Multiplying by $1$ adds a solution to an equation

Question

I have a question, which is motivated by my book's solution to finding the inverse function of $f(x)=\frac{x}{1-x^2}$ with the domain of $f(x)$ restricted the open interval $(-1,1)$. The questions are stated at the bottom, but first is the book's solution:

Solution

$f(x)$ is injective and surjective, and hence it has an inverse, $f^{-1}(x)$.

$y=\frac{x}{1-x^2}$

$x=\frac{y}{1-y^2}$

$xy^2+y-x=0$

$y=\frac{-1\pm\sqrt{1+4x^2}}{2x}$

The negative solution is the right one, and then the book notices that $y=\frac{-1\pm\sqrt{1+4x^2}}{2x}$ excludes $x=0$ from the domain, but it should be in the domain since $0$ is in the image of $f(x)$, namely $f(0)=0$. It fixes this by rationalizing the numerator by multiplying by $1$ as follows:

$y=\frac{-1+\sqrt{1+4x^2}}{2x}\cdot \frac{-1-\sqrt{1+4x^2}}{-1-\sqrt{1+4x^2}} = \frac{2x}{1+\sqrt{1+4x^2}}$

Hence $f^{-1}(x)=\frac{2x}{1+\sqrt{1+4x^2}}$. $\blacksquare$

Question

Why is it that multiplying a function by $1$ will add a value for the graph of $f^{-1}(x)$? I can see why multiplying by $1$ will sometimes remove points on the graph, namely where the denominator is zero (e.g. multiply an equation by $\frac{x}{x}$ and it won't have a value at $x=0$), but how does it work here that multiplying by $1$ adds a point to the graph? Did it just get shifted somewhere else?

Answer 1

It's not quite the "multiplying by $1$" that expands the domain. Note that after doing the multiplication, we actually have $$ y=\frac{4x^2}{2x\left(1+\sqrt{1+4x^2}\right)}. $$ Here we still can't have $x=0$. The magic happens when we declare $\frac{4x^2}{2x}=2x,$ which gets rid of the removable "hole" in the graph. So really, it's like in your own example, just in reverse: We started out with an $\frac xx$ in disguise, which we converted to a $1$. The multiplication is just a clever (and common) algebraic trick to clean up the expression.

As to why we got the hole in the first place, this happened when we used the quadratic formula. The formula is only suited to actual second degree polynomials, so we must assume that the leading term doesn't vanish when we use it. Really, we should split up in cases at that point in the proof: Either $x=0$ (which implies $y=0$), or we can solve for $y$ with the quadratic formula when $x\ne0$. Thus, another way to write the final answer would be simply $$ f^{-1}(x)=\begin{cases}\frac{-1+\sqrt{1+4x^2}}{2x}&x\ne0 \\ 0 &x=0\end{cases} .$$

Answer 2

TLDR; There is a shorter more direct answer. It took me too long to write this one. C’est la vie.

First of all let me comment on the textbook answer.

Although it derives a formula for the domain value ($f^{-1}(x)$) it does not define a function. This is because it has not specified its domain.

Furthermore, it introduces some confusion by swapping $x$ and $y$ from the very beginning, as if the domain variable had to be $x$. I suppose this is a sort of convention.

Finally, it does not encourage any understanding of why the adjustments have to be made at all, and how we should spot they are necessary when doing our own examples. [At least, from your explanation, it does not appear to do so.]

Now let’s look at this problem afresh. As a matter of course I would want to know what this function $f$ looks like. Either I would sketch it by hand, or, these days, I would use a graphing calculator to see the shape of the function quickly. Here is what the graph of $y=\frac{x}{1-x^2}$ looks like:

It is now clear why the function ($f$) is restricted to the domain $-1 < x < 1$. Moreover, the inverse looks like this (simply swap the x and y coordinates): and now it is obvious that the unrestricted domain of $f$ would fail to have a function as an inverse, since the graph shows it would have to be multi-valued — not an option for a function.

So, from these graphs we can begin to see that $f$, restricted to the domain $(-1,1)$, does have a (single-valued) inverse function and that its domain is going to be the whole of the real number line.

Now we know exactly what the inverse function is: it is that function which given a real value, call it $y$, ‘returns’ the (unique) value $x\in (-1,1)$ such that $y=\frac{x}{1-x^2}$.

In one sense this is a complete answer — I have told you exactly what the inverse is. But that is not what the textbook wants you to do. It wants you to derive an algebraic formula for calculating the value of $x$ given that real number $y$.

Now that is another kettle of fish entirely. In general it is not possible: not all (functions defined by) algebraic expressions have inverse (functions defined by) algebraic expressions. But in this case we can derive one. Let’s see how:

If $y=\frac{x}{1-x^2}$ (and $x \in (-1,1)$) then we proceed as follows:

$\begin{align*} y & = \frac{x}{1-x^2} \\ y(1-x^2) & =x \tag*{(notice that $1-x^2$ is never $0$)} \\ y-yx^2 & =x \\ \text{which rearranges to:} & {} \\ yx^2 + x -y &= 0. \end{align*}$

Now the textbook ‘solves’ this quadratic equation (in $x$) using the standard formula. You can clearly see that the value $y=0$ is excluded because the formula divides by $2y$. This is why the formula for the inverse cannot tell us the value when $y=0$.

Notice why the formula has two values for $x$: it is because the algebra (relating $x$ and $y$) doesn’t care about the restricted domain of $f$. It is telling us that there are two possible values of $x$ that the formula for $f$ maps to $y$. It is up to us to determine which one we want. In this case, we restrict ourselves to results in the interval $(-1,1)$ by choosing the formula:

$\displaystyle x = \frac{-1 + \sqrt{1+4y^2}}{2y}$

But this formula only works if $y\neq 0$ as we noted above. So a perfectly good ‘solution’ is:

$f^{-1}(y) = \begin{cases} \frac{-1 + \sqrt{1+4y^2}}{2y} & \text{if $y\neq 0$}\\ 0 & \text{if $y=0$}\end{cases}$

How do we get a formula which works even when $y=0$?

The ‘trick’ used here is to change the formula to one that doesn’t want to divide by zero when $y=0$. Obviously we don’t want to change the values it produces when $y$ isn’t zero, but maybe we can find an alternate formula which does just as well but manages not to become ‘indeterminate’ at that point.

The trick is a natural one to use when dealing with quadratic surds. We remember the formula $(a-b)(a+b) = a^2 - b^2$. So if we have something like $a-\sqrt{\text{thing}}$ then multiplying it by $a+\sqrt{\text{thing}}$ will produce something with squares of the two terms, i.e. $a^2 - \text{thing}$.

[Note: I’m assuming $\text{thing}$ is positive here.]

So let’s see what this does to the numerator of our formula. Oh, and while we’re about it, we had better divide by this term as well, otherwise the values will get mucked up. Here we go:

$\begin{align*} \frac{-1 + \sqrt{1+4y^2}}{2y} & = \frac{-1 + \sqrt{1+4y^2}}{2y}\cdot \frac{-1 - \sqrt{1+4y^2}}{-1 - \sqrt{1+4y^2}} \\ {} & = \frac{(-1 + \sqrt{1+4y^2})\cdot (-1 - \sqrt{1+4y^2})}{2y(-1 - \sqrt{1+4y^2})} \\ {} & = \frac{1-(1+4y^2)}{2y(-1 - \sqrt{1+4y^2})} \\ {} & = \frac{-4y^2}{2y(-1 - \sqrt{1+4y^2})} \\ \text{which simplifies, } & \text{when $y\neq 0$, to:} \\ {} & = \frac{2y}{1 + \sqrt{1+4y^2}} \end{align*}$

Notice that we only proved this when $y\neq 0$. But, fortunately, this formula has no problem when $y=0$. And it gives the right answer, too. Lucky us.

Multiplying by $1$ didn’t introduce any new values; we were lucky that the formula it led us to worked in the case $y=0$. That is all.

[NB: This ‘technique’ doesn’t work most of the time.]