# Thread: Inverse function, in terms with another inverse function

1. ## Inverse function, in terms with another inverse function

$f$ is a one-to-one function with inverse $f^{-1}$.
$p(x) = \frac{1}{1+f(x)}$
Caculate the inverse of the function in terms of $f^{-1}$

So I tried swapping $p(x)$ and $f(x)$.
$f^{-1}(x) = \frac{1}{1+p^{-1}(x)}$

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

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

The answer in the back of the book however, says this:
$p^{-1}(x) = f^{-1}\left(\frac{1}{x} - 1\right)$

How do I get to that answer?

2. Originally Posted by BlackBlaze
$f$ is a one-to-one function with inverse $f^{-1}$.
$p(x) = \frac{1}{1+f(x)}$
Caculate the inverse of the function in terms of $f^{-1}$

So I tried swapping $p(x)$ and $f(x)$.
$f^{-1}(x) = \frac{1}{1+p^{-1}(x)}$

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

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

The answer in the back of the book however, says this:
$p^{-1}(x) = f^{-1}\left(\frac{1}{x} - 1\right)$

How do I get to that answer?
It helps to apply the procedure correctly.

Let $y = \frac{1}{1 + f(x)}$. Swap y and x:

$x = \frac{1}{1 + f(y)}$.

Your job is to make y the subject:

$\frac{1}{x} = 1 + f(y) \Rightarrow \frac{1}{x} - 1 = f(y)$

and it should be clear what the last step is.

3. Right at the second step, I saw what a gross error I made. Don't I feel silly.

Thanks for the speedy response, as well as the formatting.