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#### checkittwice

##### Member

- Apr 3, 2012

- 37

>

[tex]f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}[/tex]

[tex]f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}[/tex]

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- #1

- Apr 3, 2012

- 37

>

[tex]f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}[/tex]

[tex]f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}[/tex]

- Feb 5, 2012

- 1,621

Hi checkittwice,>

[tex]f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}[/tex]

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,

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

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- Apr 3, 2012

- 37

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

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

[tex]y \ = \ \dfrac{(x - 1)^2}{(x + 1)^2}[/tex]

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?

- Feb 5, 2012

- 1,621

Hi checkittwice,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

[tex]y \ = \ \dfrac{(x - 1)^2}{(x + 1)^2}[/tex]

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.

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,

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

- Jan 30, 2012

- 2,541

[tex]y(1+\sqrt{x})=1-\sqrt{x}[/tex]

[tex]y+y\sqrt{x}=1-\sqrt{x}[/tex]

[tex]\sqrt{x}(1+y)=1-y[/tex]

[tex]\sqrt{x}=\frac{1-y}{1+y}[/tex]

[tex]x=\left(\frac{1-y}{1+y}\right)^2[/tex]

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

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- #6

- Apr 3, 2012

- 37

....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 [tex]f^{-1}(x) [/tex]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 [tex]f^{-1}(x) \ = \ 2x.[/tex]

>My first statement of my second post in this thread is redundant, because

the inverse will be of the form [tex]f^{-1}(x)[/tex] = 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:

[tex]f^{-1}(x) \ = \ \ \dfrac{(x - 1)^2}{(x + 1)^2}, \ \ -1< x \le 1.[/tex]