Finding inverse of a function $h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$

Question

I have a function:

$$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$

With just pen and paper, how can I determine if there exists an inverse function? Am I supposed to sketch it on paper to see if it can have an invers? Or is there another/simplier way to do it?

How would I go about solving it? This is what I did: $$h(x) = \frac{1-\sqrt{x}}{1+\sqrt{x}} = y$$ $$y(1 + \sqrt{x}) = 1-\sqrt{x}$$ $$y + y\sqrt{x} = 1-\sqrt{x}$$ $$\text{Cancel out roots:}$$ $$y^2 + y^2x = 1+x$$ $$y^2 + y^2x - x = 1$$ $$y^2 + x(y^2 - 1) = 1$$ $$x(y^2 - 1) = 1 - y^2$$ $$x = \frac{1 - y^2}{y^2 - 1}$$ $$\text{Swap x and y and we get the inverse function:}$$ $$f^{-1}(x) = \frac{1 - x^2}{x^2 - 1}$$ $$\text{But the correct answer is supposed to be:}$$ $$f^{-1}(x) = \frac{(1 - x)^2}{(1 + x)^2}, -1< x \le 1$$

What am I doing wrong?

Answer 1

Nice work! But this is how you should have done it! $$\begin{align}1-y&=y\sqrt x+\sqrt x\\ 1-y&=(y+1) \sqrt x\\ \frac{1-y}{y+1}&=\sqrt x\\ \sqrt x&=\frac{1-y}{y+1}\\ x&=\frac{(1-y)^2}{(y+1)^2}\\ \end{align} $$ Hence $f^{-1}(x)$ $$f^{-1}(x) = \frac{(1 - x)^2}{(x+ 1)^2}$$

Answer 2

It is important to remember that $$(A+B)^2=A^2+\mathbf{2AB}+B^2$$

This what I'd do: $$y(1+\sqrt x)=1-\sqrt x$$ $$y+y\sqrt x=1-\sqrt x$$ $$(y+1)\sqrt x=1-y$$ $$(y+1)^2x=(1-y)^2$$

Can you finish?

Answer 3

Alternative answer: as you are only asking if the function has an inverse, you can compute $f'(x)$ (in fact calculating $\tfrac{d}{dx}(\tfrac{1-x}{1+x} = 1-\tfrac{2}{x+1})$ is enough) and notice that $f'<0$.

So $f$ is strictly decreasing and, hence, injective. Therefore, $f$ has an inverse.

Answer 4

This is probably conceptually the simplest method.

For $\frac{a}{b}=\frac{c}{d}$, we always have $\frac{a+b}{a-b}=\frac{c+d}{c-d}$, provided that $a-b\not=0$ and $c-d\not=0$.

$$y = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$

$$\frac{y}{1} = \frac{1-\sqrt{x}}{1+\sqrt{x}}$$

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

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

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