Proving the inverse of $1/(1 + 2x)$ is $(1 - y)/2y$

Question

Let $f(x) = 1/(1 + 2x)$. I want to show that $f^{-1}(y) = (1 - y)/2y$. Is it sufficient for me to show

$$f\left(\frac{1-y}{2y}\right) = \frac{1}{1 + 2(1 - y)/2y} = \frac{1}{1 + (1 - y)/y} = \frac{1}{y}= y?$$

Or, is there another way that this assertion needs to be proved?

Answer 1

As an alternative firstly note that $f(x)$ is defined $\forall x\neq -\frac12$ and since

• $f(x)=\frac1{1+2x} \implies f'(x)=-\frac2{(1+2x)^2}<0$
• $f(x)>0$ for $x>-\frac12$
• $f(x)<0$ for $x<-\frac12$
• $\lim_{x\to \pm \infty} f(x)=0$

we have that $f(x):\mathbb{R}\setminus\{-1/2\}\to \mathbb{R}\setminus\{0\}$ is bijective and then invertible.

To find the inverse we need to solve

$$x=\frac{1}{1+2y}$$

and find $y=g(x)=f^{-1}(x)$.

Answer 2

Let $g(y) = \frac{1-y}{2y}$. When you show $f(g(y)) = y$, what you know from there is precisely the following: $f$ is surjective and $g$ is injective. You don't know that $f$ or $g$ are bijective yet.

Now, you can also show that $g(f(x)) = x$ and from there it will follow: $f$ is injective and $g$ is surjective.

Combined, you've shown $f\circ g = id$ and $g\circ f = id$, i.e. you've shown $g=f^{-1}$ by definition.

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However, if you knew that $f$ is bijective (or even just injective) beforehand, and you showed $f(g(y)) = y$, then it would be enough to conclude that $f^{-1} = g$. This is because if $f$ is injective, there exists $h$ such that $h(f(x)) = x$. Now,

$$h = h\circ id = h\circ f\circ g = id\circ g = g\implies f\circ g = id\ \text{and}\ g\circ f = id.$$