# Thread: Check whether function is invertible...?

1. ## Check whether function is invertible...?

Check whether $f(x) = [x] + \sqrt{\{x\}}$, $f:\mathbb{R}\rightarrow\mathbb{R}$, (where [.] & {.} represent greatest integral and fractional part function respectively) is invertible or not, if yes, then (i)find its inverse. (ii)solve the equation $f(x) = f^{-1}(x)$.

Spoiler:
$\bold{Answer:}\ \mbox{(i)}f^{-1} = [x] + \{x\}^2;\mbox{(ii)}\ x\in \mathbb{Z}$

I was unable to do it. Any ideas?

2. Let $n\leq x. $\therefore[x]=n$

$f(x)=n+\sqrt{x-n}$

$f'(x)=\frac{1}{2\sqrt{x-n}}>0$,when $x$ is not an integer

Incase $f(x)$ is an integer then $f(x)=x$ which is obviously invertible.

Thus $f(x)$ is clearly invertible.

Let $g(x)$ be the inverse of $f(x).$

$f(g(x))=x$

$n+\sqrt{g(x)-n}=x$

$\sqrt{g(x)-n}=x-n$

$g(x)-n=x^2-2nx+n^2$

$g(x)=n+(x-n)^2$

$g(x)=[x]+(x-[x])^2$

$f^{-1}(x)=[x]+\{x\}^2$

Since $f^{-1}(x)$ is not identically equal to $f(x)$.Therefore,solving $f(x)=f^{-1}(x)$ is same as solving for $f(x)=x$.

$[x]+\sqrt{\{x\}}=x$

$\sqrt{\{x\}}=\{x\}$

On squaring,
$\{x\}=0,1$

$\{x\}=1$ is impossible

$\{x\}=0$

i.e. $x$ is an integer

3. According to...

Nearest Integer Function -- from Wolfram MathWorld

the notation $f(x)=[x]$ indicates the 'nearest integer function' , so that ...

$[x]= n$ , $n-\frac{1}{2} \le x < n + \frac{1}{2}$

The function defined as...

$\lfloor x \rfloor = n$ , $n \le x < n+1$

is the so called 'floor function' and it is indicated as $f(x) = \lfloor x \rfloor$

Kind regards

$\chi$ $\sigma$

4. I think it is up to you to define a particular notation for any function.fardeen_gen has chosen to define floor function by [x],so let it be.

In textbooks of our country(India) this is how the floor function is defined

5. Fardeen_gen in his post first used the notation $f(x)=[x]$ and after 'verbally' described the function as 'greatest integral'... not 'floor function'... at this point some sort of doubt is legitimate...

Kind regards

$\chi$ $\sigma$

6. I agree.Actually, I was famaliar with this question and went on to write the solution right away.