# Find the domain and inverse of function $y^2-1+\log_2(x-1)=0$

Problem :

Find the domain and inverse of a function y of x : $y^2-1+\log_2(x-1)=0$

My approach : $$y^2-1+\log_2(x-1)=0 \implies y =\sqrt{\log_2\left(\frac{2}{x-1}\right)}$$ therefore, the domain of the function can be determined as:

the square root part should be $>0\implies \log_2\left(\frac{2}{x-1}\right) >0 \implies x <3$

Also $x \neq 1$ and also $\frac{2}{x-1} >0 \implies x >1$

This implies, the domain of the function is $(1,3)$. Is it correct, please suggest and also suggest how to get the inverse of this function, thanks.

• That is not a function, btw... it fails the "vertical line test" – apnorton Jun 20 '13 at 16:54
• $y=\sqrt{\log_2 \left(\frac{2}{x-1}\right)}$ is a function , $y=\pm\sqrt{\log_2 \left(\frac{2}{x-1}\right)}$ is not. – Aang Jun 20 '13 at 16:57
• As far as I know an equation isn't a function, (but I don't even know the definition of equation, so I could be wrong). This seems to be missing important information, like a neighborhood to analyze what happens. – Git Gud Jun 20 '13 at 16:58
• I think if it is a function y of x then the given equation will be : $x = 2^{y^2-1}+1$ – sultan Jun 20 '13 at 17:01
• @sultan Following up on Avatar's commment: what about $x=2^{(-y)^2-1}+1$? – Git Gud Jun 20 '13 at 17:02

If we assume function to be $y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$

Domain of $y=\sqrt{\log_2 \left(\frac{2}{x-1}\right)}$ is $(1,3]$ .

For inverse, first we need to check if it is invertible or not.

Check if $y=f(x)$ is injective,

$y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$

Since $\log_2(x-1)$ is an strictly increasing function $\implies -\log_2(x-1)$ is a strictly decreasing function $\implies y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$ is a strictly decreasing function $\implies y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$ is injective in its domain.

And every function is surjective onto its image.

Therefore, $y=f(x)=\sqrt{1-\log_2 \left({x-1}\right)}$ is invertible and its inverse is $x=f^{-1}(y)=2^{y^2-1}+1$

You need to consider positive and negative roots of $y^2$: $$y=\pm\sqrt{\log_2 \left(\frac{2}{x-1}\right)}\tag{1}$$ which is not by definition, a function. As noted in the comments, you can see that it fails the "vertical line test."

If we define $$f(x) = \sqrt{\log_2 \left(\frac{2}{x-1}\right)}\tag{2}$$

then we have a function! But note that the radicand must be greater than or equal to $0$, so the interval on which $f$ (which is not a function) is defined needs to include 3. So your domain for $(2)$ needs to be the interval $\bf (1, 3]$.

Yes: the process of finding the "inverse", or of expressing x in terms of $y$, gives you

$$x = 2^{y^2-1}+1\tag{3}$$

but with functions, (as in $(2)$), we conventionally express the inverse $f^{-1}(x)$ by swapping $x, y$:

$$f^{-1}(x) = 2^{x^2-1}+1\tag{4}$$

In your case, since the given $y:\; (1)$ is not a true function of $x$, I think you can leave the "inverse" as stated in $(3)$. In the case of $f(x)$ as in $(2)$ (restriction of $y$ to the non-negative root of $y^2$, use the conventional notation for the inverse of a function as given in $(4)$.

Graph the equations of $y$ in terms of $x$, and $x$ in terms of $y$ to see what's happening here in the first and third cases, and of the function given by $(3)$ and of $f^{-1}$. Be careful when considering the domain of $f^{-1}(x)$ if you are taking $f(x)$ equal to the non-negative root of $y$.

I would provide the above equations in your answer, but first make note of the fact that you recognize that $y$ (as given in $(1)),\;$ is not a function.