Number of answers : $f(x)=f^{-1}(x)$

Question

let $f(x)= 1+\sqrt{x+k+1}-\sqrt{x+k} \ \ k \in \mathbb{R}$ Number of answers :

$$f(x)=f^{-1}(x) \ \ \ \ :f^{-1}(f(x))=x$$

MY Try :

$$y=1+\sqrt{x+k+1}-\sqrt{x+k} \\( y-1)^2=x+k+1-x-k-2\sqrt{(x+k+1)(x+k)}\\(y-1)^2+k-1=-2\sqrt{(x+k+1)(x+k)}\\ ((y-1)^2+k-1)^2=4(x^2+x(2k+1)+k^2+k)$$

now what do i do ?

Answer 1

Hint:

Point of intersection of $f(x)$ and $f^{-1}(x)$ while same as that of $f(x)$ and the line $y=x$.

Answer 2

Let $y=f(x)=f^{-1}(x)$ and then \begin{eqnarray} y=1+\sqrt{x+k+1}-\sqrt{x+k},\tag{1}\\ x=1+\sqrt{y+k+1}-\sqrt{y+k}.\tag{2} \end{eqnarray} Subtracting (1) from (2) gives \begin{eqnarray} x-y&=&(\sqrt{y+k+1}-\sqrt{x+k+1})-(\sqrt{y+k}-\sqrt{x+k})\\ &=&\frac{y-x}{\sqrt{y+k+1}+\sqrt{x+k+1}}-\frac{y-x}{\sqrt{y+k}+\sqrt{x+k}} \end{eqnarray} and hence $$ (x-y)\left[\frac{1}{\sqrt{y+k+1}+\sqrt{x+k+1}}-\frac{1}{\sqrt{y+k}+\sqrt{x+k}}\right]=0.$$ Thus one has either $$ x=y $$ or $$ \frac{1}{\sqrt{y+k+1}+\sqrt{x+k+1}}=\frac{1}{\sqrt{y+k}+\sqrt{x+k}}\tag{3}. $$ Since (3) does holds, one must have $x=y$ and hence $$ x=1+\sqrt{x+k+1}-\sqrt{x+k}. $$ It is easy to see that $x$ is the real roots of the following equation $$x(x^3-8x^2+12x-4)-4k(x+1)^2=0.\tag{4}$$ The number of answers depends on the real solutions of (4), either 1,2,3 or 4, depending on $k$.