Find the inverse of this function . . .

checkittwice

Member
>

$$f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}$$

Sudharaka

Well-known member
MHB Math Helper
>

$$f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}$$
Hi checkittwice,

Let, $$y = \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}$$

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

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

$(y-yx-x-1)^2=4x$

$((y-1)-x(y+1))^2=4x$

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

$(y+1)^{2}x^2-\left\{2(y-1)(y+1)+4\right\}x+(y-1)^2=0$

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

$x=\frac{\left\{2(y-1)(y+1)+4\right\}\pm\sqrt{16+16(y-1)(y+1)}}{2(y+1)^{2}}$

$x=\frac{\left\{2(y-1)(y+1)+4\right\}\pm 4|y|}{2(y+1)^{2}}$

$x=\frac{y^2\pm 2|y|+1}{(y+1)^{2}}$

If we use the positive sign, $$x=1$$ whenever $$y\geq 0$$. Similarly if we use the negative sign, $$x=1$$ whenever $$y\leq 0$$. Both of these are not true considering our original equation. Therefore the only possibility is to use the positive sign when $$y<0$$ and the negative sign when $$y\geq 0$$. This gives us,

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

checkittwice

Member
$x = \frac{(y-1)^2}{(y+1)^{2}}$ ? ? ?
Sudharaka,

here are two observations/prompts:

1) Nowhere in your work did you ever interchange x any y.

2) Suppose you were to interchange x and y to get

$$y \ = \ \dfrac{(x - 1)^2}{(x + 1)^2}$$

Would that be a one-to-one function?

Or, would you have to restrict the domain (and state it as such)
to get the correct inverse?

The problem is still open.

Sudharaka

Well-known member
MHB Math Helper
Sudharaka,

here are two observations/prompts:

1) Nowhere in your work did you ever interchange x any y.

2) Suppose you were to interchange x and y to get

$$y \ = \ \dfrac{(x - 1)^2}{(x + 1)^2}$$

Would that be a one-to-one function?

Or, would you have to restrict the domain (and state it as such)
to get the correct inverse?

The problem is still open.
Hi checkittwice,

In my final answer $$y$$ is the independent variable and $$x$$ is the dependent variable. I thought it would be understood and did not interchange $$x$$ and $$y$$. I hope this is the thing that you mentioned in your first statement.

And of course I forgot to mention about the domain of the inverse in my first post. In the original function, $$x\geq 0$$. Therefore in the inverse function; $$\displaystyle y \ = \ \dfrac{(x - 1)^2}{(x + 1)^2}$$ we have to make $$y\geq 0$$. For that $$x\geq 1$$. So the inverse function would be,

$\displaystyle y \ = \ \dfrac{(x - 1)^2}{(x + 1)^2}\mbox{ where }x\geq 1$

Evgeny.Makarov

Well-known member
MHB Math Scholar
We can express x through y more simply:

$$y(1+\sqrt{x})=1-\sqrt{x}$$
$$y+y\sqrt{x}=1-\sqrt{x}$$
$$\sqrt{x}(1+y)=1-y$$
$$\sqrt{x}=\frac{1-y}{1+y}$$
$$x=\left(\frac{1-y}{1+y}\right)^2$$

Also, one can check for the original function that when $$x\in[0,\infty)$$, one has $$-1<y\le 1$$. The latter inequality determines the domain of the inverse function.

checkittwice

Member
Hi checkittwice,

In my final answer $$y$$ is the independent variable and $$x$$ is the dependent variable.
I thought it would be understood and did not interchange $$x$$ and $$y$$.

>The value of $$f^{-1}(x)$$depends on what x equals.

>You can't go by what you "understood" if the problem isn't finished as typed
by you.

>For example, if I were to ask for the inverse of y = f(x) = 0.5x, someone would
not answer with 2y = x. They would answer with $$f^{-1}(x) \ = \ 2x.$$

>My first statement of my second post in this thread is redundant, because
the inverse will be of the form $$f^{-1}(x)$$ = an expression in terms of x.

I hope this is the thing that you mentioned in your first statement.
....

And as alluded to by Evgeny.Makarov, the domain of the desired
inverse will reflect the range of the original function.

So, to clear things up, the inverse is:

$$f^{-1}(x) \ = \ \ \dfrac{(x - 1)^2}{(x + 1)^2}, \ \ -1< x \le 1.$$