I was recently given the function
$\sqrt{x}/(x-1)$
and was asked to evaluate $f^{-1}(1)$ find the inverse and draw the graph.
Firstly, in order for a function to be invertible, it must be bijective, meaning it has a one-to-one mapping and its range is the same as its co-domain.
So based on our definition, Should the inverse not be undefined as well at that point.
When searching on a calculator, I either get the result
$y=(1+2x^2\pm\sqrt{4x^2+1})/(2x^2)$
or an error telling me it is not possible
Can someone please run me through on how one can achieve this inverse? My guess is that we must use the quadratic formula however I have already tried. The furthest I have gotten is
$\sqrt{y}/(y-1) = x$
$y/(y-1)^2 = x^2$
$(y-1)^2/y = 1/x^2$
$(y^2-2y+1)/y = 1/x^2$
$y-2+1/y = 1/x^2$
$y+1/y = (1+2x^2)/x^2$
I can't seem to get any further than this for some reason.
Also if someone could explain how our definition of an invertible function still holds this would be a literal contradiction of the properties of an invertible function. I can't seem to grasp why this is possible, unless my definition for an invertible function is wrong.
Lastly I don't understand how $\pm$ is in the function, does this imply there is more than one output for each specific input, and if so I thought this was also contradictory to the definition of a function.
Best Answer
Your equation is $$\begin{align*} \frac{\sqrt{y}}{y-1} = x \\ \implies\\ \frac{y}{(y-1)^2} = x^2 \\ \iff\\ x^2y^2 -(2x^2 + 1)y + x^2 = 0 \end{align*}$$ This is a second degree polynomial in the variable $y$, so you get two roots ($\Delta = (2x^2+1)^2 - (2x^2)^2 = 4x^2+1$). $$y_1 = \frac{2x^2 + 1 + \sqrt{\Delta}}{2x^2} \qquad \text{or} \qquad y_2 = \frac{2x^2 + 1 - \sqrt{\Delta}}{2x^2}$$
You see that you have two "valid" inverses (both are positive), but why is it possible? Your fonction is not defined at $1$, it is only defined in $$[0,1[ \cup ]1,\infty[$$
But it first goes in the negative (before 1) then in the positive (after 1). If you've read this carefully, you've noticed that there is an implication (this is why you think you have two inverses) which goes from giving information from $x$ to information from $x^2$. This means we still need to do the backward step.
Since $y_1$ is always strictly positive (and greater than $1$) we can easily deduce that the right inverse for the negative is $y_2$ and the right inverse for the (strictly) positive is $y_1$!
In fact, it is bijective continuous from $\mathbb{R}_+\setminus \{1\}$ to $\mathbb{R}$and the inverse (which is unfortunately not continuous but continuous by parts) is given by $f^{-1}(x) = \frac{2x^2 + 1 + s(x) \sqrt{4x^2+1}}{2x^2}$ where $s(x) = -1$ if $x \leqslant 0$ and $s(x) = 1$ if $x >0$. It is defined by continuity on the left in $0$ but if you wish it to be clearer $f^{-1}(0) = 0$.
We deduce $f^{-1}(1) = \frac{3 + \sqrt{5}}{2}$.
For further comprehension, check the graph on any calculator/computer e.g. Wolfram.
I also suggest to draw variation tables to have a proper idea! (study the derivative).