of $cosh$. Here comes the first "sloppyness", because neither domain nor codomain is given. So, let's look at different options:

• $cosh:\mathbb{R}\rightarrow\mathbb{R}$
• $cosh:[0,\infty)\rightarrow\mathbb{R}$
• $cosh:(-\infty,0]\rightarrow\mathbb{R}$
• $cosh:[0,\infty)\rightarrow[1,\infty)$
• $cosh:(-\infty,0]\rightarrow[1,\infty)$

There are many more options. The important point is that all these functions are different, even though we are (again) sloppy and denote them by the same name.

We now find the following:

• $cosh:\mathbb{R}\rightarrow\mathbb{R}$ is neither one to one nor onto. It does not possess a left or right inverse, and thus also not an inverse.
• $cosh:[0,\infty)\rightarrow\mathbb{R}$ is one to one, but not onto. It possessess many left inverse, but no right inverse, and thus also no inverse. All of the left inverses satisfy that, for $x\geq 1$, $l(x)=log(x+\sqrt{x^2-1})$. However, by definition, $l$ also has to assign a unique value to all $x<1$. However, it doesn't matter which value(s) we choose. For example, we can set $l(x)=2$ for $x<1$, or $l(x)=|x|$, or whatever. Every possibility gives us a different left inverse $l$, thus, the left inverse is not unique.
• $cosh:(-\infty,0]\rightarrow\mathbb{R}$ again has many left inverses, but no right inverse or inverse. This time, the left inverses satisfy $l(x)=-log(x+\sqrt{x^2-1})$ for $x\geq 1$, but we can choose an arbitrary (non-positive) value for each $x<1$.
• $cosh:[0,\infty)\rightarrow[1,\infty)$ is one to one onto. It thus possesses a unique inverse, which is also (by definition) its left and right inverse. It is given by $cosh^{-1}(x)=log(x+\sqrt{x^2-1})$.
• Similarly, $cosh:(-\infty,0]\rightarrow[1,\infty)$ possesses the unique inverse $cosh^{-1}(x)=-log(x+\sqrt{x^2-1})$.

In summary: the sign of the inverse is not about clarity, convention or to avoid ambiguity. Indeed, $cosh$ might either possess or not possess an inverse, depending on its domain and codomain, but given that cosh possesses an inverse there is not the possibility that you can choose the sign. Furthermore, for different choices of domain and codomain, these inverses, if they exist, will be different functions. There is no reason why these functions have to have same signs (or for that matter, absolute values) at the intersections of their domains.

Edit

To be clear: there is ambiguity. However, this ambiguity is not coming from the definition of the inverse: if $cosh$ possesses an inverse, this inverse is unique. There is no choice of sign. The ambiguity is coming from that you didn't clearly define which function $cosh$ should refer to. It is not sufficient to say that $cosh(x)=\frac{e^x+e^{-x}}{2}$, you additionally have to state the domain and codomain of $cosh$. Without the latter, $cosh$ can refer to many different functions, each either having an inverse or not. Furthermore, each of these inverses is different. This ambiguity is however "solved by convention", i.e. one implicitly assumes that $cosh^{-1}$ refers to the inverse of $cosh:[0,\infty)\rightarrow[1,\infty)$, and thus that $cosh^{-1}:[1,\infty)\rightarrow[0,\infty)$.

Answer 2

It's just so you are actually forming an honest-to-goodness function, which requires that you specify a single value of $f^{-1}(x)$ for each $x$.

It doesn't matter which value you pick for any given $x$, you just have to pick some value to define your inverse function.

If you want the inverse to be continuous, then you would pick the values so that either $f^{-1}(x)\geq 0 $ for all $x$, or $f^{-1}(x)\leq 0$ for all $x$.

You can visualize why this is true by considering the graph. The graph of $y=\cosh x$ is a catenary, which is U-shaped and opens upwards, with the bottom at $(0,1)$. Every horizontal line above $y=1$ meets the "U" at two points.

Switching $x$ and $y$, you get a graph that is $\subset$-shaped and opens to the right, with the left side at $(1,0)$. Every vertical line to the right of $x=1$ meets the "$\subset$" at two points. This is what prevents the inverse from being a function -- you have to pick one of those two points if you want it to be a function.

It is always nice for the inverse function to be continuous, so that's why we typically pick only the upper part of the graph of the inverse when defining the inverse function. Otherwise there would be a jump discontinuity somewhere. Of course, you could pick only the lower part instead; that's perfectly acceptable, and it's just a different "branch" of the inverse.

And if you picked some points from the upper part and some from the lower part, that would be perfectly good as well, it just wouldn't be continuous everywhere (as long as you don't select two points which are vertically aligned, since that would violate the "vertical line test" for a function because $f^{-1}(x)$ would not be uniquely defined at that $x$).